Question

A 40-N crate rests on a rough horizontal floor. A 12-N horizontal force is then applied to it. If
the coefficients of friction are μs = 0.5 and μk = 0.4, the magnitude of the frictional force on
the crate is:

Answer 1

The static friction

$$F_s=\mu_sN=0.5\times 40\:\rm N=20\: N$$

The kinetic friction

$$F_k=\mu_kN=0.4\times 40\:\rm N=16\: N$$

Since the horizontal force applied to the crate is less then static friction $(F<F_s),$ so the crate is at rest.