How Does the Choice of Reference Point Affect Calculations in Torque Problems?

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Homework Help Overview

The discussion revolves around a torque problem involving a uniform plank supported by sawhorses, where participants explore how the choice of reference point affects calculations related to torque and normal forces. The original poster expresses confusion regarding the arbitrary nature of reference points in torque calculations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculating the mass of a person standing on the plank using different reference points, questioning the implications of ignoring certain forces, such as the normal force from the left sawhorse. There are attempts to clarify how torques are calculated about various points and the assumptions involved in those calculations.

Discussion Status

The discussion is active, with participants providing insights into the advantages of different reference points and the implications of ignoring certain forces. There is an ongoing exploration of the conditions under which the plank remains at rest, with no explicit consensus reached on the interpretations presented.

Contextual Notes

Participants are navigating the complexities of torque calculations, particularly in relation to the distribution of mass and the role of normal forces at the supports. The problem context includes specific measurements and conditions that may influence the calculations discussed.

henry3369
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Homework Statement


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

Homework Equations


net torque = 0

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.
 
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henry3369 said:
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).
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.
henry3369 said:
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.
What did you take the normal force from the left sawhorse to be?
 
henry3369 said:

Homework Statement


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

Homework Equations


net torque = 0

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.
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:
Nathanael said:
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?
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?
 
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.
 
henry3369 said:
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.
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.
 
henry3369 said:
If the mass is concentrated at the center of the plank
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.

henry3369 said:
wouldn't there be no normal force at the saw horse
If there was no normal force at the saw horse, wouldn't there be a net downwards force on the plank?
 
henry3369 said:
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?
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.
 
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
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
henry3369 said:
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?
Don't forget your cousin.
 
  • #12
henry3369 said:
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?
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
Zero?
 
  • #14
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
henry3369 said:
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?
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.
 

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