A 12-kg mass is placed on a 25.0° incline,and friction keeps it from sliding.The coefficient of static friction in this case is 0.580,and the coefficient of sliding friction is 0.520.What is the frictional force in this situation?
A) 55.4 N
B) 107 N
C) 49.7 N
D) 61.2 N

Question

Quizplus · Accepted Answer

Since the mass is not sliding, the frictional force is equal and opposite to the component of the weight that is parallel to the incline. The weight parallel to the incline can be found using trigonometry:

weight parallel = mg sin(25.0°) = (12 kg)(9.81 m/s^2)(sin 25.0°) = 51.0 N

The static frictional force will be equal and opposite to this, up to a maximum value of μsN, where N is the normal force. The normal force is equal to the weight component perpendicular to the incline:

N = mg cos(25.0°) = (12 kg)(9.81 m/s^2)(cos 25.0°) = 109.9 N

Thus, the maximum possible static frictional force is:

μsN = (0.580)(109.9 N) = 63.7 N

Since the weight parallel is less than this maximum, the actual static frictional force will be equal and opposite to the weight parallel:

frictional force = weight parallel = 51.0 N

Note that the coefficient of sliding friction is not relevant here since the mass is not sliding.

A 12-Kg Mass Is Placed on a 25