Question

A monkey is climbing on a rope that goes over a smooth light pulley and supports a block of equal mass at the other end (figure 5−E19). Show that whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equal acceleration. If initially both were at rest, their separation will not change as time passes

Answer 1

To Prove:

Whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equal acceleration.

Proof:

We shall use method of contradiction. First we will assume that monkey and the block are moving with different acceleration and later proof that the two accelerations are equal.

Let mass of monkey and block be m ;

According to FBD of monkey;

$T - mg = ma \: \: \: \: .......(1)$

According to FBD of block:

$mg - T = m(a') \: \: \: ......(2)$

Putting value of T in eq.(1):

$\therefore \: T - mg = ma$

$= > \: mg - m( a') - mg = ma$

$= > \: m( a') = - ma$

$= > \: a' = - a$

Since we had initially considered a' opposite to a , now it means that a and a' are in same directions and equal.

[Hence proved]

Answer 2

Answer:

Suppose the monkey acceleration upward with acceleration 'a' & the block, acceleration downward with acceleration a

1

. Let Force exerted by monkey is equal to T

From the free body diagram of monkey∴T−mg−ma=0

⇒T=mg+ma.Again from the FBD of the block

T=ma

11

−mg=0

⇒T=mg+ma+ma

1

−mg=0 [From (I)]

⇒ma=−ma

1

⇒a=a

1

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