A sphere of radius $R$ is supported by a rope attached to the wall. The rope makes an angle $\theta=45^{\circ}$ with respect to the wall. The point where the rope is attached to the wall is at a distance of $3 R / 2$ from the point where the sphere touches the wall. Find the minimum…

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A sphere of radius $R$ is supported by a rope attached to the wall. The rope makes an angle $\theta=45^{\circ}$ with respect to the wall. The point where the rope is attached to the wall is at a distance of $3 R / 2$ from the point where the sphere touches the wall. Find the minimum coefficient of friction $(\mu)$ between the wall and the sphere for this equilibrium to be possible. Use the intregral part only.

Aicharya · Accepted Answer

Net torque about $A=0$ $\begin{array}{l} f_{s} \frac{3 R}{2}=M g \cdot \frac{R}{2} \ \therefore f_{s}=\frac{M g}{3} \ {\left[\therefore P A=\frac{3 R}{2} \tan 45^{\circ}=\frac{3 R}{2}\right]} \quad \text{...(i)} \end{array}$ Horizontal equilibrium: $N=\frac{T}{\sqrt{2}}$ ...(ii) Vertical equilibrium: $\frac{T}{\sqrt{2}}+f_{s}=M g$ $T=\frac{2 \sqrt{2}}{3} M g$ ...(iii) Put in (ii) $N=\frac{2}{3} M g$ $f_{s} \leq \mu_{s} N \Rightarrow \frac{M g}{3} \leq \mu_{s} \cdot \frac{2}{3} M g \Rightarrow \frac{1}{2} \leq \mu_{s}$ Hence, $\mu_{\min }=0.50$

A sphere of radius \(R\) is supported by a rope attached to the wall. The rope makes an angle…

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