# Homework Help: Find resultant of multiple torques on a single rigid body

1. Aug 30, 2011

### WildEnergy

hi

I have multiple torques acting at different points on a rigid body
I want to find the resultant force and torque (about the body CM)

I think I can see how to "convert" an off-center torque to a torque + translation using the parallel axis theorem - but am wondering if there is a simpler way

Also why can't I find reference this common scenario anywhere? Not even in my undergrad classical mechanics textbook?

2. Aug 30, 2011

### Hootenanny

Staff Emeritus
One can often simplify matters by a clever choice of where one takes moments about. Perhaps if you posted an example problem, we could help further.

3. Aug 30, 2011

### McLaren Rulez

Maybe you could post the exact question? I suppose the way I would do it is to see it all in terms of forces and then pick the axis. Then I know the torques. It is not entirely clear if this is the problem you're facing.

4. Aug 30, 2011

### MrNerd

I think I know what you need to do.

First, you find the force at the point, and the distance to the rotational axis.
Torque is simply Fd, force times distance. You may also need to use sin or cos, if at angles.

Simply combine all the torques, and you should get the rotational result.
In combining them, make sure you get all the positives and negatives right.
A torque causing an acceleration in the same rotational direction would have the same sign.
So a force acting up on the left side and a force acting down on the right side would have the same sign.

The translational part should be easy enough. Just take the net result of the forces.

5. Aug 31, 2011

### WildEnergy

take a rigid body with CM at (0,0)
there are equal 2 torques: 1 about (4, 1) and 1 about (5, 1)
how would I determine the single torque (about 0,0) and the single translation (though 0,0) ?

6. Aug 31, 2011

### WildEnergy

take a rigid body with CM at (0,0)
there are equal 2 torques: 1 about (4, 1) and 1 about (5, 1)
how would I determine the single torque (about 0,0) and the single translation (though 0,0) ?

7. Aug 31, 2011

### K^2

Torque with no net force at point (x,y) gives you the same torque around (0,0) and no net force.

Consider net torque at (x,y). It is produced by any pair of equally opposing forces at any two points you like. The simple choice would be to place one of these forces at (0,0), then the other trivially goes to (2x, 2y). The forces are (-Fx,-Fy) and (Fx,Fy) respectively.

What is the net torque around (x,y)? It is [(2x-x)Fy-(2y-y)Fx]+[-(0-x)Fy+(0-y)Fx] = 2(xFy-yFx)

What is the net torque around (0,0)? It is (2x-0)Fy-(2x-0)Fx = 2(xFy-yFx)

The net force, in both cases, is just the sum of forces, which is zero.

Let me generalize this for you. When you take a net torque T around some (x',y') with some net force (Fx,Fy), the torque around (x,y) is T+(x'-x)Fy-(y'-y)Fx. Or more generally, in arbitrary number of dimensions, it's T+(r'-r)xF [This is a cross-product]. The net force, of course, is unchanged.

8. Aug 31, 2011

### WildEnergy

I understand placing one of the forces at CM - that is a great trick :-)

But I don't understand your above statement.

Surely an off-center torque is always equivalent to a torque about CM PLUS a translation through CM? As the CM is accelerating?

And isn't it possible that the forces in a couple that creates an off-center torque are not necessarily equal in magnitude?

Last edited: Aug 31, 2011