# If we know forces and torques applied to one point of a rigid body, how do we calculate them for another point of a rigid body

Suppose we have a rigid body with known length $$r$$ that is fixed at point $$A$$. We also have a force-torque sensor at point $$A$$. The body is fixed such that it doesn't move. We apply forces to point $$B$$ and get the values from sensor at point $$A$$. How can we translate the f-t values at $$A$$ to f-t values at $$B$$?

UPD: Decided to write balancing equations to make it a bit more clear:

The rigid body is in the equilibrium and we press with some force $$F_b$$ at point $$B$$. We can see the values of the f-t sensor at $$A$$.

Thus:

There is one balancing force at $$A$$: $$F_a$$, and one balancing torque, which will be something like $$F_{a_{t}}$$ $$\timesd$$ (depending on the design of the f-t sensor).

$$F_b = -F_a.$$

If $$F_a = (0, value, 0)$$ then opposing force $$F_b$$ will be equal to $$(0, -value, 0).$$

And if we can get values from the force-torque sensor at $$B$$, then we know the value of $$F_a$$ at $$A$$, which is equal to $$(0, -F_y, 0)$$.

Does this look correct?

• I don't know if I understood. If the body is in equilibrium, the forces and torques at A must balance the forces and torques at B. Of course the weight of the body (body force) also must be taken in account. Commented Mar 27 at 0:02
• Thanks! Am I right that, when the body is in equilibrium, forces at A balance forces at B and torques at A balance torques at B, and they do it separately, i.e. one equation for forces and one equation for torques? Commented Mar 27 at 1:05
• yes, that is it. Commented Mar 27 at 1:07
• Assume we apply force at B as shown on the picture above: F = (0, value, 0) Will the force at A be (0, -value, 0) and the torque at A be (value * len, 0, 0) ? Commented Mar 27 at 1:12
• See equipollent torque ... Commented Mar 27 at 15:04

Your notation is confusing. Each component needs to indicate both the direction that it corresponds to, and the location it is associated with.

And using the equipollent torque $$\vec{T}_A = \vec{r}_{\rm B/A} \times \vec{F}_B$$

you have the individual components of force

$$\vec{F}_A = \pmatrix{ Fx_B \\ Fy_B \\ Fz_B - m g }$$

and torque

$$\vec{T}_A = \pmatrix{ r\,Fy_B \\ \mbox{-} r\, Fx_B \\ 0 }$$

For the special case where $$\vec{F}_B = (0,Fy_B,0)$$ then $$Tx_A = r\,Fy_B$$ is the only surviving term.

Note that the weight of the part does not impart a torque on the support because the line of action of gravity goes through the support.

• Thanks for such a clear explanation and pictures! Commented Mar 27 at 16:06