# FORCES AND NEWTON S LAWS OF MOTION 4

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1 FORCES AND NEWTON S LAWS OF MOTION 4 Q4.1. Reason: If friction and air resistance are negligible (as stated) then the net force on the puck is zero (the normal force and gravitational force are equal in magnitude and opposite in direction). If the net force on the puck is zero, then Newton s first law states that it will continue on with constant velocity. So no force is needed to keep the puck moving; it will naturally keep moving unless a force acts on it to change its velocity. Assess: This question demonstrates the difference between Aristotelian thinking and Newtonian thinking. Objects do not need forces on them to keep them moving; forces are only required when we want to change the velocity of the object. The reason one has to normally keep pushing an object to keep it moving is because of friction; to keep it at constant velocity your pushing force must be equal in magnitude to the friction force. But in the case of this question, there is no friction, so there is no force needed to keep it moving. Q4.2. Reason: Even if an object is not moving forces can be acting on it. However, the net force must be zero. As an example consider a book on a flat table. The forces that act on the book are the weight of the book (a long ranger force) and the normal force exerted by the table (a contact force). There are two forces acting on the book, but it is not moving because the net force on the book is zero. Assess: The net force, which is the vector sum of the forces acting on an object, governs the acceleration of objects through Equation 4.4. Q4.3. Reason: Newton s first law does not state that there can be no forces acting on an object moving with constant velocity only that the sum of the forces must be zero (sometimes worded as no unbalanced forces ). There can be forces acting on an object with constant velocity, but the vector sum of those forces must be zero. Assess: When we say there is no net force on an object with constant velocity, we are not saying that there are no forces. Net force means the vector sum of the forces. Q4.4. Reason: The ball is initially and rest and will stay at rest unless some unbalanced (net) force acts on it. As the wagon moves forward, static friction tries to pull the ball along with the wagon. This static friction causes the ball to roll backwards in the wagon. Assess: Put a marble or some other small spherical object on your text and then push your text back and forth. The sphere will always roll in the direction opposite the direction you are moving the text. Q4.5. Reason: No. If you know all of the forces than you know the direction of the acceleration, not the direction of the motion (velocity). For example, a car moving forward could have on it a net force forward if speeding up or backward if slowing down or no net force at all if moving at constant speed. Assess: Consider carefully what Newton s second law says, and what it doesn t say. The net force must always be in the direction of the acceleration. This is also the direction of the change in velocity, although not necessarily in the direction of the velocity itself. Q4.6. Reason: The arrows have been shot horizontally and air resistance is negligible. The only force on the arrows is due to gravity, which always points straight down. There is no component of a force in the horizontal direction. Each arrow has exactly the same magnitude of horizontal force on it, 0 N. No horizontal force is required to keep the arrows moving. Rather a horizontal force is required if we wish to change the horizontal motion of the arrows. Assess: Weight always points vertically downward. Once the arrow has been fired, it is moving with a constant velocity, a force is required to change the velocity but not to maintain the velocity. Q4.7. Reason: The picture on the left is more effective at tightening the head because of the greater inertia of the head. Once moving, the head will want to continue moving (Newton s first law) after the handle hits the 4-1

