6.1 Solving Problems with Newton’s Laws

Learning objectives.

By the end of this section, you will be able to:

  • Apply problem-solving techniques to solve for quantities in more complex systems of forces
  • Use concepts from kinematics to solve problems using Newton’s laws of motion
  • Solve more complex equilibrium problems
  • Solve more complex acceleration problems
  • Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy

Applying newton’s laws of motion.

  • Identify the physical principles involved by listing the givens and the quantities to be calculated.
  • Sketch the situation, using arrows to represent all forces.
  • Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
  • Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
  • Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 6.2 (a). Then, as in Figure 6.2 (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 6.2 (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure 6.2 (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure 6.2 (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

(If, for example, the system is accelerating horizontally, then you can then set a y = 0 . a y = 0 . ) We need this information to determine unknown forces acting on a system.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Example 6.1

Different tensions at different angles.

Thus, as you might expect,

This gives us the following relationship:

Note that T 1 T 1 and T 2 T 2 are not equal in this case because the angles on either side are not equal. It is reasonable that T 2 T 2 ends up being greater than T 1 T 1 because it is exerted more vertically than T 1 . T 1 .

Now consider the force components along the vertical or y -axis:

This implies

Substituting the expressions for the vertical components gives

There are two unknowns in this equation, but substituting the expression for T 2 T 2 in terms of T 1 T 1 reduces this to one equation with one unknown:

which yields

Solving this last equation gives the magnitude of T 1 T 1 to be

Finally, we find the magnitude of T 2 T 2 by using the relationship between them, T 2 = 1.225 T 1 T 2 = 1.225 T 1 , found above. Thus we obtain

Significance

Particle acceleration.

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

Example 6.2

Drag force on a barge.

The drag of the water F → D F → D is in the direction opposite to the direction of motion of the boat; this force thus works against F → app , F → app , as shown in the free-body diagram in Figure 6.4 (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x - and y -axes are in the same direction as F → 1 F → 1 and F → 2 . F → 2 . The problem quickly becomes a one-dimensional problem along the direction of F → app F → app , since friction is in the direction opposite to F → app . F → app . Our strategy is to find the magnitude and direction of the net applied force F → app F → app and then apply Newton’s second law to solve for the drag force F → D . F → D .

The angle is given by

From Newton’s first law, we know this is the same direction as the acceleration. We also know that F → D F → D is in the opposite direction of F → app , F → app , since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as F → app , F → app , but its magnitude is slightly less than F → app . F → app . The problem is now one-dimensional. From the free-body diagram, we can see that

However, Newton’s second law states that

This can be solved for the magnitude of the drag force of the water F D F D in terms of known quantities:

Substituting known values gives

The direction of F → D F → D has already been determined to be in the direction opposite to F → app , F → app , or at an angle of 53 ° 53 ° south of west.

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

Example 6.3

What does the bathroom scale read in an elevator.

From the free-body diagram, we see that F → net = F → s − w → , F → net = F → s − w → , so we have

Solving for F s F s gives us an equation with only one unknown:

or, because w = m g , w = m g , simply

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes F s − w = − m a . F s − w = − m a . )

  • We have a = 1.20 m/s 2 , a = 1.20 m/s 2 , so that F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) yielding F s = 825 N . F s = 825 N .
  • Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because a = Δ v Δ t a = Δ v Δ t and Δ v = 0 . Δ v = 0 . Thus, F s = m a + m g = 0 + m g F s = m a + m g = 0 + m g or F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , which gives F s = 735 N . F s = 735 N .

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure 6.5 (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

Check Your Understanding 6.1

Now calculate the scale reading when the elevator accelerates downward at a rate of 1.20 m/s 2 . 1.20 m/s 2 .

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Example 6.4

Two attached blocks.

For block 1: T → + w → 1 + N → = m 1 a → 1 T → + w → 1 + N → = m 1 a → 1

For block 2: T → + w → 2 = m 2 a → 2 . T → + w → 2 = m 2 a → 2 .

Notice that T → T → is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects

When block 1 moves to the right, block 2 travels an equal distance downward; thus, a 1 x = − a 2 y . a 1 x = − a 2 y . Writing the common acceleration of the blocks as a = a 1 x = − a 2 y , a = a 1 x = − a 2 y , we now have

From these two equations, we can express a and T in terms of the masses m 1 and m 2 , and g : m 1 and m 2 , and g :

Check Your Understanding 6.2

Calculate the acceleration of the system, and the tension in the string, when the masses are m 1 = 5.00 kg m 1 = 5.00 kg and m 2 = 3.00 kg . m 2 = 3.00 kg .

Example 6.5

Atwood machine.

  • We have For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . (The negative sign in front of m 2 a m 2 a indicates that m 2 m 2 accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . Solving for a : a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 . a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 .
  • Observing the first block, we see that T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N . T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N .

Check Your Understanding 6.3

Determine a general formula in terms of m 1 , m 2 m 1 , m 2 and g for calculating the tension in the string for the Atwood machine shown above.

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

Example 6.6

What force must a soccer player exert to reach top speed.

  • We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is Δ v = 8.00 m/s Δ v = 8.00 m/s . We are given the elapsed time, so Δ t = 2.50 s . Δ t = 2.50 s . The unknown is acceleration, which can be found from its definition: a = Δ v Δ t . a = Δ v Δ t . Substituting the known values yields a = 8.00 m/s 2.50 s = 3.20 m/s 2 . a = 8.00 m/s 2.50 s = 3.20 m/s 2 .
  • Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is, F net = m a . F net = m a . Substituting the known values of m and a gives F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N . F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N .

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Check Your Understanding 6.4

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

Example 6.7

What force acts on a model helicopter.

The magnitude of the force is now easily found:

Check Your Understanding 6.5

Find the direction of the resultant for the 1.50-kg model helicopter.

Example 6.8

Baggage tractor.

  • ∑ F x = m system a x ∑ F x = m system a x and ∑ F x = 820.0 t , ∑ F x = 820.0 t , so 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . Since acceleration is a function of time, we can determine the velocity of the tractor by using a = d v d t a = d v d t with the initial condition that v 0 = 0 v 0 = 0 at t = 0 . t = 0 . We integrate from t = 0 t = 0 to t = 3 : t = 3 : d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s . d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s .
  • Refer to the free-body diagram in Figure 6.8 (b). ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N . ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N .

Recall that v = d s d t v = d s d t and a = d v d t a = d v d t . If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have d t = d s v d t = d s v and d t = d v a . d t = d v a . Now, equating these expressions, we have d s v = d v a . d s v = d v a . We can rearrange this to obtain a d s = v d v . a d s = v d v .

Example 6.9

Motion of a projectile fired vertically.

The acceleration depends on v and is therefore variable. Since a = f ( v ) , a = f ( v ) , we can relate a to v using the rearrangement described above,

We replace ds with dy because we are dealing with the vertical direction,

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

Thus, h = 114 m . h = 114 m .

Check Your Understanding 6.6

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Interactive

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

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6 Applications of Newton’s Laws

6.1 Solving Problems with Newton’s Laws

Learning objectives.

By the end of the section, you will be able to:

  • Apply problem-solving techniques to solve for quantities in more complex systems of forces
  • Use concepts from kinematics to solve problems using Newton’s laws of motion
  • Solve more complex equilibrium problems
  • Solve more complex acceleration problems
  • Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy: Applying Newton’s Laws of Motion

  • Identify the physical principles involved by listing the givens and the quantities to be calculated.
  • Sketch the situation, using arrows to represent all forces.
  • Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
  • Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
  • Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure (a). Then, as in Figure (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

This figure shows the development of the free body diagram of a piano being lifted and passed through a window. Figure a is a sketch showing the piano hanging from a crane and part way through a window. Figure b identifies the forces. It shows the same sketch with the addition of the forces, represented as labeled vector arrows. Vector T points up, vector F sub T points down, vector w points down. Figure c defines the system of interest. The sketch is shown again with the piano circled and identified as the system of interest. Only vectors T up and w down are included in this diagram. The downward force F sub T is not a force on the system of interest since it is exerted on the outside world. It must be omitted from the free body diagram. The free body diagram is shown as well. It consists of a dot, representing the system of interest, and the vectors T pointing up and w pointing down, with their tails at the dot. Figure d shows the addition of the forces. Vectors T and w are shown. We are told that these forces must be equal and opposite since the net external force is zero. Thus T is equal to minus w.

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

(If, for example, the system is accelerating horizontally, then you can then set [latex]{a}_{y}=0.[/latex]) We need this information to determine unknown forces acting on a system.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Different Tensions at Different Angles

Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in Figure . Find the tension in each wire, neglecting the masses of the wires.

A sketch of a traffic light suspended from two wires supported by two poles is shown. (b) Some forces are shown in this system. Tension T sub one pulling the top of the left-hand pole is shown by the vector arrow along the left wire from the top of the pole, and an equal but opposite tension T sub one is shown by the arrow pointing up along the left-hand wire where it is attached to the light; the left-hand wire makes a thirty-degree angle with the horizontal. Tension T sub two is shown by a vector arrow pointing downward from the top of the right-hand pole along the right-hand wire, and an equal but opposite tension T sub two is shown by the arrow pointing up along the right-hand wire, which makes a forty-five degree angle with the horizontal. The traffic light is suspended at the lower end of the wires, and its weight W is shown by a vector arrow acting downward. (c) The traffic light is the system of interest, indicated by circling the traffic light. Tension T sub one starting from the traffic light is shown by an arrow along the wire making an angle of thirty degrees with the horizontal. Tension T sub two starting from the traffic light is shown by an arrow along the wire making an angle of forty-five degrees with the horizontal. The weight W is shown by a vector arrow pointing downward from the traffic light. A free-body diagram is shown with three forces acting on a point. Weight W acts downward; T sub one and T sub two act at an angle with the vertical. A coordinate system is shown, with positive x to the right and positive y upward. (d) Forces are shown with their components. T sub one is decomposed into T sub one y pointing vertically upward and T sub one x pointing along the negative x direction. The angle between T sub one and T sub one x is thirty degrees. T sub two is decomposed into T sub two y pointing vertically upward and T sub two x pointing along the positive x direction. The angle between T sub two and T sub two x is forty five degrees. Weight W is shown by a vector arrow acting downward. (e) The net vertical force is zero, so the vector equation is T sub one y plus T sub two y equals W. T sub one y and T sub two y are shown on a free body diagram as equal length arrows pointing up. W is shown as a downward pointing arrow whose length is twice as long as each of the T sub one y and T sub two y arrows. The net horizontal force is zero, so vector T sub one x is equal to minus vector T sub two x. T sub two x is shown by an arrow pointing toward the right, and T sub one x is shown by an arrow pointing toward the left.

