Mechanics Problem 2: Friction of Box on an Accelerating Truck

A flatbed truck carries a load on its flat bed as shown in the picture above. The load is tied to the back of the truck cab with a steel cable to prevent it from sliding. Schematically it would look like below.

If we identify all forces acting on the load, then the free body diagram, whose purpose is to show these forces, would look like below.

This problem asks what should be the maximum acceleration of the truck as it accelerates from rest if we want to prevent the load from sliding, given that the load is being held by the friction force μ_s N, where N is the normal force and is equal to the load's weight m g, and the cable tension T.

1. Let us scrutinize this free body diagram. The first question is why the static friction force points to the right. The answer is conservation of momentum. Momentum p is equal to m v, where m is the object's mass and v is its velocity. The initial momentum is zero since it is initially at rest (v = 0). As the truck accelerates, it wants to conserve the zero momentum,

0 = m_truck v_truck + m_load v_load.

Since the truck velocity v_truck points to the right due to the acceleration a to the right (see the above diagram), the load tends to move to the left, i.e., v_load points to the left. Friction opposes the motion, i.e., the velocity; therefore, the friction force points to the right. The cable tension T also points to the right since it serves to prevent the load from sliding.

2. The sliding motion of the load represents a relative motion the load has with respect to the flat bed. The load accelerates with respect to the truck which is also accelerating. We say that the load moves in a noninertial frame. The Newton's second law, F = m a, does not work in a noninertial frame. F = m a has to be anchored in an inertial frame. The inertial frame is obtained when we regard the acceleration of the load is equal to the sum of the acceleration of the sliding motion and the acceleration of the truck,

a_load = a_truck + a_sliding.

Thus, the Newton's second law now reads

T + μ_s m g = m (a_truck + a_sliding).

3. We do not want the load to slide; thus, a_sliding = 0. The maximum acceleration is equal to

a_truck = T/m + μ_s g,

where T is maximum when it is equal to the tensile strength of the steel cable, which is about 400 MPa. For a cable with a cross section area of 0.25 cm², we get a maximum T of 10,000 N. Typical value of static friction coefficient is about 0.5, so that μ_s g = 4.9 m/s². For a 1000 kg load, we get T/m = 10 m/s². Hence, the maximum acceleration for these data is 14.9 m/s². 100 km/h = 27.8 m/s, so 14.9 m/s² is thus quite a high acceleration. But for a really heavy load, say a 10,000 kg load, the cable can now only bear 1 m/s² acceleration and in this case, the friction force becomes a lot more important. As a rule, therefore, the heavier the load is, the more it has to rely on friction to prevent the load from sliding.