2 4-2 Chapter 4 table, thus tightening the head, more so than in the second picture where the light handle has less inertia moving down than the head. Assess: Newton s first law, the law of inertia, says the greater the mass of an object the more it will tend to continue with its previous velocity. One can assess this by trying it with a real hammer with a loose head. Q4.8. Reason: (a) During the first collision a person in the car continues in their state of motion until a force acts to change their motion. This will happen when the seatbelt engages or when the person hits something on the inside of the car. Likewise, internal organs will continue in their state of motion until some force acts on them to change their motion. They continue moving due to Newton s first law. (b) If there is no restraint on the body after the first collision, the body will be moving at 60 mph just before the second collision, and the organs will be traveling at this speed until just before the third collision. Assess: Forces are required to change an object s state of motion. Q4.9. Reason: Kinetic friction opposes the motion, but static friction is in the direction to prevent motion. (a) Examples of motion where the frictional force on the object is directed opposite the motion would be a block sliding across a table top or the friction of the road on car tires in a skid. (b) An example of motion in which the frictional force on the object is in the same direction as the motion would be a crate (not sliding) in the back of a pickup truck that is speeding up. The static frictional force of the truck bed on the crate is in the forward direction (the same direction as the motion) because it is the net force on the crate that accelerates it forward. Assess: It is easy to think that the direction of the frictional force is always opposite the direction of motion, but static frictional forces are in the direction to prevent relative motion between the surfaces, and can be in various directions depending on the situation. Q4.10. Reason: Since there is no source of gravity, you will not be able to feel the weight of the objects. However, Newton s second law is true even in an environment without gravity. Assuming you can exert a reproducible force in throwing both objects. You could throw each and note the acceleration each obtains. Assess: Mass is independent of the force of gravity and exists even in environments with no sources of gravity. Q4.11. Reason: Both objects (Jonathan and his daughter) experience the same acceleration (i.e. the acceleration of the car). However since the objects do not have the same mass, the forces required to accelerate them will be different. The object with the greater mass (Jonathan) will require the greater force. Assess: This is a straightforward application of Newton s second law. Q4.12. Reason: Since the box is at rest, the sum of the forces acting on it must be zero. Put another way, the upward forces must equal the downward forces. If the weight is the only downward force and the normal force is the only upward force, they will have the same magnitude. However, if some other force with a vertical component (such as you pushing down on the box) acts on the box, the weight and the normal force will not have the same magnitude. Assess: For many objects in static equilibrium, the sum of the forces is equal to zero. Q4.13. Reason: The book defines weight as the gravitational pull of the earth on an object, and that doesn t change when the ball is thrown straight up. That is, if the gravitational force of the earth on the ball is 2.0 N while it sits on the scale, then the gravitational force of the earth on the ball is still 2.0 N while it is in flight even at the very top of its motion. The weight of the ball is 2.0 N all the time the ball is near the surface of the earth, regardless of its motion. Assess: Hmmm.... We previously distinguished between velocity, which is zero at the top of the trajectory (if the ball is thrown straight up), and acceleration, which is g all the while it is in free fall. But how do we explain the fact that the acceleration of the ball is g at the top of its trajectory but zero while sitting on the scale, if the gravitational force on it is the same in both cases? Doesn t Fnet = ma? Yes, Fnet = ma is true, but in the case of the ball on the scale the net force is zero because the scale is pushing up on the ball with a force of 2.0 N; since the net force is zero the acceleration is zero. While the ball is in flight the scale isn t pushing up on it, so the acceleration is g because the net force (the weight) is 2.0 N. It is also worth mentioning that not all physics texts define weight the same way. Some define weight as what the scale reads, which would be different from the definition in our text because if the scale and ball are thrown up together they would both be in free fall and the scale would read zero. That is, some books would say the ball s weight is zero at the top of its motion (because that is what the scale would read). Both definitions have merit, and it is wise to understand the difference and be prepared in future situations if someone defines weight

4 4-4 Chapter 4 There are four forces on the cart. The contact forces are the normal force of the floor on the cart and the contact force of the magnet mounting on the cart. The long range forces are the weight of the cart and the force due to magnetic attraction of the cart to the magnet. The only horizontal forces on the cart are the force due to the magnet on the cart F, magnet on cart and the horizontal component of the force of the mount on the cart, ( F mount on cart). x We will find the relationship between these two horizontal forces to determine whether the cart will move. To do that, we will need to take a look at the forces on the magnet and the mount. See the following figure for the forces on the magnet. The long range forces are the magnet s weight and the magnetic attraction between the magnet and the cart. The force of the cart on the magnet is equal and opposite to the force of the magnet on the cart since these are action/reaction pairs. F = F cart on magnet magnet on cart Since the magnet is not moving, Newton s second law gives that the net force on the magnet must be zero. We can see that the horizontal component of the force of the mount on the magnet must cancel the force of the cart on the magnet. Fcart on magnet = ( Fmount on magnet) x Consider the forces on the magnet mount in the figure below. The magnet s mounting is what provides the force that keeps the magnet in place. The force of the magnet on the mount is equal and opposite to the force of the mount on the magnet, since these are action/reaction pairs. Using Newton s third law, F = F magnet on mount mount on magnet From the free-body diagram, the x-components of the forces on the magnet mount must cancel. ( F ) = ( F ) magnet on mount x cart on mount From Newton s third law, the force of the cart on the mount must be equal and opposite to the force of the mount on the cart. F = F cart on mount mount on cart x