The system of interest is the traffic light, and its free-body diagram is shown in Figure (c). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in Figure (d). There are two unknowns in this problem ([latex]{T}_{1}[/latex] and [latex]{T}_{2}[/latex]), so two equations are needed to find them. These two equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero.

First consider the horizontal or x -axis:

Thus, as you might expect,

This gives us the following relationship:

Note that [latex]{T}_{1}[/latex] and [latex]{T}_{2}[/latex] are not equal in this case because the angles on either side are not equal. It is reasonable that [latex]{T}_{2}[/latex] ends up being greater than [latex]{T}_{1}[/latex] because it is exerted more vertically than [latex]{T}_{1}.[/latex]

Now consider the force components along the vertical or y -axis:

This implies

Substituting the expressions for the vertical components gives

There are two unknowns in this equation, but substituting the expression for [latex]{T}_{2}[/latex] in terms of [latex]{T}_{1}[/latex] reduces this to one equation with one unknown:

which yields

Solving this last equation gives the magnitude of [latex]{T}_{1}[/latex] to be

Finally, we find the magnitude of [latex]{T}_{2}[/latex] by using the relationship between them, [latex]{T}_{2}=1.225{T}_{1}[/latex], found above. Thus we obtain

Significance

Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker in Newton’s Laws of Motion .

Particle Acceleration

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

Drag Force on a Barge

Two tugboats push on a barge at different angles ( Figure ). The first tugboat exerts a force of [latex]2.7\times {10}^{5}\,\text{N}[/latex] in the x -direction, and the second tugboat exerts a force of [latex]3.6\times {10}^{5}\,\text{N}[/latex] in the y -direction. The mass of the barge is [latex]5.0\times {10}^{6}\,\text{kg}[/latex] and its acceleration is observed to be [latex]7.5\times {10}^{-2}\,{\text{m/s}}^{2}[/latex] in the direction shown. What is the drag force of the water on the barge resisting the motion? ( Note: Drag force is a frictional force exerted by fluids, such as air or water. The drag force opposes the motion of the object. Since the barge is flat bottomed, we can assume that the drag force is in the direction opposite of motion of the barge.)

(a) A view from above of two tugboats pushing on a barge. One tugboat is pushing with the force F sub 1 equal to two point seven times by ten to the five newtons, shown by a vector arrow acting toward the right in the x direction. Another tugboat is pushing with a force F sub 2 equal to three point six times by ten to the five newtons acting upward in the positive y direction. Acceleration of the barge, a, is shown by a vector arrow directed fifty-three point one degree angle above the x axis. In the free-body diagram, the mass is represented by a point, F sub 2 is acting upward on the point, F sub 1 is acting toward the right, and F sub D is acting approximately southwest. (b) The vectors F sub 1 and F sub 2 are the sides of a right triangle. The resultant is the hypotenuse of this triangle, vector F sub app, making a fifty-three point one degree angle from the base vector F sub 1. The vector F sub app plus the vector force F sub D, pointing down the incline, is equal to the force vector F sub net, which points up the incline.

The directions and magnitudes of acceleration and the applied forces are given in Figure (a). We define the total force of the tugboats on the barge as [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex] so that

The drag of the water [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}[/latex] is in the direction opposite to the direction of motion of the boat; this force thus works against [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] as shown in the free-body diagram in Figure (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x – and y -axes are in the same direction as [latex]{\mathbf{\overset{\to }{F}}}_{1}[/latex] and [latex]{\mathbf{\overset{\to }{F}}}_{2}.[/latex] The problem quickly becomes a one-dimensional problem along the direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex], since friction is in the direction opposite to [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}.[/latex] Our strategy is to find the magnitude and direction of the net applied force [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex] and then apply Newton’s second law to solve for the drag force [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}.[/latex]

Since [latex]{F}_{x}[/latex] and [latex]{F}_{y}[/latex] are perpendicular, we can find the magnitude and direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex] directly. First, the resultant magnitude is given by the Pythagorean theorem:

The angle is given by

From Newton’s first law, we know this is the same direction as the acceleration. We also know that [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}[/latex] is in the opposite direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] but its magnitude is slightly less than [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}.[/latex] The problem is now one-dimensional. From the free-body diagram, we can see that

However, Newton’s second law states that

This can be solved for the magnitude of the drag force of the water [latex]{F}_{\text{D}}[/latex] in terms of known quantities:

Substituting known values gives

The direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}[/latex] has already been determined to be in the direction opposite to [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] or at an angle of [latex]53^\circ[/latex] south of west.

The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is relatively small for a well-designed hull at low speeds, consistent with the answer to this example, where [latex]{F}_{\text{D}}[/latex] is less than 1/600th of the weight of the ship.

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

What Does the Bathroom Scale Read in an Elevator?

Figure shows a 75.0-kg man (weight of about 165 lb.) standing on a bathroom scale in an elevator. Calculate the scale reading: (a) if the elevator accelerates upward at a rate of [latex]1.20\,{\text{m/s}}^{2},[/latex] and (b) if the elevator moves upward at a constant speed of 1 m/s.

A person is standing on a bathroom scale in an elevator. His weight w is shown by an arrow near his chest, pointing downward. F sub s is the force of the scale on the person, shown by a vector starting from his feet pointing vertically upward. W sub s is the weight of the scale, shown by a vector starting at the scale pointing pointing vertically downward. W sub e is the weight of the elevator, shown by a broken arrow starting at the bottom of the elevator pointing vertically downward. F sub p is the force of the person on the scale, drawn starting at the scale and pointing vertically downward. F sub t is the force of the scale on the floor of the elevator, pointing vertically downward, and N is the normal force of the floor on the scale, starting on the elevator near the scale pointing upward. (b) The same person is shown on the scale in the elevator, but only a few forces are shown acting on the person, which is our system of interest. W is shown by an arrow acting downward, and F sub s is the force of the scale on the person, shown by a vector starting from his feet pointing vertically upward. The free-body diagram is also shown, with two forces acting on a point. F sub s acts vertically upward, and w acts vertically downward. An x y coordinate system is shown, with positive x to the right and positive y upward.

If the scale at rest is accurate, its reading equals [latex]{\mathbf{\overset{\to }{F}}}_{\text{p}}[/latex], the magnitude of the force the person exerts downward on it. Figure (a) shows the numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person is chosen to be the system of interest and a free-body diagram is drawn, as in Figure (b). Analysis of the free-body diagram using Newton’s laws can produce answers to both Figure (a) and (b) of this example, as well as some other questions that might arise. The only forces acting on the person are his weight [latex]\mathbf{\overset{\to }{w}}[/latex] and the upward force of the scale [latex]{\mathbf{\overset{\to }{F}}}_{\text{s}}.[/latex] According to Newton’s third law, [latex]{\mathbf{\overset{\to }{F}}}_{\text{p}}[/latex] and [latex]{\mathbf{\overset{\to }{F}}}_{\text{s}}[/latex] are equal in magnitude and opposite in direction, so that we need to find [latex]{F}_{\text{s}}[/latex] in order to find what the scale reads. We can do this, as usual, by applying Newton’s second law,

From the free-body diagram, we see that [latex]{\mathbf{\overset{\to }{F}}}_{\text{net}}={\mathbf{\overset{\to }{F}}}_{s}-\mathbf{\overset{\to }{w}},[/latex] so we have

Solving for [latex]{F}_{s}[/latex] gives us an equation with only one unknown:

or, because [latex]w=mg,[/latex] simply

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes [latex]{F}_{s}-w=\text{−}ma.[/latex])

which gives

The scale reading in Figure (a) is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight:

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

Check Your Understanding

Now calculate the scale reading when the elevator accelerates downward at a rate of [latex]1.20\,{\text{m/s}}^{2}.[/latex]

[latex]{F}_{\text{s}}=645\,\text{N}[/latex]

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Two Attached Blocks

Figure shows a block of mass [latex]{m}_{1}[/latex] on a frictionless, horizontal surface. It is pulled by a light string that passes over a frictionless and massless pulley. The other end of the string is connected to a block of mass [latex]{m}_{2}.[/latex] Find the acceleration of the blocks and the tension in the string in terms of [latex]{m}_{1},{m}_{2},\,\text{and}\,g.[/latex]

(a) Block m sub 1 is on a horizontal surface. It is connected to a string that passes over a pulley then hangs straight down and connects to block m sub 2. Block m sub 1 has acceleration a sub 1 directed to the right. Block m sub 2 has acceleration a sub 2 directed downward. (b) Free body diagrams of each block. Block m sub 1 has force w sub 1 directed vertically down, N directed vertically up, and T directed horizontally to the right. Block m sub 2 has force w sub 2 directed vertically down, and T directed vertically up. The x y coordinate system has positive x to the right and positive y up.

We draw a free-body diagram for each mass separately, as shown in Figure . Then we analyze each one to find the required unknowns. The forces on block 1 are the gravitational force, the contact force of the surface, and the tension in the string. Block 2 is subjected to the gravitational force and the string tension. Newton’s second law applies to each, so we write two vector equations:

For block 1: [latex]\mathbf{\overset{\to }{T}}+{\mathbf{\overset{\to }{w}}}_{1}+\mathbf{\overset{\to }{N}}={m}_{1}{\mathbf{\overset{\to }{a}}}_{1}[/latex]

For block 2: [latex]\mathbf{\overset{\to }{T}}+{\mathbf{\overset{\to }{w}}}_{2}={m}_{2}{\mathbf{\overset{\to }{a}}}_{2}.[/latex]

Notice that [latex]\mathbf{\overset{\to }{T}}[/latex] is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects

The component equations follow from the vector equations above. We see that block 1 has the vertical forces balanced, so we ignore them and write an equation relating the x -components. There are no horizontal forces on block 2, so only the y -equation is written. We obtain these results:

When block 1 moves to the right, block 2 travels an equal distance downward; thus, [latex]{a}_{1x}=\text{−}{a}_{2y}.[/latex] Writing the common acceleration of the blocks as [latex]a={a}_{1x}=\text{−}{a}_{2y},[/latex] we now have

From these two equations, we can express a and T in terms of the masses [latex]{m}_{1}\,\text{and}\,{m}_{2},\,\text{and}\,g:[/latex]

Notice that the tension in the string is less than the weight of the block hanging from the end of it. A common error in problems like this is to set [latex]T={m}_{2}g.[/latex] You can see from the free-body diagram of block 2 that cannot be correct if the block is accelerating.

