PH201 Chapter 5 Solutions 5.4. Set Up: For each object use coordinates where +y is upward. Each object has Call the objects 1 and 2, with and Solve: (a) The free-body diagrams for each object are shown in the figure below. and are replaced by their x and y components. (b) For object 2, and (c) For object 2, so 5.8. Set Up: For each muscle where Use coordinates where +y is along the Achilles tendon. The force exerted by each muscle then makes an angle of with the +y axis. Apply the equilibrium conditions to the point where the muscles attach to the tendon. The free-body diagram is given in the figure below. T is the tension in the tendon. Solve:
5.9. Set Up: Use coordinates where +y is upward and +x is to the right. The force at the foot is horizontal. The tension T in the cable that has one end attached to the leg and the other end attached to the weight is equal to W and is the same everywhere along the cable. Solve: (a) The free-body diagram for the leg is given in Figure (a) below. is the traction force on the leg. and so (b) The free-body diagram for the pulley that is attached to the foot is given in Figure (b) above. Reflect: The tractive force on the leg is greater than W. *5.15. Set Up: Apply to each block. Take +y perpendicular to the incline and +x parallel to the incline. Solve: The free-body diagrams for each block, A and B, are given in the figure below. (a) For B, and (b) For block A, and (c) for each block (d) For and For and
*5.23. Set Up: Take +y to be upward. The fish has the same upward acceleration as the elevator. Let be the upward force exerted on the fish by the spring balance; F is what the balance reads. Solve: (a) The free-body diagram for the fish is sketched in the figure below. and The weight of the fish is (b) and from part (a). The elevator is accelerating downward with (d) If the cable breaks, and The elevator is in free-fall and the balance reads zero. 5.28. Set Up: In each free-body diagram take a positive coordinate direction to be the direction of the acceleration of the object. The box accelerates to the right and the bucket accelerates downward. Their accelerations have the same magnitude. The box has mass Solve: (a) The free-body diagram for the box and bucket are shown in the figure below. (b) applied to the box applied to the bucket Combining these two equations to eliminate T (c)
*5.29. Set Up: Let and The two boxes have the same magnitude of acceleration. For the 30.0 kg box let +y be downward and for the 50.0 kg box use coordinates parallel and perpendicular to the ramp, with up the ramp. Solve: (a) The free-body diagrams are given in the figure below. (b) The downward force on the 30.0 kg box is The force pulling the system in the opposite direction is the component of directed down the ramp, and the 30.0 kg box moves downward. Therefore, the 50.0 kg box moves up the ramp. (c) applied to applied to Combining these two equations to eliminate T The 50.0 kg box accelerates up the ramp at and the 30.0 kg box accelerates downward at Reflect: Only the component of the weight of the 50 kg box that is parallel to the ramp acts to oppose the weight of the 30 kg box. Therefore, the 30 kg box pulls the 50 kg box up the ramp even though its weight is less than the weight of the 50 kg box. 5.32. Set Up: Assume that the normal force n at each hip is vertical. Solve: (a) The free-body diagram for the upper body of the person is shown in the figure below. and
(b) (c) 5.34. Set Up: Constant velocity means for each crate. Apply to each crate, with +x to the right. Solve: The free-body diagrams for each crate are shown in the figure below. (a) for each crate and and for A for B Adding these two equations (b) 5.35. Set Up: Use the information about the motion to find the acceleration of the puck and then use to relate a to the friction force. Take +x to be in the direction the puck is moving. Solve: and The free-body diagram for the puck is given in the figure below. and
5.36. Set Up: Use to find the acceleration that can be given to the car by the kinetic friction force. Then use a constant acceleration equation. Take +x in the direction the car is moving. Solve: (a) The free-body diagram for the car is shown in the figure below. and Then and (b) *5.39. Set Up: For constant speed, Apply with in the direction of motion of the box. Solve: (a) The free-body diagram for the box is given in the figure below. and
(b) and Reflect: In (b) the friction force and the acceleration are directed opposite to the motion. Since and are in opposite directions, the box slows down. 5.40. Set Up: Once the crate has started to move, friction is kinetic. The free-body diagram for the crate is given in the figure below. is the force the worker applies. The crate has mass Solve: so The force F to get the box moving is After the box is moving, friction is and 5.43. Set Up: Take +x to be down the incline. At the maximum angle, the static friction force has its maximum value, Solve: The free-body diagram for the patient is given in the figure below. and
5.48. Set Up: Use coordinates where is upward and is horizontal to the right. The applied force pushes to the left so the friction force is to the right. Solve: (a) The free-body diagram is given in the figure below. The applied force has been replaced by its x and y components.
(b) and (c) and (d) The maximum possible static friction force is (e) The vertical component of F is now upward. now 5.56. Set Up: where F is the pull on the strip or the force the strip exerts. Solve: (a) so (b) 5.58. Set Up: so Solve: (a) Yes, Hooke s law says and this is true here since the graph of F versus x is a straight line. (b) k is the slope of the graph of F versus x: (c) 5.62. Set Up: The free-body diagrams are given in the figure below. There is a kinetic friction force Solve: (a) and so (b) so and and 5.68. Set Up: The free-body diagram for one bowling ball is given in the figure below. The tension T has been resolved into its x and y components. n is the force the other ball exerts on this one.
Solve: (a) and (b) and 5.76. Set Up: Constant speed means Use Newton s third law to relate forces on A and on B. Solve: (a) Treat A and B as a single object of weight The free-body diagram for this combined object is given in Figure (a) below. (b) The free-body force diagrams for blocks A and B are given in Figures (b) and (c) above. n and are the normal and friction forces applied to block B by the tabletop and are the same as in part (a). is the friction force that A applies to B. It is to the right because the force from A opposes the motion of B. is the downward force that A exerts on B. is the friction force that B applies to A. It is to the left because block B wants A to move with it. is the normal force that block B exerts on A. By Newton s third law, and these forces are in opposite directions. Also, and these forces are in opposite directions. for block A so and for block A for block B (c) In part (a) block A is at rest with respect to B and it has zero acceleration. There is no horizontal force on A besides friction, and the friction force on A is zero. 5.82. Set Up: The tension throughout the cable connected to W is the weight W of the hanging mass. The freebody diagram for the pulley connected to the feet is given in the figure below.
Solve: (a) to support the leg. and The mass of W is 44.7 kg. (b) The tractive force along the leg is *5.87. Set Up: The block has the same horizontal acceleration a as the cart. Let +x be to the right and +y be upward. To find the minimum acceleration required, set the static friction force equal to its maximum value, Solve: The free-body diagram for the block is given in the figure below. and Reflect: The smaller is the greater a must be to prevent slipping. Increasing a increases the normal force n and that increases the maximum for a given