A monkey of mass `m` clings a rope to a slung over a fixed pulley .The opposite end of the rope is tried to a weight of mass `M` tying on a horizontal table is `mu` Find the acceleration of weight and the tension of the rope for two cases .The monkey move downward with respect to the rope with an acceleration b.

The tension of rope is

Explanation

i) If the monkey is at rest on the rope, then acceleration of both mass M and monkey are same, equal to a (say)
If T is tension in the rope, then the equation of motion of monkey is
` mg-T=ma_1`..........(1)
and equation of motion of mass M is
`T = Ma_1` ........... (2)
Adding (1) and (2) `mg =(M+m)a_1`
` :. a_1 = (m)/(M+m) g ` ......... (3)
ii) When monkey moves upward with acceleration `beta`, the reaction m `beta` acts downward, therefore equations of motion now take the form (`a_2` is acceleration of mass, M relative to plate)
` mg + m beta - T = ma_2` ( for monkey ) ............. (4)
` T= Ma_2 ` ( for mass M) ............. (5)
Adding , we get , `mg + m beta = ( M+ m)a_2`
`:. a_2 =(m (g + beta))/( M + m)`
` :.` The acceleration of mass relative to plane
` a_2 = ( m ( g + beta))/( M + m)` ............ (6)
and acceleration of monkey M relative to plate
`a_2 = a_2 - beta = (m (g + beta))/( M + m) - beta = ( m(g + beta ) - beta ( M + m))/( M+ m) = (mg + m beta - M beta - m beta)/( M + m) = ( Mg - M beta)/( M + m)`

iii) When monkey moves downward with acceleration `beta`, the reaction `(R= mbeta)` acts upward, therefore the equations of motion take the form (`a_3` is acceleration of block relative to plate)

`mg - m beta -T = ma_3` .......... (8)
`T = Ma_3 ` ............ (9)
Adding we get ` mg - m beta = (M + m) a_3`
` :. a_3 =( m (g - beta))/(M + m)`
and acceleration of monkey relative to plate
` a_3^1 = a_3 + beta = (m ( g -beta))/( M +m) + beta = (mg - m beta + M beta + m beta )/( M + m)`
`:. a_3^(1) = (mg + M beta )/( M + m)`