Calculate the acceleration of the system, and the tension in the string, when the masses are [latex]{m}_{1}=5.00\,\text{kg}[/latex] and [latex]{m}_{2}=3.00\,\text{kg}.[/latex]

[latex]a=3.68\,{\text{m/s}}^{2},[/latex] [latex]T=18.4\,\text{N}[/latex]

Atwood Machine

A classic problem in physics, similar to the one we just solved, is that of the Atwood machine , which consists of a rope running over a pulley, with two objects of different mass attached. It is particularly useful in understanding the connection between force and motion. In Figure , [latex]{m}_{1}=2.00\,\text{kg}[/latex] and [latex]{m}_{2}=4.00\,\text{kg}\text{.}[/latex] Consider the pulley to be frictionless. (a) If [latex]{m}_{2}[/latex] is released, what will its acceleration be? (b) What is the tension in the string?

An Atwood machine consists of masses suspended on either side of a pulley by a string passing over the pulley. In the figure, mass m sub 1 is on the left and mass m sub 2 is on the right. The free body diagram of block one shows mass one with force vector T pointing vertically up and force vector w sub one pointing vertically down. The free body diagram of block two shows mass two with force vector T pointing vertically up and force vector w sub two pointing vertically down.

We draw a free-body diagram for each mass separately, as shown in the figure. Then we analyze each diagram to find the required unknowns. This may involve the solution of simultaneous equations. It is also important to note the similarity with the previous example. As block 2 accelerates with acceleration [latex]{a}_{2}[/latex] in the downward direction, block 1 accelerates upward with acceleration [latex]{a}_{1}[/latex]. Thus, [latex]a={a}_{1}=\text{−}{a}_{2}.[/latex]

(The negative sign in front of [latex]{m}_{2}a[/latex] indicates that [latex]{m}_{2}[/latex] accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is

Solving for a :

  • Observing the first block, we see that [latex]\begin{array}{c}T-{m}_{1}g={m}_{1}a\hfill \\ T={m}_{1}(g+a)=(2\,\text{kg})(9.8\,{\text{m/s}}^{2}+3.27\,{\text{m/s}}^{2})=26.1\,\text{N}\text{.}\hfill \end{array}[/latex]

The result for the acceleration given in the solution can be interpreted as the ratio of the unbalanced force on the system, [latex]({m}_{2}-{m}_{1})g[/latex], to the total mass of the system, [latex]{m}_{1}+{m}_{2}[/latex]. We can also use the Atwood machine to measure local gravitational field strength.

Determine a general formula in terms of [latex]{m}_{1},{m}_{2}[/latex] and g for calculating the tension in the string for the Atwood machine shown above.

[latex]T=\frac{2{m}_{1}{m}_{2}}{{m}_{1}+{m}_{2}}g[/latex] (This is found by substituting the equation for acceleration in Figure (a), into the equation for tension in Figure (b).)

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

What Force Must a Soccer Player Exert to Reach Top Speed?

A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What is her average acceleration? (b) What average force does the ground exert forward on the runner so that she achieves this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

To find the answers to this problem, we use the problem-solving strategy given earlier in this chapter. The solutions to each part of the example illustrate how to apply specific problem-solving steps. In this case, we do not need to use all of the steps. We simply identify the physical principles, and thus the knowns and unknowns; apply Newton’s second law; and check to see whether the answer is reasonable.

Substituting the known values yields

Substituting the known values of m and a gives

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

This example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles, the knowns, and the unknowns involved in the problem. The second step is to solve for the unknown, in this case using Newton’s second law. Finally, we check our answer to ensure it is reasonable. These techniques for integrated concept problems will be useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life.

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

What Force Acts on a Model Helicopter?

A 1.50-kg model helicopter has a velocity of [latex]5.00\mathbf{\hat{j}}\,\text{m/s}[/latex] at [latex]t=0.[/latex] It is accelerated at a constant rate for two seconds (2.00 s) after which it has a velocity of [latex](6.00\mathbf{\hat{i}}+12.00\mathbf{\hat{j}})\text{m/s}\text{.}[/latex] What is the magnitude of the resultant force acting on the helicopter during this time interval?

We can easily set up a coordinate system in which the x -axis [latex](\mathbf{\hat{i}}[/latex] direction) is horizontal, and the y -axis [latex](\mathbf{\hat{j}}[/latex] direction) is vertical. We know that [latex]\Delta t=2.00s[/latex] and [latex](6.00\mathbf{\hat{i}}+12.00\mathbf{\hat{j}}\,\text{m/s})-(5.00\mathbf{\hat{j}}\,\text{m/s}).[/latex] From this, we can calculate the acceleration by the definition; we can then apply Newton’s second law.

The magnitude of the force is now easily found:

The original problem was stated in terms of [latex]\mathbf{\hat{i}}-\mathbf{\hat{j}}[/latex] vector components, so we used vector methods. Compare this example with the previous example.

Find the direction of the resultant for the 1.50-kg model helicopter.

49.4 degrees

Baggage Tractor

Figure (a) shows a baggage tractor pulling luggage carts from an airplane. The tractor has mass 650.0 kg, while cart A has mass 250.0 kg and cart B has mass 150.0 kg. The driving force acting for a brief period of time accelerates the system from rest and acts for 3.00 s. (a) If this driving force is given by [latex]F=(820.0t)\,\text{N,}[/latex] find the speed after 3.00 seconds. (b) What is the horizontal force acting on the connecting cable between the tractor and cart A at this instant?

Figure (a) shows a baggage tractor driving to the left and pulling two luggage carts. The external forces on the system are shown. The forces on the tractor are F sub tractor, horizontally to the left, N sub tractor vertically up, and w sub tractor vertically down. The forces on the cart immediately behind the tractor, cart A, are N sub A vertically up, and w sub A vertically down. The forces on cart B, the one behind cart A, are N sub B vertically up, and w sub B vertically down. Figure (b) shows the free body diagram of the tractor, consisting of F sub tractor, horizontally to the left, N sub tractor vertically up, w sub tractor vertically down, and T horizontally to the right.

A free-body diagram shows the driving force of the tractor, which gives the system its acceleration. We only need to consider motion in the horizontal direction. The vertical forces balance each other and it is not necessary to consider them. For part b, we make use of a free-body diagram of the tractor alone to determine the force between it and cart A. This exposes the coupling force [latex]\mathbf{\overset{\to }{T}},[/latex] which is our objective.

Since acceleration is a function of time, we can determine the velocity of the tractor by using [latex]a=\frac{dv}{dt}[/latex] with the initial condition that [latex]{v}_{0}=0[/latex] at [latex]t=0.[/latex] We integrate from [latex]t=0[/latex] to [latex]t=3\text{:}[/latex]

  • Refer to the free-body diagram in Figure (b). [latex]\begin{array}{ccc}\hfill \sum {F}_{x}& =\hfill & {m}_{\text{tractor}}{a}_{x}\hfill \\ \hfill 820.0t-T& =\hfill & {m}_{\text{tractor}}(0.7805)t\hfill \\ \hfill (820.0)(3.00)-T& =\hfill & (650.0)(0.7805)(3.00)\hfill \\ \hfill T& =\hfill & 938\,\text{N}.\hfill \end{array}[/latex]

Since the force varies with time, we must use calculus to solve this problem. Notice how the total mass of the system was important in solving Figure (a), whereas only the mass of the truck (since it supplied the force) was of use in Figure (b).

Recall that [latex]v=\frac{ds}{dt}[/latex] and [latex]a=\frac{dv}{dt}[/latex]. If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have [latex]dt=\frac{ds}{v}[/latex] and [latex]dt=\frac{dv}{a}.[/latex] Now, equating these expressions, we have [latex]\frac{ds}{v}=\frac{dv}{a}.[/latex] We can rearrange this to obtain [latex]{a}^{}ds={v}^{}dv.[/latex]

Motion of a Projectile Fired Vertically

A 10.0-kg mortar shell is fired vertically upward from the ground, with an initial velocity of 50.0 m/s (see Figure ). Determine the maximum height it will travel if atmospheric resistance is measured as [latex]{F}_{\text{D}}=(0.0100{v}^{2})\,\text{N,}[/latex] where v is the speed at any instant.

(a) A photograph of a soldier firing a mortar shell straight up. (b) A free body diagram of the mortar shell shows forces F sub D and w, both pointing vertically down. Force w is larger than force F sub D.

The known force on the mortar shell can be related to its acceleration using the equations of motion. Kinematics can then be used to relate the mortar shell’s acceleration to its position.

Initially, [latex]{y}_{0}=0[/latex] and [latex]{v}_{0}=50.0\,\text{m/s}\text{.}[/latex] At the maximum height [latex]y=h,v=0.[/latex] The free-body diagram shows [latex]{F}_{\text{D}}[/latex] to act downward, because it slows the upward motion of the mortar shell. Thus, we can write

The acceleration depends on v and is therefore variable. Since [latex]a=f(v)\text{,}[/latex] we can relate a to v using the rearrangement described above,

We replace ds with dy because we are dealing with the vertical direction,

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

Thus, [latex]h=114\,\text{m}\text{.}[/latex]

Notice the need to apply calculus since the force is not constant, which also means that acceleration is not constant. To make matters worse, the force depends on v (not t ), and so we must use the trick explained prior to the example. The answer for the height indicates a lower elevation if there were air resistance. We will deal with the effects of air resistance and other drag forces in greater detail in Drag Force and Terminal Speed .

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

  • Newton’s laws of motion can be applied in numerous situations to solve motion problems.
  • Some problems contain multiple force vectors acting in different directions on an object. Be sure to draw diagrams, resolve all force vectors into horizontal and vertical components, and draw a free-body diagram. Always analyze the direction in which an object accelerates so that you can determine whether [latex]{F}_{\text{net}}=ma[/latex] or [latex]{F}_{\text{net}}=0.[/latex]
  • The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating vertically, the normal force is less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force is always less than the full weight of the object.
  • Some problems contain several physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics to solve these problems.

Conceptual Questions

To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at g . Why do they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?

The scale is in free fall along with the astronauts, so the reading on the scale would be 0. There is no difference in the apparent weightlessness; in the aircraft and in orbit, free fall is occurring.

