My teacher taught me that the value of friction force can never be greater than the applied force. But recently, when I was studying rotational motion, I got a dilemma…
Suppose I made a stand (from cuboidal strips) which is fixed at one end as shown in the figure from a material whose coefficient of friction (with the surface of cylinder) is $u$; and height and breadth are very small, i.e., $dx$
Then, I inserted a hollow cylinder over this stand, which perfectly fits it as shown.
The figure would look like this:
Note---
(CROSS-SECTIONAL VIEW)
The inner radius of the cylinder is $R_1$ and outer radius is $R_2$ and $$R_2 =4R_1$$
Now suppose I applied a force $\boldsymbol{F}$ tangential to the surface of the cylinder (which would generate torque), but I observe that the cylinder does not move nor rotate. That means that net torque about Centre of mass is zero. (We know that friction will act at the opposite ends of the inner surface of the cylinder.) So, torque generated by applied force is cancelled by friction at the 2 diametrically opposite ends, right?
Let friction at the 2 ends be $F_1$ and $F_2$
Then I can write as $$F\times R_2-(F_1+F_2)R_1=0\implies F\times R_2=(F_1+F_2)R_1\\ \implies F\times 4R_1=(F_1+F_2)R_1\implies 4F=F_1+F_2$$
As you can see, the total friction generated is way more than (4 times) the value of the applied force $F$.
So, is it possible, then? Or the more right question would be: How is it possible? It really contradicts the facts stated by my teacher.
Can the value of friction force ever exceed value of applied force?