5 Forces and Newton s Laws of Motion 4-5 We can now relate the two forces on the cart to each other. Using the first two equations above, F = F = ( F ) = ( F ) ( x) magnet on cart cart on magnet mount on magnet mount on magnet Using the remaining basic equations above, ( F ) = ( F ) = ( F ) =+ ( F ) = ( F ) ( ) mount on magnet x magnet on mount x cart on mount x cart on mount x mount on cart x Putting these last two equations together, we have F magnet on cart = ( Fmount on cart) x The net horizontal force on the cart is zero. The cart will not move due to the magnet. Assess: We could have arrived at this result if we had taken the cart, mount, and magnet as the system. The force between the magnet and the cart and the cart and the magnet are then internal forces. The net external force on the system is zero, so the system will not accelerate. Note the previous distinction between uses of Newton s second and third laws. Q4.21. Reason: This is a very famous question, and one which many people miss until they understand Newton s laws. Before the string snaps, it provides a tension force toward the center of the circle (centripetal), which keeps the puck going around in a circle; in other words, the net force is toward the center, so the acceleration is toward the center. However, when the string snaps, that centripetal force disappears. There is now no net force acting on the puck, and Newton s first law tells us that its velocity must be constant (both in magnitude and direction) under those conditions. So the puck continues in the direction it was going at the instant the string snapped. The answer is C. Assess: Ask some friends, relatives, or roommates this question and then patiently explain to them the right answer. Q4.22. Reason: Since the block glued on to the original block is identical to the original block, the mass of the two together must be twice as large as the mass of the original block. If the force applied is also twice as large, the acceleration will be the same. Explicitly applying Newton s second law to the two blocks glued together gives 2 F F a = new aold 2 m = m = The correct choice is C. Assess: In Newton s second law, the acceleration is proportional to the net force and inversely proportional to the mass acccelerated. As a result if you double both the mass and the force, the acceleration will remain the same. Q4.23. Reason: We can apply Newton s second law twice to solve this problem. First to determine the force and second to determine the mass. 2 F = ma = (5.0 kg)(0.20 m/s ) = 1.0 N Then 2 m= F / a= 1.0 N / 0.10 m/s = 10 kg The correct answer is A. Assess: This problem is a straightforward application of Newton s Second Law. Q4.24. Reason: Drag points opposite to the direction of motion. As the ball is going up, the drag force acts downward. As the ball comes down, the drag force acts upward. The correct choice is D. Assess: Drag always acts opposite to the direction of motion of an object. Q4.25. Reason: The direction of the kinetic friction force will be opposite the motion, so the friction points down while the box goes up, and the friction points up while the box slides down. The answer is D. Assess: Drawing a free-body diagram (with tilted axes) and applying Newton s second law will support this conclusion. x

8 4-8 Chapter 4 Solve: There are two forces acting on the mountain climber due to her interactions with the two agents earth and rope. One of the forces on the climber is the long-range weight force by the earth. The other force is the tension force exerted by the rope. Assess: Note that the climber does not touch the sides of the crevasse so there are no forces from the crevasse walls. P4.8. Prepare: We will follow the procedure in Tactics Box 4.2. Solve: First identify the system. We are asked for forces acting on the clown, so we consider the clown as the system. A drawing of the situation is shown. The environment touches the system at the clown s hand. The force on the clown at this point is due to the spring. The only long range force acting is gravity. There are no other contact or long range forces on the clown. Assess: Use Tactics Box 4.2 along with Table 4.1 to identify forces on a system. P4.9. Prepare: Draw a picture of the situation, identify the system, in this case the baseball player, and draw a closed curve around it. Name and label all relevant contact forces and long-range forces. Solve: There are three forces acting on the baseball player due to his interactions with the two agents earth and ground. One of the forces on the player is the long-range weight force by the earth. Another force is the normal force exerted by the ground due to the contact between him and the ground. The third force is the kinetic friction force by the ground due to his sliding motion on the ground. Assess: Note that the kinetic friction force would be absent if the baseball player were not sliding. P4.10. Prepare: Draw a picture of the situation, identify the system, in this case the jet plane, and draw a closed curve around it. Name and label all relevant contact forces and long-range forces. Assume friction is negligible compared to other forces.

