Physics 16 Problem Set 4 Solutions

Physics  16  Problem  Set  4  Solutions     4.24. IDENTIFY: The reaction forces in Newton’s third law are always between a pair of objects. In Newton’s second law all the forces act on a single object. SET UP: Let be downward. . EXECUTE: The reaction to the upward normal force on the passenger is the downward normal force, also of magnitude 620 N, that the passenger exerts on the floor. The reaction to the passenger’s weight is the gravitational force that the passenger exerts on the earth, upward and also of magnitude 650 N. gives . The passenger’s acceleration is , downward. EVALUATE: There is a net downward force on the passenger and the passenger has a downward acceleration. 4.27. IDENTIFY: Identify the forces on each object. SET UP: In each case the forces are the noncontact force of gravity (the weight) and the forces applied by objects that are in contact with each crate. Each crate touches the floor and the other crate, and some object applies to crate A. EXECUTE: (a) The free-body diagrams for each crate are given in Figure 4.27. (the force on due to ) and (the force on due to ) form an action-reaction pair. (b) Since there is no horizontal force opposing F, any value of F, no matter how small, will cause the crates to accelerate to the right. The weight of the two crates acts at a right angle to the horizontal, and is in any case balanced by the upward force of the surface on them. EVALUATE: Crate B is accelerated by and crate A is accelerated by the net force . The greater the total weight of the two crates, the greater their total mass and the smaller will be their acceleration. Figure 4.27 4.32. IDENTIFY: Identify the forces on the skier and apply . Constant speed means SET UP: Use coordinates that are parallel and perpendicular to the slope. EXECUTE: (a) The free-body diagram for the skier is given in Figure 4.32. (b) with gives . . EVALUATE: T is less than the weight of the skier. It is equal to the component of the weight that is parallel to the incline. Figure 4.32 4.41. IDENTIFY: Apply Newton’s second law to calculate a. (a) SET UP: The free-body diagram for the bucket is sketched in Figure 4.41. The net force on the bucket is upward. Figure 4.41 (b) EXECUTE: gives EVALUATE: The weight of the bucket is 47.0 N. The upward force exerted by the cord is larger than this, so the bucket accelerates upward. 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 to the balloon and its passengers and cargo, both before and after objects are dropped overboard. 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