Calc Equilibrium Forces in Rigid Bodies

[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

 

[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.

 

[hellknows2008]

Thank you for your reply

So, with 4 legs of the same given length, and a given elasticity of the surface then, how do we come up with the 4th equation??

Thank you very much

 

[PhanthomJay]

It is a very difficult problem requiring advanced knowledge of plate theory, the solution of which depends on the degree of elasticity of the supporting surface, table legs, and table surface itself. See here for some info regarding its solution, but no solution is given.
https://www.physicsforums.com/showthread.php?t=138020

 

[rwisz]

Hi vc,

Thank you for your question. The equations you have provided are the correct way to calculate the equilibrium forces in a rigid body. In order to handle more than 3 table legs, we need to add more equations to the system. This can be done by considering the moments of the forces about a different point on the body, in addition to the center of mass. This will give us more equations to solve for the unknown forces.

For example, let's say we choose a point A on the table, and we know the distance from this point to each of the table legs (d0, d1, d2, d3). The equations for equilibrium will now be:

f0 + f1 + f2 + f3 = -fw
y0*f0 + y1*f1 + y2*f2 + y3*f3 = 0
-x0*f0 - x1*f1 - x2*f2 + x3*f3 = 0
-d0*f0 + d1*f1 + d2*f2 + d3*f3 = 0

Now we have 4 equations and 4 unknowns, and we can solve for the equilibrium forces using the same methods as before. This can be extended to any number of table legs by choosing additional points and finding their distances to each leg.

I hope this helps. Let me know if you have any further questions.