9 Forces and Newton s Laws of Motion 4-9 Solve: There are four forces acting on the jet plane due to its interactions with the four agents earth, ground, air, and the hot gases exhausted to the environment. One force on the plane is the long-range weight force by the earth. Second is the normal force exerted by the ground due to the plane s contact with ground. The third is the drag force by the air. The fourth is the thrust force exerted on the plane by the hot gas that is being let out to the environment. Assess: Note that we left out friction force because we assumed it to be negligible. P4.11. Prepare: We follow the outline in Tactics Box 4.2. See also Conceptual Example 4.2. The exact angle of the slope is not critical in this problem; the answers would be very similar for any angle between 0 and 90. Solve: The system is the skier. To identify forces, think of objects that are in contact with the object under consideration, as well as any longrange forces that might be acting on it. We are told to not ignore friction, but we will ignore air resistance. The objects that are in contact with the skier are the snow-covered slope and... and that s all (although we will identify two forces exerted by this agent). The long-range force on the skier is the gravitational force of the earth on the skier. One of the forces, then, is the gravitational force of the earth on the skier. This force points straight toward the center of the earth. The slope, as we mentioned, exerts two forces on the skier: the normal force (directed perpendicularly to the slope) and the frictional force (directed parallel to the slope, backward from the downhill motion). Assess: Since there are no other objects (agents) in contact with the skier (we are ignoring the air, remember?) and no other long-range forces we can identify (the gravitational force of the moon or the sun on the skier is also too small to be worth mentioning), then we have probably catalogued them all. We are not told whether the skier has a constant velocity or is accelerating, and that factor would influence the relative lengths of the three arrows representing the forces. If the motion is constant velocity, then the vector sum of the three arrows must be zero. P4.12. Prepare: The procedure in Tactics Box 4.2 can be used to identify all forces on the ball. Solve: The tennis ball is the system. A picture of the situation is shown.

10 4-10 Chapter 4 The environment touches the ball at the ball s surface. The force on the ball s surface is the drag force, and this is the only contact force on the system. The drag force is always opposite to the direction of motion of an object. The only long-range force acting is the ball s weight. Assess: Tactics Box 4.2 along with Table 4.1 can be used to identify forces on a system. P4.13. Prepare: Refer to Figure P4.13. From force = mass acceleration or mass = force/acceleration or mass = 1/(acceleration/force), mass is 1 m = slope of the acceleration-versus-force graph A larger slope implies a smaller mass. Solve: We know m 2 = 0.20 kg, and we can find the other masses relative to m 2 by comparing their slopes. Thus m1 1/slope 1 slope = = = = = 0.40 m2 1/slope 2 slope 1 5/2 5 m = 0.40 m = kg = 0.08 kg Similarly, 1 2 m3 1/slope 3 slope = = = = = 2.50 m2 1/slope 2 slope 3 2/5 2 m = 2.50 m = kg = 0.50 kg 3 2 Assess: From the initial analysis of the slopes, we had expected m 3 > m 2 and m 1 < m 2. This is consistent with our numerical answers. P4.14. Prepare: We will use the particle model for the object and use Newton s second law. Solve: (a) We are told that for an unknown force (call it F 0 ) acting on an unknown mass (call it m A ), the acceleration of the mass is 5 m/s 2. The accelerations are 3 m/s 2 and 8 m/s 2 for objects B and C. According to Newton s second law, F 0 = m A (5 m/s 2 ) = m B (3 m/s 2 ) = m C (8 m/s 2 ). This means B has the largest mass. (b) Object C has the smallest mass because it has the largest acceleration. (c) From the equation in part (a) the ratio of mass A to mass B is 3/5. Assess: Since the force is constant and Newton s second law tells us the product of the mass and acceleration is equal to the force, given the accelerations finding the mass is relatively straightforward. P4.15. Prepare: Note that an object s acceleration is linearly proportional to the net force. Solve: (a) One rubber band produces a force F, two rubber bands produce a force 2F, and so on. Because F a and two rubber bands (force 2F) produce an acceleration of 1.2 m/s 2, four rubber bands will produce an acceleration of 2.4 m/s 2. (b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2m). Using F = ma, 2F = (2m)a a = F/m = 0.60 m/s 2. Assess: Newton s second law predicts that for the same mass, doubling the force doubles the acceleration. It also says that doubling mass alone halves the acceleration. These are consistent with parts (a) and (b), respectively. P4.16. Prepare: The problem may be solved by applying Newton s second law to the present and the new situation. Solve: (a) We are told that for an unknown force (call it F o ) acting on an unknown mass (call it m o ) the acceleration of the mass is 10 m/s 2. According to Newton s second law Fo = mo(10 m/s 2 ) or Fo / m o = 10 m/s 2 1 For the new situation, the new force is Fnew = F 2 o, the mass is not changed ( mnew = mo ) and we may find the acceleration by or Fnew = mnewanew

