A monkey of mass m clings to a rope over a fixed pulley .
The opposite ends of the rope is tide to a weight of mass M lying on a horizontal plate neglecting friction, find the acceleration of both bodies relative to the plate and tension of the rope in the following case:
(i)the monkey does not move with respect to rope.
(ii)the monkey moves upwards with respect to rope with acceleration `beta`
(iii) the monkey moves downwards with respect to the rope with acceleration `beta`

Explanation

(i) If the monkey is at rest on the rope , then acceleratiion 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+mbeta-T=ma_(2)` (for monkey ) ....(4)
`T=Ma_(2)`(for mass M) ....(5)
Adding , we get , `mg+mbeta=(M+m)a_(2)`
`:. a_(2)=(m(g+beta))/(M+m)`
`:.` The acceleration of mass relative to plate
`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+mbeta-mbeta-mbeta)/(M+m)=(Mg-Mbeta)/(M+m)`
(iii) When monkey moves downward with acceleration `beta`, teh reaction `(R=mbeta)` acts upward, therefore the equations of motion take the form (`a_(3)` is acceleration of block relative to plate )
`mg-mbeta-T=ma_(3)` ....(8)
`T=Ma_(3)` ....(9)
Adding we get `mg-mbeta=(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-mbeta+Mbeta+mbeta)/(M+m)`
`:. a_(3)^(1)=(mg+Mbeta)/(M+m)`