Physics - NLM Question with Solution | TestHub

The various forces acting on the block are as shown in the figure.
When $\theta =45^{\circ },\sin \theta =\cos \theta$ The block will remain stationary and the frictional force is zero. When $\theta >45^{\circ },\sin \theta >\cos \theta$ Therefore a frictional force acts towards $Q$. When $\theta <45^{\circ },\cos \theta >\sin \theta$ Therefore a frictional force acts towards $P$.