# Reference point for torque

1. Feb 22, 2015

### henry3369

1. The problem statement, all variables and given/known data
A uniform plank of length L = 6.0 m and mass M = 90 kg rests on sawhorses separated by D = 1.5m and equidistant from the center of the plank. Cousin Throckmorton wants to stand on the right-hand end of the plank. If the plank is to remain at rest, how massive can Throckmorton be?

http://imgur.com/AVrdxIJ [Broken]

2. Relevant equations
net torque = 0

3. The attempt at a solution
In my book it says that for problems involving torque, the reference point is arbitrary. I don't understand how this is possible.

For example: when I calculate the mass of the person in this problem using the right sawhorse as the center of rotation, I find the mass to be 30 kg. But if I take the net torque about the center of mass (which is at the center of the plank), then the torque on the left will be zero with only one torque on the right (the person).
So mg x 3 = 0, which means the mass of the person has to be zero, and the normal force from the saw horse has to be zero because the mass is concentrated to the left of the it.

Last edited by a moderator: May 7, 2017
2. Feb 22, 2015

### Nathanael

Aren't you forgetting the torque from the normal force between the plank and the sawhorse?

The advantage of taking the torque around the sawhorse instead of the center of mass is that you don't need to worry about (one of) the normal force(s) because the torque from it is zero.

What did you take the normal force from the left sawhorse to be?

3. Feb 22, 2015

### haruspex

About the mass centre,there's a torque from the right hand support.
Yes, you can take any reference point,but often (as here) a smart choice avoids involving an unknownf force that you don't care about. Having taken moments about the mass centre, you need to use linear statics to get a second equation.

Last edited by a moderator: May 7, 2017
4. Feb 22, 2015

### henry3369

If the mass is concentrated at the center of the plank, wouldn't there be no normal force at the saw horse because their is no mass there?

5. Feb 22, 2015

### henry3369

When I calculated the net torque about the right saw horse, I ignored the normal force on the left saw horse and still got the correct answer because I assumed their is no normal force there.

6. Feb 22, 2015

### haruspex

That would be right, since you are interested in the case where it teeters on the right sawhorse as fulcrum. But there is a normal force at the right sawhorse.

7. Feb 22, 2015

### Nathanael

The mass of the plank is not actually located in the center. It's just that, if you calculate the force of gravity on every small piece of the planck and add it all together, it will be (mathematically) equivalent to treating the full force of gravity as if it acts on the center of mass.

If there was no normal force at the saw horse, wouldn't there be a net downwards force on the plank?

8. Feb 22, 2015

### haruspex

As Nathanael wrote, it isn't really concentrated at the centre of the plank. But even if it were the result would be the same. There has to be a normal force upward from one or both supports to balance the gravity on the plank and cousin.

9. Feb 22, 2015

### henry3369

So if I wanted to calculate the normal force on the right saw horse, would I use the entire mass of the plank? So N = mg = 90 * 9.8?

10. Feb 22, 2015

### henry3369

Also would I then ignore the normal force on the left saw horse if I used the entire mass of the plank to calculate the normal force on the right saw horse?

11. Feb 22, 2015

### haruspex

12. Feb 22, 2015

### haruspex

You are interested in the limiting case, where the plank is in danger of tipping to the right. What will the normal force at the left hand sawhorse be at that point?

13. Feb 22, 2015

### henry3369

Zero?

14. Feb 22, 2015

### henry3369

So does the fact that the left saw horse is there change anything? Wouldn't the result be the same if there were only one saw horse?

15. Feb 22, 2015

### haruspex

Same result, but rather harder to describe the problem. Besides, part of the test is that you can work out that the normal force will be zero.