# 3D equilibrium?

1. Jul 4, 2010

### hellknows2008

Hi,

Given a rigid body, to keep the body equilibrium, multiple upward forces act on the body with each a known displacement from the center of mass. How can we calculate the upward forces?

Now imagine we have a table, with 3 legs of neglible mass, the center of mass is at the center of the table, the displacements of the table legs are known. To make the table in equilibrium, we have the following equations:

f0 + f1 + f2 = -fw (F0, F1 and F2 are upward forces acted by the table legs, FW is the weight of the table)
D0 x F0 + D1 x F1 + D2 x F2 = 0 (this is the sum of torques caused by forces acted by the table legs, D0, D1 and D2 are displacements from the center of mass of the table)
the above equation turns out to be: (assuming z-axis is the vertical axis)

[x0 y0 0]^T x [0 0 f0]^T + [x1 y1 0]^T x [0 0 f1]^T + [x2 y2 0]^T x [0 0 f2], then all together gives the following system of equations:

f0 + f1 + f2 = -fw
y0*f0 + y1*f1 + y2*f2 = 0
-x0*f0 - x1*f1 - x2*f2 = 0

We can solve the above by Gaussian elimination or matrix inversion.

The problem is, how can we generalize to handle more than 3 table legs?

With 4 legs, we have

f0 + f1 + f2 + f3 = -fw
y0*f0 + y1*f1 + y2*f2 + y3*f3 = 0
-x0*f0 - x1*f1 - x2*f2 + x3*f3 = 0

However, we only have 3 equations but we have 4 unknowns here

Thanks in advance for any help

vc

2. Jul 5, 2010

### K^2

There is no way to tell. You only need 3 points of support. A 4th one is a redundancy. Realistically, on a perfect surface, one of the legs is going to be a bit short, resulting in only 3 legs used for support. On a real surface, there will be a bit of a give, so all 4 legs may touch at once. Then the exact lengths of all 4 legs and elastic properties of the surface will determine the force distribution.

If you don't account for the properties of the surface, you can't resolve the problem. If you do, there is your missing 4th equation.

3. Jul 5, 2010