This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 5-36 Chapter 5 5.87. Set Up: The block has the same horizontal acceleration a as the cart. Let 1x be to the right and 1y be upward. To ﬁnd the minimum acceleration required, set the static friction force equal to its maximum value, m sn. Solve: The free-body diagram for the block is given in Figure 5.87.
y fs 5 ms n a x n mg Figure 5.87 g . ms Reﬂect: The smaller m s is the greater a must be to prevent slipping. Increasing a increases the normal force n and that increases the maximum fs for a given m s. g Fx 5 max gives n 5 ma. g Fy 5 may gives fs 2 mg 5 0. m sn 5 mg and m sma 5 mg. a 5 5.88. Set Up: The monkey and bananas have the same mass and the tension in the rope has the same upward value
at the bananas and at the monkey. Therefore, the monkey and bananas will have the same net force and hence the same acceleration, in both magnitude and direction. Solve: (a) For the monkey to move up, T . mg. The bananas also move up. (b) The bananas and monkey move with the same acceleration and the distance between them remains constant. (c) Both the monkey and bananas are in free fall. They have the same initial velocity and as they fall the distance between them doesn’t change. (d) The bananas will slow down at the same rate as the monkey. If the monkey comes to a stop, so will the bananas. ...
View Full Document