Solution for Question 14
Consider the composite body (Pipe of mass $M$ + Rod of mass $m$).
For the system to be in stable equilibrium on the rough inclined plane, the net torque about the combined center of mass $G$ must be zero.
The forces acting on the system are:
1. Total weight $(M+m)g$ acting vertically downwards through $G$.
2. The contact force from the ground at point $P$, which resolves into Normal force $N$ and Friction $f$.
For equilibrium, the line of action of the resultant contact force $R$ (vector sum of $N$ and $f$) must pass exactly through the center of mass $G$ to balance the torque produced by gravity.
From the geometry of the forces shown in the diagram:
$$ \tan \theta = \frac{f}{N} $$
(Since $f$ balances the component of weight along the slope and $N$ balances the component perpendicular to it, the resultant $R$ is vertical).
The maximum angle $\theta_{max}$ corresponds to the limiting case where the center of mass $G$ is horizontally displaced as far as possible from the center $O$ while remaining vertically above the contact point $P$.
Let $r$ be the radius of the pipe. The distance of the combined center of mass $G$ from the geometric center $O$ is:
$$ d_{OG} = \frac{m \cdot r + M \cdot 0}{M + m} = \frac{mr}{M + m} $$
In the right-angled triangle formed by the vertical line through $P$ and $G$, and the radius $OP$:
$$ \sin \theta_{max} = \frac{d_{OG}}{r} = \frac{m}{M + m} $$
However, relating this to the coefficient of friction $\mu$:
$$ \mu = \tan \theta_{max} $$
Using the trigonometric identity $\tan \theta = \frac{\sin \theta}{\sqrt{1 – \sin^2 \theta}}$: $$ \mu = \frac{\frac{m}{M+m}}{\sqrt{1 – \left(\frac{m}{M+m}\right)^2}} $$ $$ \mu = \frac{m}{\sqrt{(M+m)^2 – m^2}} $$ Simplifying the denominator: $$ (M+m)^2 – m^2 = M^2 + 2Mm + m^2 – m^2 = M(M + 2m) $$ Thus: $$ \mu \le \frac{m}{\sqrt{M(M + 2m)}} $$