Question

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 $5\text{kg}$ and $2\text{kg}$ respectively. If $A$ can tolerate tension of $30\text{N}$ in its tail, what force should it apply on the rope in order to carry the monkey $B$ with it? Take $g=10{\text{m/s}}^2$.

• A. $30-50\text{N}$
• B. $50-60\text{N}$
• C. $70-105\text{N}$
• D. $110-130\text{N}$

Answer 1

The correct option is C $70-105\text{N}$

Given,
Mass of monkey $A,m_A=5\text{kg}$
Mass of monkey $B,m_B=2\text{kg}$
Maximum tension in tail of monkey $A=30\text{N}$

There are two situations -
(A) Both monkeys are climbing up with constant velocity.
(B) Both monkeys are climbing up with some constant acceleration.

For situation A:
$T_1\to$ Tension in rope
$T_2\to$ Tension in tail of monkey $A$
FBD of monkey $B$:
On applying $\sum F=ma$
$T_2-20=0$
$\Rightarrow T_2=20\text{N}$
FBD of monkey $A$:
On applying $\sum F=ma$
$T_1-T_2-50=0$
$\Rightarrow T_1=T_2+50$
$\Rightarrow T_1=20+50$
$T_1=70\text{N}$

For situation $B$:
$T_1\to$ Tension in rope
$T_2\to$ Tension in tail of monkey $A$

FBD of monkey $B$:
On applying $\sum F=ma$
$T_2-20=2a$
$\because$ Maximum value of tension $T_2$ in tail of monkey $A=30\text{N}$
$\therefore 30-20=2a\Rightarrow a=5{\text{m/s}}^2$

FBD of monkey $A$:
On applying $\sum F=ma$
$T_1-T_2-50=5a$
$\Rightarrow T_1=5a+50+T_2$
$\Rightarrow T_1=25+50+30=105\text{N}$
$T_1=105\text{N}$