A 34.0- packing case is initially at rest on the floor of a 1650- pickup truck. The coefficient of static friction between the case and the truck floor is 0.30, and the coefficient of kinetic friction is 0.20. Before each acceleration given below, the truck is traveling due north at constant speed.
a) Find the magnitude of the friction force acting on the case when the truck accelerates at 2.23 northward.
b) Find the magnitude of the friction force acting on the case when it accelerates at 4.72 southward.
a) Find the magnitude of the friction force acting on the case when the truck accelerates at 2.23 northward.
b) Find the magnitude of the friction force acting on the case when it accelerates at 4.72 southward.
-
weights are in kg?
all accelerations in m/s^2?
a) 2.23m/s^2*34kg = 75.82N
Max static friction force = 9.81*34*0.3 = 100.06N
Therefore, it does not slip and the force is 74.82N
b) 4.72m/s^2*34kg = 160.48N > 74.82N, therefore it slips and must use kinetic friction
kinetic friction = 9.81*34*0.2 = 66.71N
all accelerations in m/s^2?
a) 2.23m/s^2*34kg = 75.82N
Max static friction force = 9.81*34*0.3 = 100.06N
Therefore, it does not slip and the force is 74.82N
b) 4.72m/s^2*34kg = 160.48N > 74.82N, therefore it slips and must use kinetic friction
kinetic friction = 9.81*34*0.2 = 66.71N