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 $40 \,N$. If a monkey of mass $5 \,kg$ were to climb up the rope, then the maximum acceleration in the upward direction with which it can climb safely is $\left(g=10\, ms ^{-2}\right)$

• A. $2\, ms ^{-2}$
• B. $4 \,ms ^{-2}$
• C. $6 \,ms ^{-2}$
• D. $8 \,ms ^{-2}$

Answer 1

Answer: (C)

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

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