# Forces on a rigid body

1. Jul 27, 2009

### FigBug

This isn't homework, it's for a simple computer program I'm trying to write. It's been a long time since I took a physics class, but I seem to remember getting homework questions like this.

If I have a 2D rigid body, I know its mass and the location of its CG. The object is being held by two points, how can I find the forces acting on the two points?

I know there are going to be 3 force vectors that must add up to zero. The first one is easy to find, straight down with mass * 9.8. But how do I find the other two? I assume the direction of the vectors is the angle between CG and A and B? If I have one vector and two angles, is that enough to calculate everything I need?

Torque must sum to zero as well, but I don't think I need to worry about that.

Any tips for solving this in the general case? Any tutorials I should look at? I have one physics text, Physics for Scientists & Engineers, but it only covers this topic with 1 or 2 pages.

Last edited by a moderator: May 4, 2017
2. Jul 27, 2009

### rock.freak667

Well if you split the components of the force at A into x and y, and do the same for B, if the body is in equilibrium as shown, then $\Sigma F_x= 0$ and $\Sigma F_y=0$. If you know the two angles, then you'd have two equations with two unknowns.

Also if you take moments about the CG, then you'd get another equation to use.

3. Jul 27, 2009

### turin

If this is made to be so, then you can ignore the torque (because there is no applied torque in this case), and then you can solve a system of two equations and two unknowns (or four equations and four unknowns). However, this condition is not guaranteed automatically; you have to additionally require that the forces at A and B satisfy this condition.

In general, if you allow the forces at A and B to be applied in arbitrary direction, then you will have one free parameter. This simply amounts to what you probably already realize intuitively. If you just barely hold onto the rigid body, so that it does not fall, then the forces will be on the order of mg, and directed mostly upward. Alternatively, you can imagine applying opposing near-horizontal forces at A and B, in which case the forces will need to be very hard (>>mg). You can apply the forces anywhere in between these two extremes as well, giving you one free parameter.

You could also ask an interesting question: what is the minimum amount of force required at A and B?

4. Jul 28, 2009

### FigBug

I'm getting a bit confused, so I'm going to back up a bit to my actual problem.

I have a speaker array (as shown) that is hanging from two chains and I'm trying to find the forces on the pins labelled A, B, C, D

http://rabien.com/image/array.png [Broken]

Am I correct in thinking there won't be any forces in the x direction but only the y?

To find the forces on A & B, I find the mass of the entire array and the location of the CG. If the CG is left of A or right of B then the array won't be hanging from both chains, one of them will be slack and all the weight will be on the other.

If Ax < CGx < Bx then the force will be spread between the two chains.

The force on A with be something like (CGx - Ax) / (Bx - Ax) * (Fcg) in the y and 0 in the x?

I'd do something similar for C & D? But it gets more complex if A & B don't have the same y value? Then I would start getting forces in the x direction?

I think the first thing I need to do is figure out the correct way to model my problem before I can solve it.

Last edited by a moderator: May 4, 2017
5. Jul 28, 2009

### turin

I don't know if I understand your diagram. No doubt you are hoping that we can understand what you "mean" because an accurate diagram with all the necessary labels would be tedious. So, here I will tell you what my interpretation is, and if it does not match your intent, then let me know:

There are six "blocks". Each block has a circle drawn roughly in the middle, which represents the CG of that block. All of the blocks are attached consecutively, but you only had the patience to draw one set of connections, C and D, which is your paradigm for connecting the speakers. So, A and B must support the weight of the entire chain (all five speakers and the the platform at the top). There are two connections that hold the 5-speaker chain to the platform (no shown). C and D hold the bottom 4 speakers to the top speaker. Etc. The shape of the chain as you have represented it is static.

OK, so here's my analysis. The way you've drawn the connections suggests that the forces are entirely vertical. That is, A and B experience entirely vertical forces due to tensions in two vertical cables (connected to the ceiling). C and D experience approximately vertical forces due to two approximately vertically alligned brackets. And so on. So, you do need to consider torque, if you want to know how the force is dsitributed between the two connections. As an approximation, you should be able to set the x-components of all forces to zero. Then, you have two conditions to satisfy: the y-components of all forces acting on a body must add up to zero, and the torques acting on a body must add up to zero.

6. Jul 28, 2009

### FigBug

Your interpretation of my diagram is correct. As the array curves backwards, the connectors will become less vertically aligned. Until now, I've been ignoring torque. I'll incorporating torque into my calculations and I'll see where that gets me.

Thanks for the help.