Question

A block of mass m is at rest on an inclined plane that makes an angle θ with the horizontal. The force of static friction fs acting on the block must be such

that

A) fs > mg.

B) fs > mg cosθ.

C) fs > mg sinθ.

D) fs = mg cosθ.

E) fs = mg sinθ.

Answer 1

The force of static friction must be greater than the component of the gravitational force parallel to the incline (mg sinθ) to prevent the block from sliding down.

Step-by-step explanation:

The question asks which condition the force of static friction fs must satisfy to prevent a block of mass m from sliding down an inclined plane that makes an angle θ with the horizontal. The correct answer is C) fs > mg sinθ. This is because the component of the weight acting down the slope (parallel to the incline) is mg sinθ and for the block to remain at rest, static friction must overcome this component, not the entire weight of the block.

The subject of this problem is static friction which opposes the component of the gravitational force that acts parallel to the surface of the incline. The maximum static friction that can occur is given by fs = μs N, where μs is the coefficient of static friction and N is the normal force, which equals mg cosθ. Thus, to prevent sliding, static friction must be equal to or greater than the parallel component of gravity.