11 Forces and Newton s Laws of Motion 4-11 a = F / m = ( F / 2) / m = ( F / m ) / 2 = (10 m/s ) / 2 = 5.0 m/s new new new o o o o 2 2 (b) For the new situation, the force is unchanged F = F, the new mass is 1 new o mnew = m 2 o and we may find the acceleration by or Fnew = mnewanew a = F / m = F / ( m / 2) = 2( F / m ) = 2(10 m/s ) = 20 m/s new new new o o o o 2 2 (c) A similar procedure gives a = 10 m/s 2. (d) A similar procedure gives a = 2.5 m/s 2. Assess: Knowing Newton s second law, one may obtain the answers in a qualitative manner. However the quantitatine method used here usually results in fewer errors. P4.17. Prepare: The problem may be solved by applying Newton s second law to the present and the new situation. Solve: (a) We are told that for an unknown force (call it F o ) acting on an unknown mass (call it m o ) the acceleration of the mass is 8.0 m/s 2. According to Newton s second law Fo = mo(8.0 m/s 2 ) or Fo / m o = 8.0 m/s 2 For the new situation, the new force is Fnew = 2 Fo, the mass is not changed ( mnew = mo ) and we may find the acceleration by or Fnew = mnewanew a = F / m = 2 F / m = 2( F / m ) = 2(8 m/s ) = 16 m/s new new new o o o o 2 2 (b) For the new situation, the force is unchanged F = F, the new mass is half the old mass ( m = m /2) new o new o and we may find the acceleration by or Fnew = mnewanew a = F / m = F / 2 m = ( F / m ) / 2 = (8.0 m/s ) / 2 = 4.0 m/s new new new o o o o 2 2 (c) A similar procedure gives a = 8.0 m/s 2. (d) A similar procedure gives a = 32 m/s 2. Assess: From the algebraic relaionship a= F / mwe can see that when (a) the force is doubled, the acceleration is doubled; (b) the mass is doubled, the acceleration is halved; (c) both force and mass are doubled, the accelertaion doesn t change; and (d) force is doubled and mass is halved, the acceleration will be four times larger. P4.18. Prepare: We can use Newton s second law, Equation 4.4, to find the acceleration of the wagon with the child. Solve: We do not know the mass of the wagon, but we do know that the same force is applied to the empty wagon as to the wagon with the child in it. We also know that the mass of the child is three times that of the wagon, so the child and wagon together have a mass of four times the mass of the empty wagon. Using Newton s second law for both the empty wagon and the wagon containing the child we have F 2 aempty wagon = = 1.4 m/s m F 1 F a = wagon and child (1.4 m/s ) 0.35 m/s 4 m = 4 m = 4 = Assess: We could have also used the fact that acceleration is inversely proportional to mass as discussed in Section 4.5. Since the mass is four times larger, the acceleration must be a quarter the original acceleration.

