# [Curiosity, not homework] How do you compute the balance point for an asymmetrical hanging object?

Asked by Jeruba (44758) November 2nd, 2009

This is a question for those types who do mathematical puzzles for fun (not I!). I do not have any use for the answer. I’m just curious about how you would solve this.

I have a picture calendar hanging on my wall. I suspend it not from the little punched holes but from a small metal clamp attached to the top (so the holes won’t tear). The clamp hangs over a nail. So I can position the clamp anywhere across the top. Normally I have it centered, and the calendar hangs straight.

Right now I have a large clamp attached to the lower left of the calendar page, holding an envelope with my November 29th opera tickets in it. So the calendar hangs crooked.

By moving the suspending clamp a little off center, I can straighten the calendar again.

What kind of formula describes the distance and direction off center that I have to move the top clamp so that the calendar hangs level once again with the extra weight on the lower left side?

Observing members: 0 Composing members: 0

I’m a little rusty in Physics 1, but I think this works. When you calendar is suspended draw a virtual (or physical) line from the point of suspension (the clip) directly towards the ground (it has to be perfect, or close). Now re-position the attachment point and repeat the line. The intersection of the two lines is the centre of gravity. Now, detach the calendar, place in on the wall in the desired position and attach the clip so that it is directly above the centre of gravity. Voila?

gggritso (5443)

Reposition the attachment point to what?

Jeruba (44758)

Doesn’t matter, any other point on the edge of the calendar.

gggritso (5443)

I would use the good old eyeball combined with trial and error method!

rooeytoo (26924)

Well, I’ve already successfully rehung the calendar by exactly the method you suggest, @rooeytoo. That is not the question. The question is purely academic, not practical. What I’m curious about is the method of predicting where the right point of attachment would be. It was closer to the center than I had expected.

I don’t see how it could not matter where the second attachment point is. I could place it anywhere on all four sides and get different answers. How is that going to find the center of gravity?

Jeruba (44758)

The centre of mass (in this case the centre of gravity) is a point in the object which you can use for the purposes of modeling. One would make an assumption that the entire mass of the object is actually located in that small point in order to make calculations easier. Following that logic, if all of the mass of the calendar is in that point, that point would always be directly below the point where the object is suspended from. It doesn’t matter what point on the calendar is attached to the wall, the centre of gravity is always on a vertical line directly below it. Therefore, if you suspect the calendar from a number of different points on said calendar all the lines would intersect in one point, the centre of mass. In order to find it, you only need two lines and their intersection.

More details here

gggritso (5443)

Oh. The second line should be plumb from the point of suspension. Sorry, I didn’t get that from “repeat the line.”

So—this is not going to be a formula but rather a—what? Are those lines vectors?

Jeruba (44758)

Sorry if I wasn’t clear before, yes they start at the point of suspension. I don’t think the method requires a more formal definition. It wouldn’t be wrong to call the lines vectors, but really — they’re just lines. Their only use is to quickly find the centre of mass; no more, no less. It’s a quickie geometry problem, really.

I’m quite happy that I still remember something from last year :)

gggritso (5443)

or