The monkey B shown in figure is holding on to the tail of the monkey A which is climbing up a rope. The masses of the monkeys A and B are 5 kg and 2kg respectively. If A can tolerate a tension of 30 N in its tail what force should it apply on the rope in order to carry the monkey B with it? Take g=10 ,m/`s^2`.

Answer

105N

Explanation

Let `T_(1)` be the tension in rope and `T_(2)` the tension in the tail of monkey A. Let the force exerted by monkey A on the rope of F.
Let `M_(A)` and `M_(B)` be masses of monkey A and B, respectively. The free-body diagram of monkeys A and B are shown in fig. (a) and (b) . `F=T_(1)`

The equation of motion of monkeys A and B are
`T_(1)=M_(A)g=T_(2)=M_(A)a` ..(i)
`T_(2)=M_(B)g=M_(B)a` ...(ii)
For maximum tension in tail.
From, Eq (ii), we get
`a=(T_(2)-M_(B)g)/(M_(B)) = (30-2xx10)/(2)= 5ms^(-2)`
Then from Eq. (i) we get,
`T_(1)=T_(2)=M_(A)(g+a)=30+5(10+5)=105N`.