When the coefficient of friction between the weight and the palte is `mu`.

Answer

(1) `a_(1)=((m-muM)g)/(m+M), F_(1)=(mMg(1+mu))/(m+M)`;
(2) `a_(2)=(m(g+b)-muM g)/(m+M), F_(2)=(mM[g(1+mu)+b])/(m+M)`;
(3) `a_(3)=(m(g-b)-muMg)/(m+M), F_(3)=(mM[g(1+mu)-b])/(m+M)`

Explanation

(1) If the monkey is at rest with respect to the rope, the equations of motion are of the form
`F_(1) - mu Mg = Ma_(1), mg - F_(1)= ma_(1)`
whence the answer sought. The solution is meaningful if `m gt muM`.
When `m le muM, a_(1)=0 and F_(1)=mg`.
(2) If the monkey moves upwards with respect to the rope with an acceleration b, the equations of motion assume the form
`F_(2)-muMg= Ma_(2), mg-F_(2) = m(a_(2)-b)`
The solution is meaningful if `muMg lt m(g+b)`. In the contrary case
`a_(2)=0, a._(2)= -b, F_(2)= m(g+b)`.
(3) If the monkey moves downwards with respect to the rope with an acceleration b, the sign of b in the equations of motion for the previous case should be reversed.