Question

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

• A. Between $50\text{N}$ and $80\text{N}$
• B. Between $30\text{N}$ and $80\text{N}$
• C. Between $70\text{N}$ and $105\text{N}$
• D. Between $60\text{N}$ and $90\text{N}$

Answer 1

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

Let the acceleration of monkey $A$ upwards be $a$ so that a maximum tension of $30\text{N}$ is produced in its tail.


$T-5g-T_1-5a=\mathrm{0.......}\left(i\right)$
$T-5g-30-5a=0$


From equation (i) and (ii), we have


$T_1-2g-2a=\mathrm{0.......}\left(ii\right)$
$30-2g-2a=0$
$\Rightarrow a=5{\text{m/s}}^2$
So, putting $(a=5{\text{m/s}}^2)$ in equation $\left(i\right)$
we get that $A$ can apply a maximum force of $105\text{N}$ on the rope to carry monkey $B$, hence $F_{\text{max}}=105\text{N}$

Now for the minimum force, monkey $A$ can carry monkey $B$ with constant velocity. So, $a=0$
From equation $\left(ii\right)$

$\Rightarrow T_1=20$ N
Rewriting equation $\left(i\right)$ for monkey A, we get:
$T-5g-20=0\Rightarrow T=70\text{N}$

$\therefore$ To carry monkey $B$ with it, monkey $A$ should apply a force of magnitude between $70\text{N}$ and $105\text{N}$

Hence option C is the correct answer