How is the torque shared between two points?

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Homework Help Overview

The discussion revolves around a system in equilibrium involving torque and forces at two points, labeled a and b. The original poster presents a scenario with a mass and distances, seeking to understand how torque is shared between these two points.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to derive equations for the forces at points a and b based on torque and equilibrium conditions. They question whether the torques at the two points are equal or if the forces are equal. Other participants suggest using the equilibrium condition of total torque being zero to form equations for the forces.

Discussion Status

Some participants have provided guidance on using equilibrium equations to analyze the forces, while the original poster expresses difficulty in forming a solvable system of equations. There is acknowledgment of the problem being statically indeterminate, leading to discussions about potential approaches to resolve the issue.

Contextual Notes

The original poster mentions that this problem is part of an independent project, which may influence the constraints and assumptions being discussed. There is also a note about the complexity of the problem, indicating that it may require advanced understanding of beam mechanics.

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vLvKC60.png


1. Homework Statement

We know system is in equilibrium and we are given the mass m, and the distances d, r1, and r2.
We assume that the rotational inertia of the bar is negligible.

What are the forces at points a and b?

Homework Equations


Torque = radius × Force
Force = mass * acceleration
ΣFi = 0
ΣTi =0
where i = {x,y,z}

The Attempt at a Solution


The force of the mass is F=mg, therefore the torque produced is T = d×(mg) = d*mg*sin(pi / 2) = dmg. (Where g is 9.81 m/s^2.)

Since we are in equilibrium we know r1*Fa+r2*Fb=T. Where Fa and Fb are the forces on the bar at the points a and b respectively.

Now how is the opposing torque shared between the two points?
Perhaps the torques r1*Fa and r2*Fb are equal, or perhaps Fa and Fb are equal?
 
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Welcome to PF!

In equilibrium you know that the total torque around any point is zero. You can use that to your advantage here to directly get an equation for Fa and Fb.

Alternatively, you can use the equation for torque around the left support point as you wrote it and combine it with the equation for translational equilibrium.
 
Filip Larsen said:
Welcome to PF!

In equilibrium you know that the total torque around any point is zero. You can use that to your advantage here to directly get an equation for Fa and Fb.

Alternatively, you can use the equation for torque around the left support point as you wrote it and combine it with the equation for translational equilibrium.

Oh right, fantastic! Let me give that a try.
 
Filip Larsen said:
In equilibrium you know that the total torque around any point is zero. You can use that to your advantage here to directly get an equation for Fa and Fb.

Been working on it for a few hours and I'm quite stuck.
I'm trying the first method.
The following is what I think the free body diagram is.

pXcqzpG.png


From which I get the following equations for the torque around each point
Torque around midpoint between 2 and a: (d + 0.5*r1)F1 - 0.5*r1F2 - 0.5*r1Fa - (0.5*r1 + r2)Fb = 0
Torque around 2: dF1 - r1Fa - r2Fb = 0
Torque around 1: dF2 - (d + r1)Fa - (d + r2)Fb = 0
Torque around a: (d + r1)F1 - r1F2 - (r2 - r1)Fb = 0
Torque around b: (d + r2)F1 - r2F2 + (r2 - r1)Fa = 0

However the matrix formed with all the combinations I have tried are singular.
Where am I going wrong?

This is for an independent project I'm doing rather than a course, is this in the right section?
If not where do I got to get direct help?
Can you show me how to solve this?
 
My apologies for giving you a wrong late night knee-jerk response for what I assumed to be a textbook problem. I should of course have analysed it better before answering - you even provided an excellent diagram. My bad for giving a false impression that this was easy to solve.

You are correct about your system of equations being singular. The problem you have stated is statically indeterminate [1] and thus have no single solution for Fa and Fb. To make your system determinate you will either have to remove a constraint (like a or b), or, if you are up to more advanced beam mechanics, expand the model to include a bending moment from the beam itself in which case you may want to skim some of the search results in [2] or get a textbook on beam mechanics.

[1] http://en.wikipedia.org/wiki/Statically_indeterminate
[2] http://www.google.com/search?q=statically+indeterminate+beams
 
Filip Larsen said:
My apologies for giving you a wrong late night knee-jerk response for what I assumed to be a textbook problem. I should of course have analysed it better before answering - you even provided an excellent diagram. My bad for giving a false impression that this was easy to solve.

You are correct about your system of equations being singular. The problem you have stated is statically indeterminate [1] and thus have no single solution for Fa and Fb. To make your system determinate you will either have to remove a constraint (like a or b), or, if you are up to more advanced beam mechanics, expand the model to include a bending moment from the beam itself in which case you may want to skim some of the search results in [2] or get a textbook on beam mechanics.

[1] http://en.wikipedia.org/wiki/Statically_indeterminate
[2] http://www.google.com/search?q=statically+indeterminate+beams

No worries! That seems to be everyone's first impression, including mine.

Thank you for the help! The more advanced beam mechanics is exactly what I'm looking for, do you recommend any books in particular?
 

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