Chapter 6
Force and Motion
Solutions to Homework Problems: 13, 20, 21, 26, 29, 37, 41, 47, 49 HRW06.13 A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15 above the horizontal. If the coefficient of static fr
53 The free-body diagram for the plane is shown on the right. F is the magnitude of the lift on the wings and m is the mass of the plane. Since the wings are tilted by 40 to the horizontal and the lift force is perpendicular to the wings, the angle is 50
23 The free-body diagrams for block B and for the knot . . T2 . just above block A are shown on the right. T1 is the . . . . T1. . . . . magnitude of the tension force of the rope pulling . . . . . . . . . . . . . . . . . . . . . FN . . . . . . . . . . .
1 (a) The free-body diagram for the bureau is shown on the right. F is the applied force, f is the force of friction, FN is the normal force of the floor, and mg is the force of gravity. Take the x axis to be horizontal and the y axis to be vertical. Assu
53 (a) The free-body diagrams are shown to the right. F is the applied force and f is the force of block 1 on block 2. Note that F is applied only to block 1 and that block 2 exerts the force f on block 1. Newtons third law has thereby been taken into acc
35 The free-body diagram is shown at the right. FN is the normal force of the plane on the block and mg is the force of gravity on the block. Take the positive x axis to be down the plane, in the direction of the acceleration, and the positive y axis to b
19 (a) The free-body diagram is shown in Fig. 516 of the text. Since the acceleration of the block is zero, the components of the Newtons second law equation yield T mg sin = 0 and FN mg cos = 0. Solve the first equation for the tension force of the strin