A 12-kg mass is placed on a 25.0 $\degree$ 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

The frictional force is equal to the component of the gravitational force parallel to the incline, which is m*g*sin(theta). For a 12-kg mass on a 25-degree incline, this is 12*9.8*sin(25) = 49.7 N. Since the object is not sliding, the static frictional force is equal to this value.

A 12-Kg Mass Is Placed on a 25 °\degree° Incline, and Friction Keeps It from Sliding

A 12-Kg Mass Is Placed on a 25 $\degree$ Incline, and Friction Keeps It from Sliding