A block of mass M lies stationary on a rough plane inclined at an angle x to the horizontal. Find a general expression relating the coeffecient of friction between the block and the plane and the angle x. At what angle does the box begin to slide?

The first step to any problem is to draw a suitable diagram. Here we draw a free body diagram, marking the angle x, the friction force F, normal reaction force N and weight of the block Mg. "Stationary" implies equilibrium, allowing us to resolve parallel and perpendicular forces, and equate their sum to zero, givingParallel components: Mgsin(x) - F = 0 ---- (1)Perpendicular components: Mgcos(X) - N = 0. ---- (2)Substituting (1) and (2) in to the familiar expression F = uN where u is the coefficient of friction between the block and the plane, and rearranging gives the resultu = (Mgsin(x)) / (Mgcos(x)) = tan(x),which is the general expression relating u to x. The block clearly begins to slide as u decreases, or x increases, leading to the inequalityx > tan^-1(u).