12 4-12 Chapter 4 P4.19. Prepare: The principle involved is that the acceleration is inversely proportional to the mass on which the force acts. We will assume that the same maximum force accelerates the two-car system as was available to the one car alone. Solve: Doubling the mass halves the acceleration if the force is the same, so the new maximum acceleration would be 2 a 5.0 m/s a = = = 2.5 m/s 2 2 Assess: Hopefully the second law has had enough time to sink into your psyche so that this just feels right. P4.20. Prepare: Inspect the graph carefully and read off some selected values of forces. 2 Solve: Newton s second law is F = ma. When F = 2 N, we have 2 N = (0.5 kg)a, hence a = 4 m/s 2. After repeating this procedure at various points, the previous graph is obtained. Assess: Because F and a are linearly related, we expected a force-versus-acceleration graph to be a straight line. P4.21. Prepare: Refer to Figure P4.21. Solve: Newton s second law is F = ma. We can read a force and an acceleration from the graph, and hence find the mass. Choosing the force F = 1 N gives us a = 4 m/s 2. Newton s second law then yields m = 0.25 kg. Assess: Slope of the acceleration-versus-force graph is 4 m/n s 2, and therefore, the inverse of the slope will give the mass. P4.22. Prepare: We can use Newton s second law to find the acceleration of the bear. Solve: The only forces on the bear are exerted by the girl and boy. A free-body diagram is shown. (a) From the free-body diagram shown, the net force in the x-direction is F = F + F = 15 N 17 N = 2 N Net Boy Girl The net force acting on the bear is 2 N to the left. Since the net force on the bear is not zero, the bear is accelerating. Since at this instant, we know nothing about the rate at which the bear s position is changing nothing can be said about the velocity of the bear. (b) The bear is accelerating, since there is a net force on the bear. From part (a), the net force is 2 N to the left. We can use Newton s second law to find the acceleration of the bear given the mass of the bear. a F m 2N 0.2 kg net = = = The acceleration is in the same direction as the force, to the left. 10 m/s 2

13 Forces and Newton s Laws of Motion 4-13 Assess: Knowing the mass of an object and the net force acting on it, Newton s second law may be used to determine its acceleration. The acceleration is always in the direction of the net force acting on an object. P4.23. Prepare: We can use our knowledge of kinematics and our conceptual understanding of acceleration to determine the acceleration of the car and then Newton s second law to determine the force acting on the car. Solve: First let s determine the acceleration of the car. m m 2 a= v/ t = ( vf vi) / t = / 2.0 s = 0.50 m/s s s Next, let s determine the force acting on the car. 2 2 F = ma = (1500 kg)(0.50 m/s ) = N Assess: It should be noted that even though we are in chapter 4 and dealing primarily with forces, it is critical that we stay up on the content of previous chapters. P4.24. Prepare: We can use Newton s second law, Equation 4.4, to find the acceleration of the electron. Solve: Applying Equation 4.4, F a = = = m kg N m/s Assess: As expected, this is a huge acceleration. The mass of the electron is numerically extremely small compared to the numerical value for the weight of a penny. P4.25. Prepare: This is a straightforward application of Newton s second law: Fnet = ma, where 8 m = kg, and F net = F thrust (because we are told the supertanker is subject to no other forces), and 6 F thrust = 5 10 N (as given in Table 4.2). The vectors F net and a must point in the same direction (because m is never negative), so we now compute the magnitude. Solve: Solve the second law for a. a F m F m kg 6 net thrust 5 10 N 2 2 = = = = m/s 0.02 m/s 8 Assess: It appears that there is only one significant figure in the data in Table 4.2, hence the rounding to m/s.if we assume one more significant figure in the table then we can report a = m/s. However, the point of the problem is clear: The rocket motor (the largest force in Table 4.2) was able to give the supertanker only an impressively small acceleration. P4.26. Prepare: The free-body diagram shows two equal and opposite forces such that the net force is zero. The force directed down is labeled as a weight, and the force directed up is labeled as a tension. With zero net force the acceleration is zero. Draw as shown a picture of a real object with two forces to match the given freebody diagram. Solve: A possible description is: An object hangs from a rope and is at rest. Or, An object hanging from a rope is moving up or down with a constant speed. Assess: This problem and the following two problems make it clear how important it is to know all forces (and their direction) acting on an object in order to determine the net force acting on the object.