A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force [latex]\mathbf{\overset{\to }{F}}[/latex] so that the swing ropes are [latex]30.0^\circ[/latex] with respect to the vertical. (a) Calculate the tension in each of the two ropes supporting the swing under these conditions. (b) Calculate the magnitude of [latex]\mathbf{\overset{\to }{F}}.[/latex]

a. 170 N; b. 170 N

Find the tension in each of the three cables supporting the traffic light if it weighs 2.00 × 10 2 N.

A sketch of a traffic light suspended by a cable that is in turn suspended from two other cables is shown. Tension T sub 3 is the tension in the cable connecting the traffic light to the upper cables. Tension T sub one is the tension in the upper cable pulling up and to the left, making a 41 degree angle with the horizontal. Tension T sub two is the tension pulling up and to the right, making a 63 degree angle with the horizontal. Force vector w equal to 200 Newtons pulls vertically downward on the traffic light.

Three forces act on an object, considered to be a particle, which moves with constant velocity [latex]v=(3\mathbf{\hat{i}}-2\mathbf{\hat{j}})\,\text{m/s}\text{.}[/latex] Two of the forces are [latex]{\mathbf{\overset{\to }{F}}}_{1}=(3\mathbf{\hat{i}}+5\mathbf{\hat{j}}-6\mathbf{\hat{k}})\,\text{N}[/latex] and [latex]{\mathbf{\overset{\to }{F}}}_{2}=(4\mathbf{\hat{i}}-7\mathbf{\hat{j}}+2\mathbf{\hat{k}})\,\text{N}\text{.}[/latex] Find the third force.

[latex]{\mathbf{\overset{\to }{F}}}_{3}=(-7\mathbf{\hat{i}}+2\mathbf{\hat{j}}+4\mathbf{\hat{k}})\,\text{N}[/latex]

A flea jumps by exerting a force of [latex]1.20\times {10}^{-5}\,\text{N}[/latex] straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of [latex]0.500\times {10}^{-6}\,\text{N}[/latex] on the flea while the flea is still in contact with the ground. Find the direction and magnitude of the acceleration of the flea if its mass is [latex]6.00\times {10}^{-7}\,\text{kg}[/latex]. Do not neglect the gravitational force.

Two muscles in the back of the leg pull upward on the Achilles tendon, as shown below. (These muscles are called the medial and lateral heads of the gastrocnemius muscle.) Find the magnitude and direction of the total force on the Achilles tendon. What type of movement could be caused by this force?

An Achilles tendon is shown in the figure with two forces exerted on it by the lateral and medial heads of the gastrocnemius muscle. F sub one, equal to two hundred Newtons, is shown as a vector making an angle twenty degrees to the right of vertical, and F sub two, equal to two hundred Newtons, is shown making an angle of twenty degrees left of vertical.

After a mishap, a 76.0-kg circus performer clings to a trapeze, which is being pulled to the side by another circus artist, as shown here. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.

A circus performer hanging from a trapeze is being pulled to the right by another performer using a rope. Her weight is shown by a vector w acting vertically downward. The trapeze rope exerts a tension, T sub one, up and to the left, making an angle of fifteen degrees with the vertical. The second performer pulls with tension T sub two, making an angle of ten degrees above the positive x direction.

A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow the first dolphin if it was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.)

When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b) How far does he travel?

A large rocket has a mass of [latex]2.00\times {10}^{6}\,\text{kg}[/latex] at takeoff, and its engines produce a thrust of [latex]3.50\times {10}^{7}\,\text{N}.[/latex] (a) Find its initial acceleration if it takes off vertically. (b) How long does it take to reach a velocity of 120 km/h straight up, assuming constant mass and thrust?

a. [latex]7.70\,{\text{m/s}}^{2}[/latex]; b. 4.33 s

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110.0 kg.

A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110.0 m. (a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.

a. 46.4 m/s; b. [latex]2.40\times {10}^{3}\,{\text{m/s}}^{2}\text{;}[/latex] c. 5.99 × 10 3 N; ratio of 245

A 0.500-kg potato is fired at an angle of [latex]80.0^\circ[/latex] above the horizontal from a PVC pipe used as a “potato gun” and reaches a height of 110.0 m. (a) Neglecting air resistance, calculate the potato’s velocity when it leaves the gun. (b) The gun itself is a tube 0.450 m long. Calculate the average acceleration of the potato in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the potato in the gun? Express your answer in newtons and as a ratio to the weight of the potato.

An elevator filled with passengers has a mass of [latex]1.70\times {10}^{3}\,\text{kg}[/latex]. (a) The elevator accelerates upward from rest at a rate of [latex]1.20\,{\text{m/s}}^{2}[/latex] for 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of [latex]0.600\,{\text{m/s}}^{2}[/latex] for 3.00 s. What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity?

a. [latex]1.87\times {10}^{4}\,\text{N;}[/latex] b. [latex]1.67\times {10}^{4}\,\text{N;}[/latex] c. [latex]1.56\times {10}^{4}\,\text{N;}[/latex] d. 19.4 m, 0 m/s

A 20.0-g ball hangs from the roof of a freight car by a string. When the freight car begins to move, the string makes an angle of [latex]35.0^\circ[/latex] with the vertical. (a) What is the acceleration of the freight car? (b) What is the tension in the string?

A student’s backpack, full of textbooks, is hung from a spring scale attached to the ceiling of an elevator. When the elevator is accelerating downward at [latex]3.8\,{\text{m/s}}^{2}[/latex], the scale reads 60 N. (a) What is the mass of the backpack? (b) What does the scale read if the elevator moves upward while slowing down at a rate [latex]3.8\,{\text{m/s}}^{2}[/latex]? (c) What does the scale read if the elevator moves upward at constant velocity? (d) If the elevator had no brakes and the cable supporting it were to break loose so that the elevator could fall freely, what would the spring scale read?

a. 10 kg; b. 90 N; c. 98 N; d. 0

A service elevator takes a load of garbage, mass 10.0 kg, from a floor of a skyscraper under construction, down to ground level, accelerating downward at a rate of [latex]1.2\,{\text{m/s}}^{2}[/latex]. Find the magnitude of the force the garbage exerts on the floor of the service elevator?

A roller coaster car starts from rest at the top of a track 30.0 m long and inclined at [latex]20.0^\circ[/latex] to the horizontal. Assume that friction can be ignored. (a) What is the acceleration of the car? (b) How much time elapses before it reaches the bottom of the track?

a. [latex]3.35\,{\text{m/s}}^{2}[/latex]; b. 4.2 s

The device shown below is the Atwood’s machine considered in Figure . Assuming that the masses of the string and the frictionless pulley are negligible, (a) find an equation for the acceleration of the two blocks; (b) find an equation for the tension in the string; and (c) find both the acceleration and tension when block 1 has mass 2.00 kg and block 2 has mass 4.00 kg.

An Atwood machine consisting of masses suspended on either side of a pulley by a string passing over the pulley is shown. Mass m sub 1 is on the left and mass m sub 2 is on the right.

Two blocks are connected by a massless rope as shown below. The mass of the block on the table is 4.0 kg and the hanging mass is 1.0 kg. The table and the pulley are frictionless. (a) Find the acceleration of the system. (b) Find the tension in the rope. (c) Find the speed with which the hanging mass hits the floor if it starts from rest and is initially located 1.0 m from the floor.

Block m sub 1 is on a horizontal table. It is connected to a string that passes over a pulley at the edge of the table. The string then hangs straight down and connects to block m sub 2, which is not in contact with the table. Block m sub 1 has acceleration a sub 1 directed to the right. Block m sub 2 has acceleration a sub 2 directed downward.

Shown below are two carts connected by a cord that passes over a small frictionless pulley. Each cart rolls freely with negligible friction. Calculate the acceleration of the carts and the tension in the cord.

Two carts connected by a string passing over a pulley are on either side of a double inclined plane. The string passes over a pulley attached to the top of the double incline. On the left, the incline makes an angle of 37 degrees with the horizontal and the cart on that side has mass 10 kilograms. On the right, the incline makes an angle of 53 degrees with the horizontal and the cart on that side has mass 15 kilograms.

A 2.00 kg block (mass 1) and a 4.00 kg block (mass 2) are connected by a light string as shown; the inclination of the ramp is [latex]40.0^\circ[/latex]. Friction is negligible. What is (a) the acceleration of each block and (b) the tension in the string?

Block 1 is on a ramp inclined up and to the right at an angle of 40 degrees above the horizontal. It is connected to a string that passes over a pulley at the top of the ramp, then hangs straight down and connects to block 2. Block 2 is not in contact with the ramp.

6.1 Solving Problems with Newton’s Laws Copyright © 2016 by OpenStax. All Rights Reserved.

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  • Newton's Law of Motion

Newton's Third Law of Motion: Solved Problems

In this article, you learn the concepts of Newton's third law of motion by problem-solving approach. Here, we are going to find pairs of forces appearing in Newton's third law using some worked problems. 

Introduction

Consider the following everyday experiment: Imagine applying a force with a hammer to drive a nail into a wooden block. As a result, the hammer comes to a complete stop. 

What forces caused the hammer's speed to be reduced to zero? 

This is the question that Newton speculated on. 

According to Newton's idea, the nail exerts a force of the same magnitude but in the opposite direction on the hammer, causing it to rapidly come to a complete stop upon contact. 

This is the essence of Newton's third law, which states that if two objects 1 and 2 interact, the force $\vec{F}_{12}$ that object 1 exerts on object 2 is equal in magnitude but opposite in direction to the force $\vec{F}_{21}$ that object 2 exerts on object 1. \[\vec{F}_{12}=-\vec{F}_{21}\] One of the most important consequences of this law is that the forces cannot exist in isolation and always occur in pairs. 

Solved Problems:

Problem (1):  A box stands on a table. Identify all action-reaction forces between the box and the table. 

Solution : If object 1 exerts a force of $\vec{F}_{12}$ on object 2, object 2 also exerts a force $\vec{F}_{21}$ on object 1. This force is equal in magnitude and opposite in direction and can be written as $\vec{F}_{12}=-\vec{F}_{21}$. This simple relation represents Newton's third law of motion.

In this definition, one of the forces, regardless of which one, is referred to as the action force, while the other force is known as the reaction force.

When dealing with problems involving Newton's third law, the main objective is to identify pairs of forces that comply with this law.

The box sitting on a chair does not fall through due to its weight $\vec{w}$. An upward force called the normal force $\vec{n}$ is exerted on the box to balance out the downward weight force and keep the box at rest on the table. These two forces are actual external forces.

