A block of mass (m) is placed on a rough slope inclined at an angle (a) to the horizontal, find an expression in terms of (a) for the smallest coefficient of friction (x), such that the block does not fall down the slope.

The first thing to do as with any mechanics question is draw a force diagram with the weight (mg), the friction (F) and the normal reaction (R). As we are finding the smallest coefficient of friction possible, the block must be in limiting equilibrium, so the friction (F) = the coefficient of friction (x) multiplied by the normal reaction (R), so F = xR. Now we can resolve forces parallel and perpendicular to the slope to obtain two simultaeneous equations. Resolving perpendicular, we find that R = mgcos(a), and resolving parallel we find that xR = mgsin(a). By substituting the first equation into the second, we get xmgcos(a) = mgsin(a). By cancelling out the "mg" term from both sides, and dividing through by cos(a), we end up with x = sin(a) / cos(a), which of course leads to our final answer that x = tan(a).