14 4-14 Chapter 4 P4.27. Prepare: The free-body diagram shows three forces with a net force (and therefore net acceleration) upward. There is a force labeled w directed down, a force F thrust directed up, and a force D directed down. Now, draw a picture of a real object with three forces to match the given free-body diagram. Solve: A possible description is, A rocket accelerates upward. Assess: It is given that the net force is pointing up. Then, Fnet = Fthrust w D must be greater than zero. In other words, F must be larger than ( w+ D ). thrust P4.28. Prepare: The free-body diagram shows three forces. There is a weight force w, which is down. There is a normal force labeled n, which is up. The forces w and n are shown with vectors of the same length so they are equal in magnitude and the net vertical force is zero. So we have an object on the ground that is not moving vertically. There is also a force f k to the left. This must be a frictional force and we need to decide whether it is static or kinetic friction. The frictional force is the only horizontal force, so the net horizontal force must be f k. This means there is a net force to the left producing an acceleration to the left. This all implies motion and therefore the frictional force is kinetic. Draw a picture of a real object with three forces to match the given free-body diagram. Solve: A possible description is, A baseball player is sliding into second base. Assess: On the free-body diagram, kinetic friction force is the only horizontal force, and it is pointing to the left. This tells us that the baseball player is sliding to the right. P4.29. Prepare: Draw a picture of the situation, identify the system, in this case the car, and draw a closed curve around it. Name and label all relevant contact forces and long-range forces. Solve: There are two forces acting on the car due to its interactions with the two agents the earth and the ground. One of the forces on the car is the long-range weight force by the earth. Another force is the normal force exerted by the ground due to the contact between the car and the ground. Since the car is sitting in the parking lot, acceleration is zero, and therefore the net force must also be zero. The free-body diagram is shown on the right.

16 4-16 Chapter 4 Solve: There are three forces acting on the physics textbook due to its interactions with the two agents the earth and the table. One of the forces on the book is the long-range weight force by the earth. Forces exerted on the book by the table are the normal force and the force of kinetic friction. The normal force exerted by the surface of the table is due to the contact between the book and the table. The force of kinetic friction exerted by the surface of the table is due to the sliding contact between the book and the table. Since the textbook is slowing down, it has an acceleration and hence net force. The free-body diagram is shown on the right. Assess: The problem uses the word sliding. Any real sliding situation involves kinetic friction with the surface the object is sliding over. The force of kinetic friction always opposes the motion of the object. P4.33. Prepare: Follow the steps outlined in Tactics Boxes 4.2 and 4.3. Draw a picture of the situation, identify the system, in this case the elevator, and draw a closed curve around it. Name and label all relevant contact forces (the tension) and long-range forces (weight). Solve: There are two forces acting on the elevator due to its interactions with the two agents earth and the cable. One of the forces on the elevator is the long-range weight force by the earth. Another force is the tension force exerted by the cable due to the contact between the elevator and the cable. Since the elevator is coming to a stop while ascending, it has a negative acceleration, hence a negative net force. As a result, we know that the length of the vector representing the weight of the elevator must be greater than the length of the vector representing the tension in the cable. The free-body diagram is shown in the previous diagram on the right. Assess: There are only two forces on the elevator. The weight is directed down and the tension in the cable is directed up. Since the elevator is slowing down (has a negative acceleration) the tension must be less than the weight. P4.34. Prepare: Follow the steps outlined in Tactics Boxes 4.2 and 4.3. Draw a picture of the situation, identify the system, in this case the skier, and draw a closed curve around it. Name and label all relevant contact forces (normal and kinetic friction) and long-range forces (weight). For the force diagram, identify an appropriate coordinate system (parallel and perpendicular to the motion of the skier), draw and label all forces and then the net force acting on the skier. Solve: There are three forces acting on the skier due to interactions with the two agents earth and the snow. One of the forces is the long-range weight force by the earth. Forces exerted on the skier by the snow are the normal force and the force of kinetic friction. The normal force exerted by the snow is due to the contact between the snow and the skier. The force of kinetic friction exerted by the snow is due to the sliding contact between the snow and skier. Since the skier is traveling down the slope at a constant speed, there is no acceleration and hence no net force. The free-body diagram is shown on the right. Assess: There is a lot more to this problem than we are presently able to appreciate. After you master the content of chapter 5, you will be able to appreciate your physics growth by looking at this problem again. The language and the ideas you will use then as opposed to now will be considerably more sophisticated.

