Question

Three monkeys A,B,C with masses of 10,15,8 kg respectively are climbing up and down the rope suspended from $D$. At the instant represented, $A$ is descending the rope with an acceleration of $2m/s^2$ and $C$ is pulling himself up with an acceleration of $1.5m/s^2$. Monkeys $B$ is climbing up with a constant speed of $0.8$m/s. Treat the rope and monkeys as a complete system and calculate the tension $T$ in the rope at $D$. $(g=10m/s^2)$

Answer 1

For $A$,

It is descending with acc. $2m/s^2$ which requires a force of $F=md=10\times 2=20N$. It is moving downward which means it is applying force on the rope in the upward direction as it will be getting a force of in downward direction both the forces will have same magnitude. [Newton's third law]

For $B$, it is moving up with constant velocity so, no other force except force of gravity.

Similar to the case of $A$, in case $C$ it will has a force of $8\times 1.5=5N$ on the rope along the downward direction.

Thus $T-100-150-80-6+20=0$

$T=316N$.