Question

2.57 In Figure P2.57 the masses of blocks_4, B, and C are 50, 20, and 30 kg, respectively. Find the accelerations of each if the table is removed. Which block will hit the floor first? How long will it take? Light pulley 0.8 m Figure P2.57

Answer 1

The free-body diagram for the system of the blocks is as follows, # 2T Light pulley mg mg 0.8m As the weight of the block A is equal to the weight of the blocks B and C, the block A will be in rest which has zero acceleration, that is, a=0. The net force on the block A is, 2T =m 8 Rearrange the above equation for T and substitute the values in the equation. T-m,& _ (50 kg) (9.81 m/s) = 245.25 N 2 (2) The net force on the block C is, m.8-T = m a Rearrange the above equation for a and substitute the values in the equation. a= mcg-T mc (30 kg) (9.81 m/s?)-(245.25 N) (30 kg) =1.635 m/s

The blocks B and C will have the same magnitude of the accelerations, but their directions are different, the block B will accelerate upwards and the block C will accelerate downwards. The acceleration of block A is, a=0 The acceleration of block B is, az =1.635 m/s upwards. The acceleration of the block C is, a, =1.635 m/s? downwards. As the block C will accelerate downwards, the block C will reach the ground first. Therefore, the block C will reach the ground first. The block C will starts from rest so it initial velocity is zero, that is, u=0 Use the following kinematic equation. h=ut+zat? Substitute 0 for u in above equation. h=(0)1 + Rearrange the above equation for t and substitute the values in the equation. ta 2(0.8 m) V(1.635 m/s) = 0.98 s Therefore, the time taken by the block C to reach to the ground is 0.98 s.