Why Is Frictional Force Less Than Weight on an Inclined Plane?

[StephenDoty]

If a block is at rest on an inclined plane with a rough surface, what is the frictional force compared to the weight of the block?

Well if the block is at rest that means that the frictional force is the static frictional force, and since the block is not moving shouldn't the static frictional force be greater than the weight of the block?

According to my professor the correct answer is that the frictional force acting on the block is less than the weight of the block. WHY?

Even if the force is kinetic frictional force, since the block is not moving shouldn't the frictional force be equal to the weight of the block?

Thank you.

Stephen

 

[Doc Al]

If the block is at rest--in equilibrium--what must be the net force on it? Analyze the forces on the block that act parallel to the incline. (One of those forces is friction. What's the other?)

 

[StephenDoty]

weight equaling mgsin(theta)

so using Newton's laws shouldn't the fictional force equal the weight?

Why is the correct answer that the fictional force is less than the weight?

 

[Doc Al]

weight equaling mgsin(theta)

That's the component of the weight parallel to the incline. The weight itself is mg.

so using Newton's laws shouldn't the fictional force equal the weight?

The net force must be zero, thus the friction force must equal that component of the weight.

Why is the correct answer that the fictional force is less than the weight?

If friction = mg sin(theta), how must it compare to the full weight (mg)?

 

[rsala]

since it is at rest, and static frictional force varies, the weight downwards only the inclined plane equals the frictional force, and will remain to equal it, until the maximum static friction is reached then it jerks loose