A monkey of mass m climbs up to a rope hung over a fixed pulley. The opposite end of the rope is tied to a weight of mass M lying on a horizontal plane. Neglecting the friction, find the acceleration of both the bodies (relative to the plane) and the tension of the rope for the three cases:
(i) The monkey does not move with respect to the rope
(ii) The monkey moves upwards with respect to the rope with acceleration a.
(iii) The monkey moves downwards with respect to the rope, with acceleration a.

Explanation

(i) if the monkey does not move on the rope, the acceleration of both bodies will be the same, equal to `a_(0)` The equations of motion are:
For mass M, `T=Ma_(0)` ….(i)
For monkey `mg-T=ma_(0)` ….(ii)
From Eqns. (i) and (ii)
`a_(0)=m/(m+M)g,T=(Mm)/(M+m)g`
(ii) The monkey's actual acceleration is superposition of acceleration of rope and acceleration of monkey relative to rope. Let the monkey move upwards with respect to the rope with an acceleration `vec(a)` MR=a The weight mvoes with an acceleration `a_(0)` which is the acceleration of rope relative to fixed pulley

`vec(a)_(MR)=vec(a)_(M)-vec(a)_(R)`
`vec(a)_(M)=vec(a)_(MR)+vec(a)_(R)=(a_(0)-a)`
[As monkey moves upwards and rope slides downwards overall the monkey will be accelerated downwards]
For mass `M,T=Ma_(0)` ....(i)
For monkey `mg-T=m(a_(0)-a)` ....(ii)
Solving Eqns (i) and (ii), we will get
`a_(0)=(m(g-a))/(M+m) and T=(Mm(g+a))/(M+m)`
(iii) For downward motion of the monkey, we have to change sign of a to get
`a_(0)=(m(g-a))/(M+m) and T=(Mm(g-a))/(M+m)`