# Inclined plane, circular motion and the friction

Suppose we have a cylinder on an inclined plane of mass $m$ and radius $R$ moving without sliding (so that $\varepsilon = a/R$). Why is the friction $F$ causing the circular motion sometimes lower than $F' = f \cdot N$, where $N$ is a normal force and $f$ the friction coefficient? ($F \le f \cdot N$)

• this sounds like a 'homework' question. – theo Dec 7 '14 at 13:18
• But it is not. It simply wasn't sufficiently explained to us. – marmistrz Dec 7 '14 at 18:22
• Thats OK, either way! Do you have a diagram for it? What are the parameters "$\epsilon$" and "$a$" ? – theo Dec 7 '14 at 19:37
• You mean some drawing of the situation? As for parameters, well, as ordinarily - circular and linear acceleration. – marmistrz Dec 8 '14 at 6:07
• Usually an $\alpha$ and not an $\varepsilon$ is used to represent angular (circular) acceleration – Steeven Sep 7 '16 at 8:34

It's like that because the rotation without sliding is caused by static friction, not the kinetic one. Thus $T \le f N$. On the other hand, while if the cylinder is sliding, we take the kinetic friction.
Draw a free body diagram for the object. You will see that this friction $F$ is the only force that attacks at the edge. This is the only force that gives a torque.
The size of this force $F$ doesn't matter - as long as it exists, it will cause a torque.