A monkey of massm clings to a rope slung over a fixed pulley. The opposite end of the rope is tied to a weight of

mass M lying on a horizontal plat (Fig.). Neglecting friction find the acceleration of both bodies (relative to the plate) and the tension of the rope for the three cases:
(1) the monkey does not move with respect to the rope,
(2) the monkey moves upwards with respect to the rope with acceleration b,
(3) the monkey moves downwards with respect to the rope with an acceleration b.

Answer

(1) `a_(1)=a'_(1) = (mg)/(m+M), F_(1)=Ma_(1);`
(2) `a_(2) = (m(g+b))/(m+M), a'_(2)=a_(2)-b, F_(2) =Ma_(2);`
(3) `a_(3)=(m(g-b))/(m+M), a'_(3)=a_(3)+b, F_(3)=Ma_(3)`.

Explanation

(1) If the monkey is at rest on the rope, the acceleration of both bodies will be the same, equal to `a_(1)`. The equations of motion are
`F_(1) =Ma_(1), mg - F_(1)=ma_(1)`
where `F_(1)` is the tension of the rope.
(2) If the monkey upwards with respect to the rope with an acceleration b, the motion of both bodies will be different: the weight will move with an acceleration `a_(2)` and the monkey with an acceleration `a._(2)=a_(2)-b`. The equation of motions will assume the form
`F_(2)= Ma_(2), mg-F_(2)=ma._(2)`,
(3) The downward motion of the monkey with respect to the rope with acceleration b is described by the same equations, one has only to change the sign of b.
The downward acceleration of the monkey with respect to the rope cannot exceed the acceleration due to gravity. (Why?) Therefore `a_(3) ge 0 and F_(3) ge 0`.