Question

a container of mass 200 kg rests of the back of an open truck. if the truck accelerates at 1.5 m/sec2, what is the minimum coefficient of static friction between the container and the bed of the truck required to prevent the container from the sliding off the back of the truck.

Answer 1

Fsf = ma, Fsf = k * m * g.
k * m * g = m * a
k * g = a
k = a / g
k =
1.5 / 9.8 = 0.1531