Physics 16 Problem Set 4 Solutions Part 2

Physics 16 Problem Set 4 Solutions 4.43. IDENTIFY: Use Newton’s 2nd law to relate the acceleration and forces for each crate. (a) SET UP: Since the crates are connected by a rope, they both have the same acceleration, (b) The forces on the 4.00 kg crate are shown in Figure 4.43a. EXECUTE: Figure 4.43a (c) SET UP: Forces on the 6.00 kg crate are shown in Figure 4.43b The crate accelerates to the right, so the net force is to the right. F must be larger than T. Figure 4.43b (d) EXECUTE: EVALUATE: mass gives We can also consider the two crates and the rope connecting them as a single object of The free-body diagram is sketched in Figure 4.43c. This agrees with our answer in part (d). Figure 4.43c 4.56. IDENTIFY: Apply objects are dropped overboard. to the balloon and its passengers and cargo, both before and after SET UP: When the acceleration is downward take to be downward and when the acceleration is upward take to be upward. EXECUTE: (a) The free-body diagram for the descending balloon is given in Figure 4.56. L is the lift force. (b) gives and . (c) Now is upward, so , where m is the mass remaining. , so . Mass must be dropped overboard. EVALUATE: In part (b) the lift force is greater than the total weight and in part (c) the lift force is less than the total weight. Figure 4.56 5.15. IDENTIFY: Apply Newton’s first law to the ball. Treat the ball as a particle. SET UP: The forces on the ball are gravity, the tension in the wire and the normal force exerted by the surface. The normal force is perpendicular to the surface of the ramp. Use x and y axes that are horizontal and vertical. EXECUTE: (a) The free-body diagram for the ball is given in Figure 5.15. The normal force has been replaced by its x and y components. (b) gives (c) gives and . and . EVALUATE: Note that the normal force is greater than the weight, and increases without limit as the angle of the ramp increases towards . The tension in the wire is , where is the angle of the ramp and T also increases without limit as . Figure 5.15 5.63. IDENTIFY: Apply to the rope. SET UP: The hooks exert forces on the ends of the rope. At each hook, the force that the hook exerts and the force due to the tension in the rope are an action-reaction pair. EXECUTE: (a) The vertical forces that the hooks exert must balance the weight of the rope, so each hook exerts an upward vertical force of on the rope. Therefore, the downward force that the rope T sin θ = w/2 exerts at each end is end , so Tend = mg/(2 sin θ). (b) Each half of the rope is itself in equilibrium, so the tension in the middle must balance the horizontal force that each hook exerts, which is the same as the horizontal component of the force due to the tension at the end; Tend cos θ = Tmiddle . so Tmiddle = mg/(2 tan θ) = mg cot θ/2 . (c) Mathematically speaking, because this would cause a division by zero in the equation for or . Physically speaking, we would need an infinite tension to keep a non-massless rope perfectly straight. (d) At 90º, there is no tension in the middle of the rope because all forces are in the up/down direction. Each end of the rope supports half the weight, so Tend=mg/2. EVALUATE: The tension in the rope is not the same at all points along the rope. 5.67. IDENTIFY: Apply to each block. Use Newton’s 3rd law to relate forces on A and on B. SET UP: Constant speed means . EXECUTE: (a) Treat A and B as a single object of weight diagram for this combined object is given in Figure 5.67a. . . The free-body gives . gives (b) The free-body force diagrams for blocks A and B are given in Figure 5.67b. 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. downward force that A exerts on B. block B wants A to move with it. is the is the friction force that B applies to A. It is to the left because is the normal force that block B exerts on A. By Newton’s third law, and these forces are in opposite directions. Also, opposite directions. for block A gives , so . and these forces are in , and for block A gives . for block B gives EVALUATE: 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. A larger force F is needed in part (b), because of the friction force between the two blocks. Figure 5.67a–c 5.92. IDENTIFY: Apply to each block. SET UP: Use coordinates where is directed down the incline. EXECUTE: (a) Since the larger block (the trailing block) has the larger coefficient of friction, it will need to be pulled down the plane; i.e., the larger block will not move faster than the smaller block, and the blocks will have the same acceleration. For the smaller block, or and similarly for the larger, . Adding these two relations, (b) Substitution into either of the above relations gives (c) The string will be slack. The 4.00-kg block will have and the 8.00-kg block will have until the 4.00-kg block overtakes the 8.00-kg block and collides with it. EVALUATE: If the string is cut the acceleration of each block will be independent of the mass of that block and will depend only on the slope angle and the coefficient of kinetic friction. The 8.00-kg block would have a smaller acceleration even though it has a larger mass, since it has a larger .