Q.Define the angle of repose. By considering a block on the verge of sliding down a rough inclined plane under gravity alone, show that the angle of repose is numerically equal to the angle of friction.

Consider a block of mass $m$ resting on an incline tilted at the critical angle $\theta$, i.e. the angle of repose -- the steepest tilt at which the block, relying on friction alone, is just on the verge of sliding down under its own weight.\n\nResolving the weight $mg$ along and perpendicular to the incline: the component along the incline (tending to slide the block down) is $mg\sin\theta$, and the component perpendicular to the incline (balanced by the normal reaction $N$) is $mg\cos\theta$, so $N = mg\cos\theta$.\n\nAt this critical angle, the block is on the verge of sliding, so friction is acting at its limiting value, $f_{s(\text{max})} = \mu_s N$, directed up the slope, exactly balancing the down-slope component of gravity: $mg\sin\theta = \mu_s N = \mu_s\, mg\cos\theta$ Dividing both sides by $mg\cos\theta$: $\tan\theta = \mu_s$\n\nSeparately, the angle of friction, $\phi$, is defined as the angle the resultant of $N$ and $f_{s(\text{max})}$ makes w …

Resolve gravity along and perpendicular to the incline at the critical sliding angle, set the down-slope component equal to the limiting friction, and compare the re …

• Treating the angle of repose and the angle of friction as two unrelated angles that happen to be equal by coincidence, rather than showing both reduce algebraically to the same expression, $\tan^{-1}(\mu_s)$. …