# Rigid body object - translative and rotational forces?

If I have free 2D rigid body with 3 point masses, m1, m2 and m3, and I have 3 forces F1, F2 and F3 acting on these three masses, is it correct to say that the net translative force on the Centre of Mass the vector sum of all forces acting in the direction of the COM, or am I missing something? In this case, is the torque of the overall body derived from the sum of the tangential component of these forces x distance of each mass from COM?

I'm trying to model this, and while the rotational equation seems to work, the translation is not. Hopefully someone can help.

• the vector sum of all forces acting in the direction of the COM: what do you mean with this. The translation force is just the vector sum of the three forces applied to the CoM. Are you doing any other computation? – rodrigo Aug 20 '16 at 19:16
• your statement concerning the torque with respect to the center of mass is correct. (If by tangential you mean perpendicular to the vector from the center of mass to the respective mass). Concerning the translational force I agree with rodrigo – Crimson Aug 20 '16 at 19:24
• On second read, now I understand that you are projecting the forces to the line from the point of application to the CoM. That is wrong, you don't need any transformations to the forces, just add them all as-is and apply to the CoM. – rodrigo Aug 20 '16 at 19:28
• Thank you rodrigo, yes - originally I only included the forces in the direction of the COM from each mass, thinking that you could split the forces into a translational component and a rotational component. However the maths didn't work out and the only way it seemed it could work was to simply add the forces together. And you have also anticipated my second question! Why are we double-counting when it comes to the rotational component of the force.. Hmm. Thanks very much!! – ColmR Aug 20 '16 at 20:34

It seems to me that you think that those forces are divided into two pieces:

• One, perpendicular to the vector from the point of application (PoA) to the CoM, that will apply torque but not change linearm momentum.
• Two, along the same vector, will change linear momentum, but not apply torque.

Well, the One is right, the Two is wrong. To compute the linear force, just vector-sum all the forces and apply them directly to the CoM. Easy.

"But wait", you say, "if I do this, I'm accounting for the same piece of force twice".

Yes, it is right. A force applied off the CoM will increase both the angular and the linear momentum. It may look unintuitive (it looked to me the first time I realized it), but that is how it works.

The equipollent system is the vector sum of all the forces acting through the center of mass as well as the vector sum of all torques about the center of mass.

$$\sum \vec{F}_i = m \vec{a}_{cm}$$ $$\sum \left(\vec{r}_i - \vec{r}_{cm}\right) \times \vec{F}_i = {\rm I}_{cm} \alpha$$

With $\times$ the planar cross product $(x,y)\times(F_x,F_y) = x F_y - y F_x$