We consider this normal force as the action force. The reaction to $\vec{n}$ is the force exerted by the box on the table, denoted as $\vec{n}'$. This reaction force has equal magnitude but opposite direction to $\vec{n}$.

Pairs of forces due to applying Newton's third law on a box sitting on a table.

Thus, we can express this pair of forces as $\vec{n}=-\vec{n}'$. Since these forces are exerted on two different objects, they can be considered an action-reaction pair according to Newton's third law.

Problem (2): Identify the action-reaction force pair in each of the following situations: (a) An apple falls freely.  (b) A hammer hits a nail. (c) The blades of a helicopter push air downward.

Solution : Newton's third law states that when two objects interact, the force that object 1 applies on object 2, called $\vec{F}_{12}$ is equal in magnitude but in the opposite direction to the force that object 2 exerts on object 1, i.e., $\vec{F}_{21}$. \[\vec{F}_{12}=-\vec{F}_{21}\] In all Newton's third law problems, first of all, identify the system under study. 

(a) Here, the apple and the Earth create our system. The Earth applies a downward force on the apple, known as weight $\vec{w}$, and correspondingly, the apple exerts a force on the Earth by $\vec{w}'$. These two are third law force pairs. \[\vec{w}=-\vec{w}'\]  (b) In this case, our system is hammer-nail, two different interacting bodies. The force exerted by the hammer on the nail $\vec{F}_{hn}$ is equal in magnitude but in the opposite direction to the force exerted by the nail on the hammer, $\vec{F}_{nh}$. \[\vec{F}_{hn}=-\vec{F}_{hn}\]  (c) The blades and air form a system as a combined two interacting objects. In this case, the blades exert a force downward on the air and conversely, the air also exerts a force upward on the blades. These two forces act on different objects, not the same object, so it can be considered as Newton's third law pair force. 

Problem (3): A car is skidding to a complete stop on a horizontal road. First, identify all action-reaction pairs of forces, then show them on a free-body diagram.

Solution : The car slows down and eventually comes to a complete stop, experiencing deceleration. According to Newton's third law, action-reaction pairs of forces act between two interacting objects. In this case, the car and the road (Earth) form our system. The gravitational force on the car due to Earth is represented by $\vec{w}$, while the reaction force exerted by the car on Earth is represented by $\vec{w}'$. Therefore, these two forces act on different objects. \[\vec{w}=-\vec{w}'\] When the car is in contact with the road, an upward force is exerted by the road's surface on the car, known as the normal force ($\vec{n}$). Conversely, the car also applies a downward force on the road with the same magnitude but in the opposite direction ($-\vec{n}'$). These two forces form another pair of action-reaction forces according to Newton's third law.

Applying Newton's third law on a car skidding across a road.

Problem (4): A girl exerts a force of 40 N upward to hold a box. Identify the reaction force by determining (a) its magnitude and direction, (b) on what object it is exerted, and (c) by what object it is exerted. 

Solution : Here, three objects are interacting, the girl, the box, and the Earth. The Earth exerts a gravitational force $\vec{F}_g$ on the box. The reaction to this force is the force exerted by the box on the Earth, $\vec{F}'_g$. These two forces are acting on two different objects and are equal in magnitude and opposite in direction, so they form an action-reaction pair force. \[\vec{F}_g=-\vec{F}'_g\] On the other hand, the box is held by the girl. The girl exerts an upward force to hold it up and the reaction to this force is the force exerted by the box downward on the girl's hand. 

Problem (5): A girl stands on a spring scale while riding in an elevator accelerating at $2\,\rm m/s^2$ upward. How does Newton's third law apply to find the scale reading? 

Solution : The scale reads the value of the downward force that the girl exerts on the spring scale. By Newton's third law, the reaction to this force is the upward force exerted by the scale on the woman, known as the normal force $\vec{n}$. These two forces are depicted in the following figure. 

Applying Newton's second law gives us the magnitude of the apparent weight that the girl feels while riding in a moving elevator. To see how to solve such elevator problems, read this. 

Problem (6): A box is pushing on a rough horizontal surface at constant velocity with a force of $\vec{F}$. The friction is also applied by $-\vec{F}$. Are the pushing force and friction the action-reaction pair of forces?

Solution : It is important to remember that according to Newton's third law, force pairs appear between two interacting bodies. In this scenario, three objects are interacting: the person applying a push on the box, the box itself, and the surface over which the box is moving. 

In the person-box system, an external force $\vec{F}$ is applied to the box by the person. As a reaction to this force, the box exerts an equal and opposite force $-\vec{F}$ on the person. 

In the box-surface (Earth) system, friction $-\vec{F}$ is exerted on the box by the rough surface. Conversely, as a reaction to this frictional force, an equal and opposite force $\vec{F}$ is exerted by the box on the surface. 

As you can see, both the push force and friction are applied separately to the same object (the box). Therefore, they cannot be considered as third law pair forces.

Problem (7): A person is standing motionless on the level ground.  (a) Identify all forces acting on the person. (Hewit 49, p.120) (b) Are these forces equal in magnitude and opposite? (c) Do these forces make an action-reaction force pair? Why or why not? (d) Determine all other pairs of forces not shown here.

Solution : (a) The Earth pulls down the person by the weight force, $\vec{w}$ and the surface prevents the person from sinking into the Earth by the upward support force, called normal force $\vec{n}$.  (b) The person is motionless on the level ground, so by the second law, the net force on it must be zero which yields \begin{gather*} F_{net}=ma\\ \vec{w}+\vec{n}=0 \\ \Rightarrow \boxed{\vec{w}=-\vec{n}}\end{gather*} This expression tells us that these two above forces are equal in magnitude and opposite in direction.  (c) Every two equal and opposite forces do not constitute an action-reaction pair of forces. Newton's third law states that these forces must be applied to two different objects.  In this problem, the Earth exerts a downward force (weight) on the person, and the person also exerts an upward force on the Earth with the same magnitude.  Additionally, the Earth's surface applies an upward force (normal force) on the person, and the person exerts an equal and opposite force on the Earth. As you can see, both weight and normal forces are exerted separately between the Earth and the person and do not interact with each other. Therefore, these forces do not form a pair according to Newton's third law.

(d) In the previous section, we mentioned that the weight and normal forces are actual external forces and not paired forces. In this context, the downward weight acts as an action, with its reaction being the force that pulls the Earth towards the person. Similarly, the upward normal force acts as an action, with its reaction being the downward force exerted by the person on the surface.

Problem (8):  An astronaut performing a spacewalk with a combined mass of 165 kg applies a force of 280 N to a satellite that is freely floating in space and has a mass of 850 kg. Determine the magnitude and direction of the reaction force exerted by the satellite on the astronaut. Additionally, calculate the accelerations experienced by both the astronaut and the satellite.

Solution : The satellite and the astronaut are two distinct objects that interact with each other by the forces that make the action-reaction pair force, according to Newton's third law of motion. 

(a) The astronaut applies a force $\vec{F}_{as}$ to the satellite and, by Newton's third law, the satellite also exerts an equal force in the opposite direction on the astronaut as $-\vec{F}_{sa}$. Thus, the satellite exerts a $280\,\rm N$ force on the astronaut in the negative $x$-direction or in vector notation as \[\vec{F}_{sa}=-280\,\rm N\]  (b) The second law gives us the acceleration of an object when a force $F$ is applied to it as $\vec{a}=\frac{\vec{F}}{m}$. Therefore, the astronaut's acceleration is \[a=\frac{-280}{165}=-1.70\,\rm m/s^2\] and for the satellite can be found as \[a=\frac{280}{850}=0.33\,\rm m/s^2\] The negative indicates the direction of the acceleration, which is toward the negative $x$-axis. 

Author : Dr. Ali Nemati Published : August 27, 2023

© 2015 All rights reserved. by Physexams.com

law of motion problem solving with solution

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Solution Guides to College Physics by Openstax Chapter 4 Banner

Chapter 4: Dynamics: Force and Newton’s Laws of Motion

Newton’s Second Law of Motion: Concept of a System

Newton’s Third Law of Motion: Symmetry in Forces

Normal, Tension, and Other Example of Forces

Problem Solving Strategies

Further Applications of Newton’s Laws of Motion

Newton’s second law of motion – problems and solutions

Solved problems in Newton’s laws of motion – Newton’s second law of motion 

1. A 1 kg object accelerated at a constant 5 m/s 2 . Estimate the net force needed to accelerate the object.

Wanted : net force (∑F)

2. Mass of an object = 1 kg, net force ∑F = 2 Newton. Determine the magnitude and direction of the object’s acceleration….

Mass (m) = 1 kg

Mass (m) = 2 kg

∑ F = F 1 – F 2 = 5 – 3 = 2 Newton

Wanted : The magnitude and direction of the acceleration (a)

∑ F = F 1x – F 2 = 5 – 1 = 4 Newton

The direction of the acceleration = the direction of the net force = direction of F 1

Force (F) = 200 N

∑ F = net force, m = mass, a = acceleration

200 – F g = 120

7. Block A with a mass of 100-gram place above block B with a mass of 300 gram, and then block b pushed with a force of 5 N vertically upward. Determine the normal force exerted by block B on block A.

Mass of block B (m B ) = 300 gram = 0.3 kg

Weight of block B (w B ) = (0.3 kg)(10 m/s 2 ) = 3 kg m/s 2 = 3 Newton

w B = weight of block B (act on block B)

Apply Newton’s second law of motion on both blocks :

F – w A – w B = (m A + m B ) a

5 – 4 = (0.4) a

N A – w A = m A a

8. An object with weight of 4 N supported by a cord and pulley. A force of 2 N acts on the block and one end of the cord pulled by a force of 9 N. Determine the net force acts on object X.

Weight of X (w X ) = 4 Newton

The tension force has the same magnitude in all part of the cord. So the tension force is 9 N.

The net force act on the object X is 3 Newton, vertically upward.

The correct answer is A.

B. 1.70 N and its direction same as force acted by Andrew

Minus sign indicated that (F 2 ) is opposite with push force act by Andrew (F 1 ).

