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 2.20 m/s2 northward and (b) when it accelerates at 3.40 m/s2 southward.

Solution 85P Mass of the case is = 40.0 kg Coefficient of static friction = 0.30 2 Static friction force is F s 0.30 × 40.0 kg × 9.8 m/s F =s117.6 N (a) When the truck accelerates northward at 2.20 m/s , the force on the case is = 40 kg × 2.20 m/s = 88.0 N This force lower than the static friction value of 117.6 N. Therefore, at the given 2 acceleration of 2.20 m/s the force of friction is 88.0 N. The direction of this friction force will be northward. (b) The coefficient of kinetic friction is k= 0.20 When the truck accelerates at 3.40 m/s southward, the force on the case is = 40.0 kg × 3.40 m/s = 136 N This value exceeds the static friction F = s17.6 N value. Therefore, the case will be in motion. So, the friction force on the case can be calculated as shown below. F k = k 40.0 kg × 9.8 m/s , is tke coefficient of kinetic friction. F = 0.20 × 392 N k F = 78.4 N k 2 Therefore, the friction force of the case when it accelerates at 3.40 m/s is 78.4 N. Since it is in motion now, the acceleration will be southward.