A 40.0-kg packing case is initially at rest on the floor of a 1500-kg 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. Find the magnitude and direction of the friction force acting on the case (a) when the truck accelerates at 2.20m/s22.20m/s 2 northward and
(b) when it accelerates at 3.40m/s23.40m/s 2 southward.

Before providing an answer to the question, the values for acceleration given in questions A and B were written twice. So correction would go like this: For (a) when the truck accelerates at 2.20m/s2 northward, and for (b) when it accelerates at 3.40m/s2 southward.

The

(a) 88N, northward.

(b) 78.4‬, southward.

Explanation:

(a) Maximum frictional force acting on the packing case= (coefficient of static friction) X (Normal force)

Normal force = mass X acceleration due to gravity

Maximum static frictional force acting on the packing case = (coefficient of static friction) X (mass of packing case  X acceleration due to gravity)

Maximum static frictional force = (0.30) X (40.0-kg) X (9.8m/s 2) = 117.6N

While Reaction force acting on the packing case = (mass of packing case) x (acceleration generated by the pickup truck)

Reaction force acting on the case = (40.0-kg) X (2.20m/s2) = 88N

With these values, one can conclude that the packing case is at rest since the reaction force of the case acting in the opposite direction is lesser than the frictional force. Making the magnitude and direction of the friction force acting on the case still move northward, and the static frictional force acting on equals the reaction force.

The answer is 88N, northward.

(b)  Here too, we need to still compare the reaction force with the value of the already determined Maximum static frictional force (117.6N) above. This is necessary to know the frictional force between the pickup truck"s floor and the packing case.

Reaction force acting on the case when acceleration is 3.40m/s2 = (40.0-kg) X (3.40m/s2) = 136‬ N

We can conclude that the reaction force (136‬ N) is greater than the maximum static frictional force (117.6N), suggesting that the packing case is in motion and the frictional force is no longer static.

This means a kinetic force is now acting on the pickup truck"s floor causing the packing case to also move. This kinetic force can be calculated as:

kinetic force = (coefficient of kinetic friction) X  (mass of packing case  X acceleration due to gravity)

= (0,20) X (40.0-kg) X (9.8m/s 2) = 78.4‬N

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