ΣF = 3 N + 4 N + 2 N = 9 Newton, rightward

a = ΣF / m = 40 N / 20 kg = 2 N/kg = 2 m/s 2

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law of motion problem solving with solution

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UsingKinEqns1ThN.png

Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep) 

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

t = 32.8 s

v = 0 m/s

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

t = 5.21 s

v = 0 m/s

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

t = 2.6 s

v = 0 m/s

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

v = 18.5 m/s

v = 46.1 m/s

t = 2.47 s

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

v = 0 m/s

d = -1.40 m

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

v = 0 m/s

v = 444 m/s

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

v = 0 m/s

v = 7.10 m/s

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

v = 0 m/s

v = 65 m/s

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

v = 22.4 m/s

v = 0 m/s

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

a = -9.8 m/s

v = 0 m/s

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

v = 0 m/s

v = 521 m/s

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

  • (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use:  v f  = v i  + a*t

0 m/s = v i  + (-9.8  m/s 2 )*(3.13 s)

0 m/s = v i  - 30.7 m/s

v i  = 30.7 m/s  (30.674 m/s)

Now use:  v f 2  = v i 2  + 2*a*d

(0 m/s) 2  = (30.7 m/s) 2  + 2*(-9.8  m/s 2 )*(d)

0 m 2 /s 2  = (940 m 2 /s 2 ) + (-19.6  m/s 2 )*d

-940  m 2 /s 2  = (-19.6  m/s 2 )*d

(-940  m 2 /s 2 )/(-19.6  m/s 2 ) = d

Return to Problem 13

v = 0 m/s

d = -370 m

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

v = 367 m/s

v = 0 m/s

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

t = 3.41 s

v = 0 m/s

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

a = -3.90 m/s

v = 0 m/s

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

v = 0 m/s

v = 88.3 m/s

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

v = 0 m/s

v = m/s

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20

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Applications of Newton’s Laws

Solving Problems with Newton’s Laws

Learning objectives.

By the end of the section, you will be able to:

  • Apply problem-solving techniques to solve for quantities in more complex systems of forces
  • Use concepts from kinematics to solve problems using Newton’s laws of motion
  • Solve more complex equilibrium problems
  • Solve more complex acceleration problems
  • Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

  • Identify the physical principles involved by listing the givens and the quantities to be calculated.
  • Sketch the situation, using arrows to represent all forces.
  • Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
  • Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
  • Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in (Figure) (a). Then, as in (Figure) (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

\stackrel{\to }{T}

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See (Figure) (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. (Figure) (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in (Figure) (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

\sum {F}_{x}=m{a}_{x},\phantom{\rule{0.5em}{0ex}}\sum {F}_{y}=m{a}_{y}.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Different Tensions at Different Angles Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in (Figure) . Find the tension in each wire, neglecting the masses of the wires.

A sketch of a traffic light suspended from two wires supported by two poles is shown. (b) Some forces are shown in this system. Tension T sub one pulling the top of the left-hand pole is shown by the vector arrow along the left wire from the top of the pole, and an equal but opposite tension T sub one is shown by the arrow pointing up along the left-hand wire where it is attached to the light; the left-hand wire makes a thirty-degree angle with the horizontal. Tension T sub two is shown by a vector arrow pointing downward from the top of the right-hand pole along the right-hand wire, and an equal but opposite tension T sub two is shown by the arrow pointing up along the right-hand wire, which makes a forty-five degree angle with the horizontal. The traffic light is suspended at the lower end of the wires, and its weight W is shown by a vector arrow acting downward. (c) The traffic light is the system of interest, indicated by circling the traffic light. Tension T sub one starting from the traffic light is shown by an arrow along the wire making an angle of thirty degrees with the horizontal. Tension T sub two starting from the traffic light is shown by an arrow along the wire making an angle of forty-five degrees with the horizontal. The weight W is shown by a vector arrow pointing downward from the traffic light. A free-body diagram is shown with three forces acting on a point. Weight W acts downward; T sub one and T sub two act at an angle with the vertical. A coordinate system is shown, with positive x to the right and positive y upward. (d) Forces are shown with their components. T sub one is decomposed into T sub one y pointing vertically upward and T sub one x pointing along the negative x direction. The angle between T sub one and T sub one x is thirty degrees. T sub two is decomposed into T sub two y pointing vertically upward and T sub two x pointing along the positive x direction. The angle between T sub two and T sub two x is forty five degrees.  Weight W is shown by a vector arrow acting downward. (e) The net vertical force is zero, so the vector equation is T sub one y plus T sub two y equals W. T sub one y and T sub two y are shown on a free body diagram as equal length arrows pointing up. W is shown as a downward pointing arrow whose length is twice as long as each of the T sub one y and  T sub two y arrows. The net horizontal force is zero, so vector T sub one x is equal to minus vector T sub two x. T sub two x is shown by an arrow pointing toward the right, and T sub one x is shown by an arrow pointing toward the left.

Solution First consider the horizontal or x -axis:

{F}_{\text{net}\phantom{\rule{0.2em}{0ex}}x}={T}_{2x}-{T}_{1x}=0.

Thus, as you might expect,

{T}_{1x}={T}_{2x}.

This gives us the following relationship:

{T}_{1}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}30\text{°}={T}_{2}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}45\text{°}.

Now consider the force components along the vertical or y -axis:

{F}_{\text{net}\phantom{\rule{0.2em}{0ex}}y}={T}_{1y}+{T}_{2y}-w=0.

This implies

{T}_{1y}+{T}_{2y}=w.

Substituting the expressions for the vertical components gives

{T}_{1}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}30\text{°}+{T}_{2}\text{sin}\phantom{\rule{0.2em}{0ex}}45\text{°}=w.

which yields

1.366{T}_{1}=\left(15.0\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(9.80\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right).

Significance Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker in Newton’s Laws of Motion .

Particle Acceleration

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

2.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}

The angle is given by

\theta ={\text{tan}}^{-1}\left(\frac{{F}_{2}}{{F}_{1}}\right)={\text{tan}}^{-1}\left(\frac{3.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}}{2.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}}\right)=53.1\text{°}.

However, Newton’s second law states that

{F}_{\text{net}}=ma.

Substituting known values gives

{F}_{\text{D}}=\left(4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}\right)-\left(5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(7.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)=7.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{N}\text{.}

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

1.20\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2},

which gives

{F}_{\text{s}}=735\phantom{\rule{0.2em}{0ex}}\text{N}\text{.}

Significance The scale reading in (Figure) (a) is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight:

{F}_{\text{net}}=ma=0={F}_{\text{s}}-w

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In (Figure) (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

1.20\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}.

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

{m}_{1}

Strategy We draw a free-body diagram for each mass separately, as shown in (Figure) . Then we analyze each one to find the required unknowns. The forces on block 1 are the gravitational force, the contact force of the surface, and the tension in the string. Block 2 is subjected to the gravitational force and the string tension. Newton’s second law applies to each, so we write two vector equations:

\stackrel{\to }{T}+{\stackrel{\to }{w}}_{1}+\stackrel{\to }{N}={m}_{1}{\stackrel{\to }{a}}_{1}

Solution The component equations follow from the vector equations above. We see that block 1 has the vertical forces balanced, so we ignore them and write an equation relating the x -components. There are no horizontal forces on block 2, so only the y -equation is written. We obtain these results:

\begin{array}{cccc}\mathbf{\text{Block 1}}\hfill & & & \mathbf{\text{Block 2}}\hfill \\ \sum {F}_{x}=m{a}_{x}\hfill & & & \sum {F}_{y}=m{a}_{y}\hfill \\ {T}_{x}={m}_{1}{a}_{1x}\hfill & & & {T}_{y}-{m}_{2}g={m}_{2}{a}_{2y}.\hfill \end{array}

Solving for a :

a=\frac{{m}_{2}-{m}_{1}}{{m}_{1}+{m}_{2}}g=\frac{4\phantom{\rule{0.2em}{0ex}}\text{kg}-2\phantom{\rule{0.2em}{0ex}}\text{kg}}{4\phantom{\rule{0.2em}{0ex}}\text{kg}+2\phantom{\rule{0.2em}{0ex}}\text{kg}}\left(9.8\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)=3.27\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}.

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

What Force Must a Soccer Player Exert to Reach Top Speed? A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What is her average acceleration? (b) What average force does the ground exert forward on the runner so that she achieves this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

Strategy To find the answers to this problem, we use the problem-solving strategy given earlier in this chapter. The solutions to each part of the example illustrate how to apply specific problem-solving steps. In this case, we do not need to use all of the steps. We simply identify the physical principles, and thus the knowns and unknowns; apply Newton’s second law; and check to see whether the answer is reasonable.

\text{Δ}v=8.00\phantom{\rule{0.2em}{0ex}}\text{m/s}

Substituting the known values yields

a=\frac{8.00\phantom{\rule{0.2em}{0ex}}\text{m/s}}{2.50\phantom{\rule{0.2em}{0ex}}\text{s}}=3.20\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}.

Substituting the known values of m and a gives

{F}_{\text{net}}=\left(70.0\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(3.20\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)=224\phantom{\rule{0.2em}{0ex}}\text{N}\text{.}

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Significance This example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles, the knowns, and the unknowns involved in the problem. The second step is to solve for the unknown, in this case using Newton’s second law. Finally, we check our answer to ensure it is reasonable. These techniques for integrated concept problems will be useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life.

Check Your Understanding The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

5.00\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{m/s}

Solution We have

a=\frac{\text{Δ}v}{\text{Δ}t}=\frac{\left(6.00\stackrel{^}{i}+12.00\stackrel{^}{j}\text{m/s}\right)-\left(5.00\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)}{2.00\phantom{\rule{0.2em}{0ex}}\text{s}}=3.00\stackrel{^}{i}+3.50\stackrel{^}{j}{\text{m/s}}^{2}

The magnitude of the force is now easily found:

F=\sqrt{{\left(4.50\phantom{\rule{0.2em}{0ex}}\text{N}\right)}^{2}+{\left(5.25\phantom{\rule{0.2em}{0ex}}\text{N}\right)}^{2}}=6.91\phantom{\rule{0.2em}{0ex}}\text{N}\text{.}

Check Your Understanding Find the direction of the resultant for the 1.50-kg model helicopter.

49.4 degrees

F=\left(820.0t\right)\phantom{\rule{0.2em}{0ex}}\text{N,}

Significance Since the force varies with time, we must use calculus to solve this problem. Notice how the total mass of the system was important in solving (Figure) (a), whereas only the mass of the truck (since it supplied the force) was of use in (Figure) (b).

v=\frac{ds}{dt}

Strategy The known force on the mortar shell can be related to its acceleration using the equations of motion. Kinematics can then be used to relate the mortar shell’s acceleration to its position.