17 Forces and Newton s Laws of Motion 4-17 P4.35. Prepare: We follow the steps outlined in Tactics Boxes 4.2 and 4.3. Solve: The system is the picture. The objects in contact with the picture are the wall and your hand. The wall exerts a normal force (opposite the pushing force of the hand) and a static friction force, which is directed upward and prevents the picture from falling down. The important long-range force is the gravitational force of the earth on the picture (i.e., the weight). Assess: The net force is zero, as it should be for an object which is motionless (isn t accelerating). P4.36. Prepare: Draw a picture of the situation, identify the objects and agents, and the pair-wise interactions between them. There are four objects, namely the weightlifter, a barbell, a rough surface (floor), and the earth. Name and label all relevant contact forces and long-range forces.

18 4-18 Chapter 4 Solve: Figure (i) shows a weightlifter (WL) holding a heavy barbell (BB) across his shoulders. He is standing on a rough surface (S) that is a part of the earth (E). We distinguish between the surface (S), which exerts a contact force, and the earth, which exerts the long range force of gravity. Figure (ii) shows the barbell, the weightlifter, and the earth separated from one another. This separation helps to indicate the forces acting on each object. Figures (iii) and (iv) are free-body diagrams for the barbell and the weightlifter, respectively. Altogether there are four interactions. There is the interaction between the barbell and the weightlifter, the weightlifter and the surface of the earth (contact force), the barbell and the earth, and the weightlifter and the earth. The interaction between the barbell and the weightlifter leads to two forces: F WL on BB and F. BB on WL This is an action/reaction pair of forces. The interaction between the weightlifter and the surface of the earth is a contact interaction leading to n S on WL and n. WL on S These two forces also constitute an action/reaction pair. The weight force w E on BB has its action/reaction force w BB on E, and the weight force w E on WL has its action/reaction pair force w. WL on E Assess: Note that if we are studying a particular object, say the weightlifter, all the other three objects (barbell, rough surface, and earth) will be the agents interacting with the chosen object (in this case the weightlifter). In other words, the terms object and agent are interchangeable depending on which one is chosen to be the object. P4.37. Prepare: Knowing that for every action there is an equal and opposite reaction and that these forces are exerted on different objects, we can identify, draw and label all the action-reaction pairs. Knowing all the forces acting on skater 2, we can construct a free-body diagram for skater 2. Solve: Figure (i) shows all three skater and the pair-wise interactions (connected by the dashed line) between them. Figure (ii) identifies skater 2 as the object of interest, shows the three contact forces that act on her, and the one long-range force that acts on her. Finally Figure (iii) shows a free body diagram for skater 2. Assess: We have been informed that there is no friction and that the skaters are standing. If this is the case the net force acting on skater 2 should be zero. Also note that the action reaction pairs act on different objects. P4.38. Prepare: Knowing that for every action there is an equal and opposite reaction and that these forces are exerted on different objects, we can identify all the action-reaction pairs. Solve: When the girl stands on the sofa there is only one action-reaction pair of forces acting between the girl and the sofa. This action pair consist of the force the girl exerts on the sofa and the force the sofa exerts on the girl. Assess: The two forces involved in an action pair act on two different objects. The girl exerts a force on the sofa (one object) and the sofa exerts a force on the girl (the other object).

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