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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., $dx$

enter image description here

Then, I inserted a hollow cylinder over this stand, which perfectly fits it as shown.

enter image description here

The figure would look like this:

enter image description here

Note---

CROSS-SECTIONAL VIEW

(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?

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3 Answers 3

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No.

You've discovered the Class 2 Lever, which places the load between the input force and the fulcrum. You have correctly calculated that the friction force is 4 times the input force. The reason for this is because your machine is a lever which applies 4 times the input force at a point on the outer ring to the corresponding point on the inner ring.

Your fulcrum is the fixed axle on which your outer cylinder rotates. (Otherwise the whole cylinder would accelerate away at $a=F/m$ by Newton's 2nd, and you've stipulated that it doesn't move when you apply the force.)

Your load is a friction force equal to the applied (tangent) force at the radius where the friction force is applied.

For any lever:

Let $F_a$ be the input force, applied a displacement from the fulcrum $a$.

Let $F_b$ be the resultant force at displacement $b$.

then:

$\dfrac{F_b}{F_a} = \dfrac{a}b \to F_b = \frac{a}{b} F_a$

Image from wikipedia: enter image description here

(Incidentally, your machine is pretty close to how a drum brake works, except with the friction force as the input force and inertia as the load.)

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First, there is an important qualification to "the value of friction force can never be greater than the applied force". This refers to static friction. Indeed, you pose a scenario where "the cylinder does not move nor rotate". It should be emphasized that the relationship is limited to this static case, where the friction force is equal and opposite to the applied tangential force, up to the maximum friction force given by the static friction coefficient times the normal force. (Kinetic friction, on the other hand, is opposite to the relative velocity of the surfaces and is independent of the applied tangential force, which could be zero.)

Now, to the rest of your question, the "applied force" means the external tangential force acting at the surfaces in contact, where friction is occurring. It is irrelevant whether a larger or smaller force is being applied elsewhere as part of a mechanical equilibrium, as in your cylinder scenario. The upshot is that static friction will prevent relative motion by fully counteracting the applied force at the surfaces in contact, as long as this applied force does not exceed the static friction limit.

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To answer the question in the title, the friction force can be larger than the normal force that is producing it. There is no restriction. Coefficients of friction larger than 1 are not common in intro physics problems but they are encountered in practice, especially when people want them to be high. Like ruber rubbing on rubber or rough wood.

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    $\begingroup$ I think OP means that the static friction force cannot be greater than the externally applied tangential force (not normal force). In other words, friction can prevent sliding, but cannot cause sliding in the reverse direction. $\endgroup$
    – nanoman
    Jun 18 at 4:31

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