question_answer

A monkey of mass ' m' climbs up to a rope hung over a fixed pulley with an acceleration relative to the rope g/4. The opposite end of the rope is tied to a block of mass \[M\]lying on a rough horizontal plane. The coefficient of friction

Between the block and horizontal plane is \[\mu .\]Find the tension in the rope.

A) \[\frac{M\left( 5m-4\mu M \right)g}{4\left( M+m \right)}+\mu Mg\]

B) \[\frac{m\left( 5m-4\mu M \right)g}{4\left( M+m \right)}+\mu Mg\]

C) \[\frac{M\left( 5m-5\mu M \right)g}{4\left( M+m \right)}+\mu Mg\]

D) None of these

Correct Answer: A

Solution :

\[{{a}_{\operatorname{monkey}/rope}}=\frac{g}{4}\]

Let acceleration of \[\operatorname{M}={{a}_{0}}\]

So, acceleration of rope \[={{a}_{0}}\]

\[{{\overset{\to }{\mathop{\operatorname{a}}}\,}_{monkey}}={{\overset{\to }{\mathop{\operatorname{a}}}\,}_{monkey/rope}}-{{\overset{\to }{\mathop{\operatorname{a}}}\,}_{\operatorname{rope}}}\]

\[=\left( {{a}_{0}}-\frac{g}{4} \right)\]

Now for mass \[M,T-\mu Mg=M{{a}_{0}}..(i)\]

For monkey, \[Mg-T=m\left( {{a}_{0}}-\frac{g}{4} \right)..(ii)\]

From equation (i) and (ii)\[mg-\mu Mg={{a}_{0}}\left( M+m \right)-\frac{mg}{4}\]\[\Rightarrow {{a}_{0}}=\left( \frac{5m}{4}-\mu M \right)g/\left( M+m \right)\]\[=\frac{\left( 5m-4\mu M \right)g}{4\left( M+m \right)}\]

\[\therefore T=\frac{M\left( 5m-4\mu M \right)g}{4\left( M+m \right)}+\mu Mg\]