Question

A block of weight $W$ rests on a horizontal plane with which the angle of friction is $\lambda .$ A force P inclined at an angle to the plane is applied to the plane until it is one the point of moving. Find the value of $\theta$ for which the value of P will be least.

Answer 1

In the limiting case contact force $F$ is inclined at $\lambda$ to the normal. Only three forces act on the block.

Applying Lami's theorem, we get:
$\frac{P}{\sin ({180}^{\circ }-\lambda )}=\frac{W}{\sin ({90}^{\circ }-\theta +\lambda )}$
$\frac{F}{\sin ({90}^{\circ }+\theta )}$
or $\frac{P}{\sin \lambda }=\frac{w}{\cos (\theta -\lambda )}$
or $P=\frac{W}{\cos (\theta -\lambda )}$
P will be least when $\cos (\theta -\lambda )$ is greatest because $W$ and $\lambda$ are constant.
i.e., when $\cos (\theta -)=1$ and $\theta \lambda =0^{\circ }$
or $\theta =\lambda$