Question

One end of a light string is fixed to a clamp on the ground and the other end passes over a fixed frictionless pulley as shown in the figure. It makes an angle of $30^{\circ }$ with the ground. The clamp can tolerate a vertical force of $40N$. If a monkey of mass $5kg$ were to climb up the rope, then the maximum acceleration in the upward direction with which it can climb safely is $\left(g=10m{s}^{-2}\right)$

• A. $2m{s}^{-2}$
• B. $4m{s}^{-2}$
• C. $6m{s}^{-2}$
• D. $8m{s}^{-2}$

Answer 1

Answer: (C)

Let $T$ is tension in the string.
Maximum vertical force on clamp, $T\sin 30^{\circ }=40$
$\Rightarrow T\cdot \frac{1}{2}=40$
$\Rightarrow T=80N$
Let maximum acceleration of monkey for sate
climbing is $a$. Then, FBD of monkey

$T-mg=ma$
$\Rightarrow a=\frac{T-mg}{m}$
$a=\frac{80-5\times 10}{5}$
$a=6m/s^2$