In the arrangement shown in figure the cylinder of mass $M=9 \mathrm{~kg}$ is at rest on an incline. The plane is inclined at angle $37^{\circ}$ with horizontal. The string between the cylinder and the pulley $(P)$ is horizontal. Find the minimum the mass of the block (in $\mathrm{kg}$) to keep the system in equilibrium.

Question

Aicharya · Accepted Answer

Let the mass of the block be $m$ For equilibrium of the block, $T=m g$ ...(i) For rotational equilibrium of the cylinder, net torque about its centre must be zero. $\begin{array}{l} \therefore \quad T R=f R \quad \text{[R= radius of cylinder]}\ \Rightarrow m g=f \quad \text{...(ii)} \end{array}$ For translational equilibrium of the cylinder $\begin{aligned} & N=M g \cos \theta+T \sin \theta \ \Rightarrow \quad N &=M g \cos \theta+m g \sin \theta \quad \text{...(iii)} \end{aligned}$ and $f+T \cos \theta=M g \sin \theta$ ...(iv) $\begin{aligned} \Rightarrow & f(1+\cos \theta)=M g \sin \theta\ \Rightarrow f &=\frac{M g \sin \theta}{(1+\cos \theta)} \Rightarrow m g=\frac{M g \sin \theta}{1+\cos \theta} \ & m=\frac{M \sin \theta}{1+\cos \theta} \quad \text{...(v)} \end{aligned}$ On substituting the values of $M$ and $\theta$ we get $m=3 \mathrm{~kg}$

In the arrangement shown in figure the cylinder of mass \(M=9 \mathrm{~kg}\) is at rest on an incline. The…

Solution