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

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Aicharya · Accepted Answer

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

$\begin{aligned} & T-m g=m a \Rightarrow a=\frac{T-m g}{m} \\ & a=\frac{80-5 \times 10}{5} \\ a= & 6 \mathrm{~m} / \mathrm{s}^2\end{aligned}$

One end of a light string is fixed to a clamp on the ground and the other end passes over a fixed…

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