# Statics Question about Friction on an Incline

So for an object, say a block, on an incline, there hold three cases in which minimum force $$P$$ is dependent:

a.) To start the block moving upwards.This is the case where static friction is at the maximum, i.e., $$f_s = (μ_s)(N).$$ b.) To keep the object moving upwards. The case where kinetic friction is in charge, i.e., $$f_k = (μ_k)(N)$$ c.) To keep the object from moving downwards.

This last one is what I don't get. I think this is the minimum static friction required. But I don't know how to obtain this in order to get the minimum force $$P$$.

Take note that the required force $$P$$ here is way lower than the force required to keep the object moving upwards (case b), which led me to think that the minimum static friction is what's in charge here.

• To better understand what you are dealing with draw free body diagrams of the block for a and c Feb 8, 2020 at 10:43

c.) To keep the object from moving downwards.

This last one is what I don't get. I think this is the minimum static friction required.

It' not clear what you are asking regarding case (c), but no upward force $$P$$ would be needed to prevent the block from moving downwards as long as (1) the upward static friction force equals the downward force of gravity parallel to the plane and (2) the downward force of gravity parallel to the plane is less than the maximum possible static friction force.

So downward motion will not occur if

$$f_{s}$$ = $$mg$$ sin θ

and

$$mg$$ sin θ < $$f_{max}$$ = $$μ_s$$N

If $$P$$ is applied down the plane then $$P$$ works with gravity to oppose the upward static friction force. Therefore, for impending motion down the plane due to applied $$P$$ down the plane we have

$$P$$ + $$mg$$ sin θ = $$μ_s$$N

or

$$P$$ = $$μ_s$$N - $$mg$$ sin θ

Regarding (a), in order for impending upward motion to occur, the upward pulling force $$P$$ has to equal the downward maximum static friction force plus the downward force of gravity, or

$$P$$ =$$μ_s$$N + $$mg$$ sin θ

In short, less pulling force is needed to pull the block down because gravity works with the pulling force to overcome the maximum static friction force that opposes it, whereas more pulling force is needed to pull the block up because the upward pulling force has to overcome both the downward maximum static friction force and the downward force of gravity that oppose it.

Finally, keep in mind that the static friction force only exists in opposition to an externally applied force, and when an externally applied force is applied the magnitude of the static friction force will match the applied force, up until the applied force equals the maximum possible static friction force. So the static friction force is a variable up until the maximum possible static friction force occurs.

Hope this helps.