Question

A monkey climbs up and another monkey climbs down a rope hanging from a tree with same uniform acceleration separately. If the respective masses of monkeys are in the ratio 2:3, the common acceleration must be

1)g. 2)6g
3)g/2. 3) g/5​

Answer 1

Given :

The masses of monkeys are in the ratio m1:m2 = 2 : 3

To Find :

The common acceleration

Solution :

Let first monkey climb the rope with acceleration = a

Then,

T - m1g = m1a ( T = Tension of a rope) ______ (1)

Again, Let the second monkey climb the rope with acceleration = a

Then,

m2g - Tm2a ( T = Tension of a rope) ______(2)

Adding (1) & (2) , we get

(m2 - m1)g = ( m1 + m2)a

$(1 - \frac{m1}{m2} )g \: = ( \frac{m1}{m2} + \: 1)a$

or,

$(1 \: - \frac{2}{3} )g \: = \: ( \frac{2}{3} + \: 1)a$

or,

$\frac{1}{3}g \: = \: \frac{2}{3}a$

$a = \: \frac{g}{5}$

Thus, the required acceleration, a = g/5 .