# A monkey is climbing up on a rope that goes over a smooth pulley and supports a block of equal mass at other end. Show that whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equal acceleration

148 Points
14 years ago

Dear Sh Ha

Let the acceleration of monkey is a

for this acceleration to achieve by monkey it exerts a force T on the rope and  same reaction force is exerted by rope on the hand of monket

which cause monket to climb

So T-mg =ma

Now let the mass is acceleration by a'

force balance on mass

T-mg=ma'

or  ma=ma'

a=a'

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ankit singh
3 years ago

## Suppose the monkey acceleration upward with acceleration 'a' & the block, acceleration downward with acceleration a1​. Let Force exerted by monkey is equal to T

From the free body diagram of monkeyTmgma=0
T=mg+ma.Again from the FBD of the block
T=ma11mg=0
T=mg+ma+ma1mg=0 [From (I)]
ma=ma1a=a1
Acceleration a downward i.e. a upward.
The block & the monkey move in the same direction with equal acceleratio.If initially they are rest (no force is exertied by monkey) no motion of monkey of block occurs as they have same weight (same mass). Their separation will not changed as time passes.