# Homework Help: Applied friction point between two disks

1. Dec 21, 2011

### AxioWolf

1. The problem statement, all variables and given/known data
Okay, so I have a homework-like question that I need help with. It seems so simple, but I'm stumped and can't find anything online. A similar real world example if my description below isn't clear would be a clutch mechanism where one disk is pressed up against another and the friction prevents the disks from slipping.

I have two disks that have a radius, r.
The disks are assumed to be perfectly rigid.
One disk is fixed and the other can rotate about an axis centered in the fixed disk.
The non-fixed disk has a torque, T, applied of say 200 in-lbs
There is static friction, $$μ_s$$ between the two disks
These disks are sandwiched together with a force, N, that is applied along the axis of rotation (nut and bolt basically).

Goal: Plot N vs $$μ_s$$ with varying r values, which will result in multiple curves. This will be at the equilibrium point when the floating disk is just about to rotate.

2. Relevant equations
$$f_s = μ_s*N$$
$$T = F*x$$
...?

3. The attempt at a solution

The set-up seems nearly complete except for the missing piece of the distance, x, away from the center of the disk that the friction force acts. The distance is needed to calculate the resulting moment that resists the applied moment.

$$f_s = μ_s*N$$
$$T = f_s*x$$

$$T = μ_s*N*x$$

Thus, what is x?:
x = r/2
x = r
or maybe the circle that divides the disk's contact area equally in half... (see image)
x = r*√(2)/2

My suspicion is that I need to integrate to solve this, but I'm not sure how to set up the integration.

AxioWolf

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Last edited: Dec 21, 2011
2. Dec 21, 2011

### Delphi51

Welcome to PF!
What a weird question! Are you sure it says "N vs μs"? If so, that is crazy. μs is a fixed number so there would be only one point on the graph. It must be a mistake somewhere. Torque vs N isn't very meaningful, either though it would vary with r. Your instinct to integrate the torque over x from 0 to r would give you the total torque, which is a very useful thing to know.

It does look like you paraphrased the question. Can you give us the exact wording?

3. Dec 22, 2011

### AxioWolf

Thanks for the welcome!
Yes, it sounds weird, but it's only a homework-like question. This is actually for a project I'm working on, not a homework problem for a class. According to the guidelines, this was the best forum to place it in. I just tried to list out all the knowns, unknowns and assumptions that are relevant.

The end equation will be:
T = 200 in-lbs
$$μ_s = T/(N*x)$$

I'm trying to determine suitable materials for varying applied normal forces, so with different normal forces, the μs will change to keep the final equation balanced. In addition, each radius will alter the curves on the graph and I expect it'll bring the μs values lower as the radius increase.
In the end, I will select a few different radii, plug in various normal forces and solve for μs. Then, I'll have a nice collection of curves.

So, I could use some help with creating the integral to solve for the total torque. My initial stab at it is
sum of the moments = 0
$$T - T_f = 0$$
$$T_f = ∫^r_0 μ_s*N*dx$$
$$T_f = μ_s*N*∫^r_0 dx$$
but this is merely x = r. I'm not confident this is correct.
Hmm, does the Normal force, N, depend on x and change as it moves further away from the center even though it's a completely rigid body? Yeah, I'm just not sure.
Thanks!
AxioWolf

Last edited: Dec 22, 2011
4. Dec 22, 2011

### AxioWolf

I think I should instead look at the force as a distribution instead of a simple point source... duh!
Okay, so the distributed force is over the area:

$$N_{total} = \frac{N}{\pi r^2}$$
Thus, my new integral is:
$$T_f = ∫^r_0 \frac{μ_s*N*dx}{\pi x^2}$$
$$T_f = \frac{μ_s*N}{\pi }∫^r_0 \frac{dx}{x^2}$$
$$T_f = \frac{μ_s*N}{\pi }*\frac{1}{-1*x} |(r,0)$$
hmm... not looking ideal due to a zero in the denominator:
$$T_f = \frac{μ_s*N}{\pi }*(\frac{1}{0}-\frac{1}{r})$$
Thus, the integral is unbounded and I must perform a limit on the integral instead as 's' approaches 0. This will result in the value not converging, & approach infinity.

Hmm, the results are actually counter-intuitive because it basically is saying that as the radius increases... there will need to be a larger μs or N to balance the equation.
Do I need to instead somehow adapt the "first moment of the area" equation?
$$M_y = ∫_A x*dA$$
This doesn't seem right either.

Hmm, after much thought I'm stuck again.
- AxioWolf

btw, I like the Latex stuff! Takes a little getting used to, but pretty cool.

Last edited: Dec 22, 2011
5. Dec 22, 2011

### PhanthomJay

None of these answers are correct.
Setting the calculus aside as a last resort, keep it simple by recognizing that the force distribution is triangular varying from zero at the center of the disk to to a maximum radially outward at its circumferential edge. The torque can be represented by a couple whose equal and opposite forces act at the centroids of the triangles. What's the centroid location of a triangle?

6. Dec 23, 2011

### Delphi51

Your idea that the force F is spread equally over the area πr² looks good. But the dimensions are not correct on your integral. The dF on a ring area dA at radius x would be (F/πr²)*dA = (F/πr²)*2πx*dx. I'm sure you can work out the element of torque for that area and integrate it.

Last edited: Dec 23, 2011