A 150-N block,on a 25° incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.52.The force due to friction is:
A) 0 N.
B) 136 N.
C) 63 N.
D) 150 N.
E) 96 N.

Question

A 150-N block,on a 25° incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.52.The force due to friction is:
A) 0 N.
B) 136 N.
C) 63 N.
D) 150 N.
E) ​96 N.

Quizplus · Accepted Answer

First, draw a diagram of the situation to help visualize the forces acting on the block.

The force of gravity (weight) can be split into components parallel and perpendicular to the incline.

F_gravity_parallel = m*g*sin(theta) 
= 150 N * sin(25°) 
= 63 N

This is the force that would cause the block to slide down the incline if there were no friction.

The force of static friction must be equal and opposite to this force in order to keep the block from moving.

F_friction = F_gravity_parallel 
= 63 N

Using the given coefficient of friction:

F_friction = friction_coefficient * F_normal 
where F_normal is the normal force acting perpendicular to the incline.

Since the block is not accelerating vertically (it's being held motionless), the normal force must be equal and opposite to the perpendicular component of the weight:

F_normal = m*g*cos(theta) 
= 150 N * cos(25°) 
= 136 N

Plugging this back into the friction equation:

F_friction = friction_coefficient * F_normal 
= 0.52 * 136 N 
= 70.72 N

This is larger than the force required to keep the block motionless (63 N), which means the block is not actually held in place by the maximum possible static friction force. The coefficient of friction tells us how large the friction force can be before the block starts to slide, not how large it must be to keep the block stationary. Therefore, the force due to friction is 63 N, which is Answer C.

A 150-N Block,on a 25° Incline,is Being Held Motionless by Friction.The