# Homework Help: Using a prony brake to determine motor torque

1. Oct 10, 2014

### dotmatrix

1. The problem statement, all variables and given/known data
Hey guys, not much of a physicist. But, I am fairly certain I am right.
We are using a "prony brake" at school to calculate torque in a motor. To me the amount of force supplied should equal the amount of force the motor can turn ie: when the motor stops the force applied on the "prony brake" is 33 oz.
Now my instructor has given us an equation that says that the radius of the shaft of the motor has something to do with the calculation of the force applied.
My problem is a) we did not apply the force to the shaft but instead a coupler which was roughly 4" in diameter (as opposed to a one inch shaft)
b) the force applied was by a belt which squeezed half of the circumference of the shaft
c) it varied with friction (we poured water on the belt to keep it cool and it reduced the amount of "force"}

So I know enough to say that the equation isn't quite right. And I would like to say that the force that I am exerting on the motor is equivalent to the torque (as it is a frictional force that opposes the rotation) but I don't actually know anything for sure. I think that it shouldn't have anything to do with mass times the radius of the shaft or whatever.

2. Relevant equations

3. The attempt at a solution

2. Oct 10, 2014

### Staff: Mentor

Hi dotmatrix. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

Torque = force x radial distance (r being measured perpendicular to the line of action of F)

Your instructor is correct. (Edit: thread title changed from "Help prove prof wrong!")

If the coupler expanded the shaft radius to a large enough dimension, you would be able to stall the motor by lightly pressing on the shaft with just one finger (and non-destructively, at that!).

F alone is not an adequate measure of a motor's "strength".

Last edited by a moderator: May 7, 2017
3. Oct 10, 2014

### dotmatrix

The radial distance of the shaft is 1'' the force applied at the coupler is at a radial distance of 4'' (this is where we are applying it)
Furthermore the belt that is applying the force is not perpendicular to the coupler it in fact encompasses the lower 180 degrees of the rotating coupler.

4. Oct 10, 2014

### dotmatrix

And P in=P out so the motors "strength" to run something plus the "strength" to run its self for all intensive purposes is fairly accurate representation.

5. Oct 10, 2014

### Staff: Mentor

At any point of contact, belt friction is perpendicular to the radial distance. It is the torque produced by this frictional force which opposes the shaft's rotation.

6. Oct 11, 2014

### OmCheeto

When you said; "So I know enough to say that the equation isn't quite right.", to which equation were you referring to?

According to wiki, the de Prony brake measures the force on a tension belt, from which you can derive both torque and power, knowing other variables.

7. Oct 11, 2014

### dotmatrix

Well that's what I thought?!?
But the force is exerted on a coupling which is much larger than the shaft (4" vs 1") The shaft of the motor is what they're using for the "calculation"
So therefore the force exerted is across a belt which rests upon 180 degrees of the lower circumference of the coupling.
Would this not be a measurement of frictional force in direct opposition to torque?
*Note electrician, not physicist

8. Oct 11, 2014

### OmCheeto

I'm fairly certain that if you had done the measurement with the belt around the 1" shaft, the force would have been much higher than the 33oz you measured from the 4" coupling.

I will leave that to you as a homework assignment, to determine what the force would be on the 1" shaft.

9. Oct 11, 2014

### Staff: Mentor

Last edited by a moderator: May 7, 2017
10. Oct 11, 2014

### CWatters

Any chance of a diagram or photo? Is it similar to...

http://en.wikipedia.org/wiki/De_Prony_brake

11. Oct 11, 2014

### dotmatrix

More or less, take the diagram, but instead of using weight, your actually tightening the device and measuring the output with a spring scale. Does that make sense?
Rotary power (in newton-meters per second, N·m/s) = 2π × lever length (in meters, m) × rotational speed (in revolutions per second) × measured force (in newtons, N)
Rotary Power being equal to torque

12. Oct 11, 2014

### dotmatrix

i just read that description, feeling a little slow at this hour

13. Oct 11, 2014

### Staff: Mentor

Last edited by a moderator: May 7, 2017