{y}_{0}=0

We replace ds with dy because we are dealing with the vertical direction,

ady=vdv,\text{ }\phantom{\rule{0.5em}{0ex}}\left(-0.00100{v}^{2}-9.80\right)dy=vdv.

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

h=114\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}

Significance Notice the need to apply calculus since the force is not constant, which also means that acceleration is not constant. To make matters worse, the force depends on v (not t ), and so we must use the trick explained prior to the example. The answer for the height indicates a lower elevation if there were air resistance. We will deal with the effects of air resistance and other drag forces in greater detail in Drag Force and Terminal Speed .

Check Your Understanding If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

  • Newton’s laws of motion can be applied in numerous situations to solve motion problems.

{F}_{\text{net}}=ma

  • The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating vertically, the normal force is less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force is always less than the full weight of the object.
  • Some problems contain several physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics to solve these problems.

Conceptual Questions

To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at g . Why do they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?

The scale is in free fall along with the astronauts, so the reading on the scale would be 0. There is no difference in the apparent weightlessness; in the aircraft and in orbit, free fall is occurring.

\stackrel{\to }{F}

a. 170 N; b. 170 N

Find the tension in each of the three cables supporting the traffic light if it weighs 2.00 × 10 2 N.

A sketch of a traffic light suspended by a cable that is in turn suspended from two other cables is shown. Tension T sub 3 is the tension in the cable connecting the traffic light to the upper cables. Tension T sub one is the tension in the upper cable pulling up and to the left, making a 41 degree angle with the horizontal. Tension T sub two is the tension pulling up and to the right, making a 63 degree angle with the horizontal. Force vector w equal to 200 Newtons pulls vertically downward on the traffic light.

Two muscles in the back of the leg pull upward on the Achilles tendon, as shown below. (These muscles are called the medial and lateral heads of the gastrocnemius muscle.) Find the magnitude and direction of the total force on the Achilles tendon. What type of movement could be caused by this force?

An Achilles tendon is shown in the figure with two forces exerted on it by the lateral and medial heads of the gastrocnemius muscle. F sub one, equal to two hundred Newtons, is shown as a vector making an angle twenty degrees to the right of vertical, and F sub two, equal to two hundred Newtons, is shown making an angle of twenty degrees left of vertical.

376 N pointing up (along the dashed line in the figure); the force is used to raise the heel of the foot.

After a mishap, a 76.0-kg circus performer clings to a trapeze, which is being pulled to the side by another circus artist, as shown here. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.

A circus performer hanging from a trapeze is being pulled to the right by another performer using a rope. Her weight is shown by a vector w acting vertically downward. The trapeze rope exerts a tension, T sub one, up and to the left, making an angle of fifteen degrees with the vertical. The second performer pulls with tension T sub two, making an angle of ten degrees above the positive x direction.

A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow the first dolphin if it was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.)

When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b) How far does he travel?

2.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{kg}

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110.0 kg.

A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110.0 m. (a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.

2.40\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\text{;}

a. 10 kg; b. 90 N; c. 98 N; d. 0

1.2\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}

The device shown below is the Atwood’s machine considered in (Figure) . Assuming that the masses of the string and the frictionless pulley are negligible, (a) find an equation for the acceleration of the two blocks; (b) find an equation for the tension in the string; and (c) find both the acceleration and tension when block 1 has mass 2.00 kg and block 2 has mass 4.00 kg.

An Atwood machine consisting of masses suspended on either side of a pulley by a string passing over the pulley is shown. Mass m sub 1 is on the left and mass m sub 2 is on the right.

Two blocks are connected by a massless rope as shown below. The mass of the block on the table is 4.0 kg and the hanging mass is 1.0 kg. The table and the pulley are frictionless. (a) Find the acceleration of the system. (b) Find the tension in the rope. (c) Find the speed with which the hanging mass hits the floor if it starts from rest and is initially located 1.0 m from the floor.

Block m sub 1 is on a horizontal table. It is connected to a string that passes over a pulley at the edge of the table. The string then hangs straight down and connects to block m sub 2, which is not in contact with the table. Block m sub 1 has acceleration a sub 1 directed to the right. Block m sub 2 has acceleration a sub 2 directed downward.

Shown below are two carts connected by a cord that passes over a small frictionless pulley. Each cart rolls freely with negligible friction. Calculate the acceleration of the carts and the tension in the cord.

Two carts connected by a string passing over a pulley are on either side of a double inclined plane. The string passes over a pulley attached to the top of the double incline. On the left, the incline makes an angle of 37 degrees with the horizontal and the cart on that side has mass 10 kilograms. On the right, the incline makes an angle of 53 degrees with the horizontal and the cart on that side has mass 15 kilograms.

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Saturday, November 12, 2016

Laws of motion problems with solutions six.

law of motion problem solving with solution

Laws of Motion Problems with Solutions Two

Laws of motion problems with solutions three, laws of motion problems with solutions four, laws of motion problems with solutions five, newton law of inertia and newton law of force an introduction   impulse, change of momentum, applications and problems with solutions acceleration due to gravity and one dimensional motion  equations   free body diagrams for newton laws of motion problems with solutions conservation of linear momentum applications problems and solutions, no comments:, post a comment.

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Laws of Motion Numericals

  • Newton's First Law of Motion
  • Newton's Third Law of Motion
  • Motion in a Vertical Circle
  • Laws of Logarithms
  • Kepler's Laws of Planetary Motion
  • Applications of Fluid Dynamics
  • Uniform Circular Motion
  • Applications of Charles Law
  • Motion Under Gravity
  • Equation of Tangents and Normals
  • Motion of Center of Mass
  • Applications of Kepler's Laws
  • Law of Inertia
  • What is Motion?
  • Uniform and Non-Uniform Motion
  • Oscillatory Motion Formula
  • Kinematics of Rotational Motion
  • Dynamics of Circular Motion
  • Secant Method of Numerical analysis

Laws of Motion describe how objects move under the influence of different types of forces. These forces can be due to any physical phenomenon, but their effect is the same. All the forces change the momentum of the system on which they are acting. Newton gave three laws, these laws describe the interaction between two objects and the forces between them. These laws become the building block for the further theory of mechanics and motion. Let’s look at these concepts and some problems with them.

Table of Content

Newton’s Laws of Motion

  • Newton’s First Law

Newton’s Second Law

  • Newton’s Third Law
  • Solved Examples

Numericals on Laws of Motion

Sir Isaac Newton was an English physicist and mathematician who gave the three laws of motion called Newton’s Law of Motion which formed the base for classical mechanics. These laws still keep giving accurate predictions except for bodies traveling with speeds comparable to light or the size of an electron. These were the first laws that described the forces acting on the bodies and the motion of the body which is governed by these forces. There are three laws of motion that are given by Sir Issac Newton that include,

  • Newton’s First Law of Motion
  • Newton’s Second Law of Motion
  • Newton’s Third Law of Motion

Now let’s learn about these laws further in this article.

Newton’s First Law (Law of Inertia)

Newton’s First Law of motion also called the law of interia states that if a body is at rest or is moving in a straight line with constant speed. It will keep moving in a straight line at constant speed or will remain at rest until it is acted upon by an external force. This property of any object to resist a change in its state is called inertia and thus this law is known Law of Inertia.  

Newton’s Second Law is a quantitative description of the changes that take place when an external force acts on the body. The momentum of the body is defined as the product of the mass and velocity of that body. When a force acts on the body, it brings about changes in the momentum of the body or its direction, or both. It is one of the most important laws in the field of classical mechanics. Assuming the mass of the body is “m”, the law is given by,

Here, F is the force acting on the particle, and “a” denotes the acceleration produced in the body. The direction of acceleration is the same as the direction of motion.

Newton’s Third Law (Law of Action and Reaction)

Newton’s Third Law also called law of action and reaction states that when two bodies interact with each other, they apply forces to one another which are equal in magnitude and opposite in direction. This law is also known as action-reaction law. It allows us to explain phenonmenon such as static equilibrium, where all the forces are balanced, but it also applies to bodies in uniform or accelerated motion. If the net forces acting on the body are equal, the body is said to be in equilibrium.

Also, Check

  • Acceleration

Examples on Newton Laws of Motion

Example 1: Calculate the momentum of a ball thrown at a speed of 10 m/s and weighing 800 g.

Given, M = 800 g V = 10 m/s Momentum is given by, p = MV Plugging in the values in the formula p = MV p = (800)(10)  p = 8000 gm/s p = 8 × 10 3 gm/s

Example 2: Calculate the momentum of a ball thrown at a speed of 10 m/s and weighing 20 g.

Given, M = 20 g V = 10 m/s Momentum is given by, p = MV Plugging in the values in the formula p = MV p = (20)(10) p = 200 gm/s p = 2 x 10 2 gm/s

Example 3: A force of 20N is acting on a body of mass 2Kg. Find the acceleration produced.

Given, m = 2 Kg F = 20 N Acceleration will be given by, F = ma Plugging in the values, F = ma 20 = (2)(a) 10 m/s 2 = a

Example 4: A force of 100N is acting on a body of mass 5Kg. Find the acceleration produced.

Given, m = 5 Kg F = 100 N Acceleration will be given by, F = ma Plugging in the values, F = ma 100 = (5)(a) 20 m/s 2 = a

Example 5: A body of 2 kg is moving at a velocity of 50m/s. A force starts acting on it and the velocity becomes 20m/s in a time of 5 seconds. Find the force applied to the body.

Given, m = 5 Kg v i = 50 m/s v f = 20 m/s t = 5 s Force is defined as the rate of change of momentum. F = m(v f  – v i )/t F = (5)(50 – 20)/(5) F = 30N

Example 6: A body of 10 kg is moving at a velocity of 100m/s. A force starts acting on it and the velocity becomes 20m/s in a time of 10 seconds. Find the force applied to the body.

Given, m = 10 Kg v i = 100 m/s v f = 20 m/s t = 10 s Force is defined as the rate of change of momentum F = m(v f  – v i )/t F = (10)(80 – 20)/(10) F = 80N

Example 7: The momentum of the body is given by the equation below,

p(t) = 3t 2 + 4t + 5

Find the force acting on the body at t = 5.

Force rate of change of momentum, F = dp/dt Given, p(t) = 3t 2 + 4t + 5 F = dp/dt = d/dt(3t 2 + 4t + 5) F = 6t + 4 At t = 5 F = 6(5) + 4 F = 34 N Thus, force acting on the body at t = 5 sec is 34 N.

Example 8: The momentum of the body is given by the equation below,

p(t) = e t + t 2 + 20

Find the force acting on the body at t = 0.

Force rate of change of momentum, F = dp/dt Given, p(t) = e t + t 2 + 20 F = dp/dt F = d/dt (e t + t 2 + 20) F = e t + 2t At t = 0 F = e 0 + 2×1    = 1 + 2    = 3 F = 3 N Thus, the force acting on the body at t = 0 is 3 N.

Lets learn Law of Motion Numericals for class 9 and class 11

1. If the momentum of any body is, p(t) = 3t 3 + 5t 2 + t. Find the force acting on the body at t = 2.

2. If the distance covered by an object is given by, d(t) = t 3 + t. Find the acceleration on the body at t = 3.

3. If the distance covered by an object is given by, d(t) = 4t 4 + 3t + 5. Find the velocity on the body at t = 0.

4. A body of 19 kg is moving at a velocity of 100 m/s. A force starts acting on it and the velocity becomes 120 m/s in a time of 10 seconds. Find the force applied to the body.

5. Force acting on body is 120 N, the mass of the body is 12 Kg. Find the acceleration produced produced by the body.

FAQs on Laws of Motion

1. who discovered the laws of motion.

The laws of motion were discovered by Sir Isaac Newton an English Mathematician.

2. What is Law of Inertia?

Law of Inertia is also called the Newton First Law of Motion. This law states that an object at the state of rest or at the state of motion stays in its state until an external force is applied.

3. What is Law of Action and Reaction?

Law of Action and Reaction is the other name of Newton Third Law of Motion. This law states that, “Every action has its equal and opposite reaction.”

4. What is Formula for Force Acting on a Body?

The force acting on a body is given using the formula, F = ma where, m is Mass of Object a is Acceleration of Object

5. What is Formula for Momentum of a Body?

The momentum of an object is given using the formula, F = mv where, m is Mass of Object v is Velocity of Object

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A General Approach

Basic problem-solving techniques can aid in the solution of problems involving motion (i.e., the laws of motion).

learning objectives

  • Assess the laws of motion through practiced problem solving techniques

When dealing with the laws of motion, although knowledge of concepts and equations is important, understanding basic problem solving techniques can simplify the process of solving problems that may appear difficult. Your approach to problem solving can involve several key steps.

image

Free body diagram : An example of a drawing to help identify forces and directions.

First, gather all relevant information from the problem. Identify all quantities that are given (the knowns ), then do the same for all quantities needed (the unknowns ). Also, identify the physical principles involved (e.g., force, gravity, friction, etc. ).

Next, a drawing may be helpful. Sometimes a drawing can even help determine the known and unknown quantities. It need not be a work of art, but it should be clear enough to illustrate proper dimension, (meaning one, two, or three dimensions). You can then use this drawing to determine which direction is positive and which is negative (making note of this on the drawing).

A next step is to use what is known to find the appropriate equation to find what is unknown. While it is easiest to find an equation that leaves only one unknown, sometimes this is not possible. In these situations, you can solve multiple equations to find the right answer. Remember that equations represent physical principles and relationships, so use the equations and drawings in tandem.

You may then substitute the knowns into the appropriate equations and find a numerical solution.

Check the answer to see if it is reasonable and makes sense. Your judgment will improve and fine tune as you solve more problems of this nature. This “judgement” step helps intuit the problem in terms of its conceptual meaning. If you can judge whether the answer is reasonable, you have a deeper understanding of physics than simply the mechanics of problem solving.

When solving problems, we tend to perform these steps in different order, as well as do several steps simultaneously. There is no rigid procedure that will work every time. Creativity and insight grow with experience. In time, the basics of problem solving can become relatively automatic.

  • Gathering all relevant information and identifying knowns and unknowns is an important first step.
  • Always make a drawing to help identify directions of forces and to establish \(\mathrm{x, y}\), and \(\mathrm{z}\) axes.
  • Choose the correct equations, solve the problem, and check that the answer fits expectations numerically.
  • equation : An assertion that two expressions are equal, expressed by writing the two expressions separated by an equal sign; from which one is to determine a particular quantity.

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CC LICENSED CONTENT, SHARED PREVIOUSLY

  • Curation and Revision. Provided by : Boundless.com. License : CC BY-SA: Attribution-ShareAlike

CC LICENSED CONTENT, SPECIFIC ATTRIBUTION

  • OpenStax College, College Physics. September 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m42125/latest/?collection=col11406/latest . License : CC BY: Attribution
  • equation. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/equation . License : CC BY-SA: Attribution-ShareAlike
  • Free body diagram. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Free_body_diagram . License : Public Domain: No Known Copyright

IMAGES

  1. Laws of Motion Problems with Solutions Six

    law of motion problem solving with solution

  2. Newton's Second Law Of Motion

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  3. Newtons Laws Of Motion With Examples Problems Solutions And Images

    law of motion problem solving with solution

  4. Laws of Motion Problems with Solutions One

    law of motion problem solving with solution

  5. Laws of Motion Problems with Solutions Two

    law of motion problem solving with solution

  6. LAWS OF MOTION PROBLEM AND THEIR SOLUTION

    law of motion problem solving with solution

VIDEO

  1. Problem 3.11

  2. Lagrangian Mechanics

  3. Problem Solving Law Of Motion

  4. 1D Kinematics (Linear Motion)

  5. Laws of Motion

  6. Class 11th Physics

COMMENTS

  1. 6.2: Solving Problems with Newton's Laws (Part 1)

    Determine the system of interest. The result is a free-body diagram that is essential to solving the problem. Apply Newton's second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line. Check the solution to see whether it is reasonable.

  2. 6.1 Solving Problems with Newton's Laws

    Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton's laws in Newton's Laws of Motion; in this chapter, we continue to discuss these strategies and apply a step-by-step process. Problem-Solving Strategies

  3. Newton's first law of motion

    Newton's first law of motion - problems and solutions. 1. A person is in an elevator that moving upward at a constant velocity. The weight of the person is 800 N. Immediately the elevator rope is broke, so the elevator falls. Determine the normal force acted by elevator's floor to the person just before and after the elevator's rope broke.

  4. Newton's Law Problem Sets

    Newton's Laws of Motion: Problem Set Problem 1: An African elephant can reach heights of 13 feet and possess a mass of as much as 6000 kg. Determine the weight of an African elephant in Newtons and in pounds. (Given: 1.00 N = .225 pounds) Audio Guided Solution

  5. Newton's Laws

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  6. Newton's Law Problem Sets

    Problem Set NL11 - Friction and Mu 1. Solve problems involving the use of a coefficient of friction value. Most problems are highly scaffolded. Includes 7 problems. Problem Set NL12 - Friction and Mu 2. Solve problems involving the use of a coefficient of friction value. Some of the problems are scaffolded.

  7. 6.1 Solving Problems with Newton's Laws

    Determine the system of interest. The result is a free-body diagram that is essential to solving the problem. Apply Newton's second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line. Check the solution to see whether it is reasonable.

  8. Newton's Third Law of Motion: Solved Problems

    In this article, you learn the concepts of Newton's third law of motion by problem-solving approach. Here, we are going to find pairs of forces appearing in Newton's third law using some worked problems. ... Solution: Newton's third law states that when two objects interact, the force that object 1 applies on object 2, called $\vec{F}_{12} ...

  9. Chapter 4: Dynamics: Force and Newton's Laws of Motion

    Access step-by-step solutions to the problem exercises of Chapter 4: Dynamics, Force, and Newton's Laws of Motion from the College Physics textbook by OpenStax. Get a better understanding of the principles behind force, motion, and the laws of physics.

  10. Newton's second law of motion

    Solution : We use Newton's second law to get the net force. ∑ F = m a. ∑ F = (1 kg) (5 m/s2) = 5 kg m/s2 = 5 Newton. See also Capacitors in series - problems and solutions. 2. Mass of an object = 1 kg, net force ∑F = 2 Newton. Determine the magnitude and direction of the object's acceleration…. Known :

  11. Newton's Law of Motion

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  12. PDF Solving Problems Involving Newton's Laws of Motion

    Identify the Acceleration Identify (and if necessary calculate) the acceleration of the object(s) of interest. Apply Newton's Laws of Motion Apply the vector equations: a = Fnet / m. Solve For The Unknown Quantities. Check the Results There is often more than one way to solve the problem!

  13. PDF 6.1

    Problem-Solving Strategy: Applying Newton's Laws of Motion 1. Identify the physical principles involved by listing the givens and the quantities to be calculated. 2. Sketch the situation, using arrows to represent all forces. 3. Determine the system of interest. The result is afree-body diagramthat is essential to solving the problem. 4.

  14. Kinematic Equations: Sample Problems and Solutions

    A useful problem-solving strategy was presented for use with these equations and two examples were given that illustrated the use of the strategy. Then, the application of the kinematic equations and the problem-solving strategy to free-fall motion was discussed and illustrated. In this part of Lesson 6, several sample problems will be presented.

  15. Solving Problems with Newton's Laws

    Determine the system of interest. The result is a free-body diagram that is essential to solving the problem. Apply Newton's second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line. Check the solution to see whether it is reasonable.

  16. PDF Problem Workbook Solutions Forces and the Laws of Motion

    II Ch. 4-4 Holt Physics Solution Manual II Copyright © by Holt, Rinehart and Winston. All rights reserved. Givens Solutions 4. ma = 54.0 kg mw = 157.5 kg anet = 1. ...

  17. All of Newton's laws of motion (practice)

    Report a problem. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

  18. Laws of Motion Problems with Solutions Six

    Laws of Motion Problems with Solutions Six. Newton Laws of motion deals with the impact of force on the motion of the body. This force can be defined as the rate of change of momentum. Here in this post we are going to discuss regarding connected bodies. The bodies are connected with the help of string or in contact with each other.

  19. Laws of Motion Questions: Numericals and FAQs

    Numericals on Laws of Motion. Lets learn Law of Motion Numericals for class 9 and class 11. 1. If the momentum of any body is, p (t) = 3t3 + 5t2 + t. Find the force acting on the body at t = 2. 2. If the distance covered by an object is given by, d (t) = t3 + t. Find the acceleration on the body at t = 3.

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  21. 4.5: Problem-Solving

    When dealing with the laws of motion, although knowledge of concepts and equations is important, understanding basic problem solving techniques can simplify the process of solving problems that may appear difficult. Your approach to problem solving can involve several key steps.. Free body diagram: An example of a drawing to help identify forces and directions.