I am trying to develop a little simulation (for the sake of learning a little bit about physics) involving a circle (mass m, radius r, regular density) and two forces f1, f2 that can be applied on any two points within the circle.

I have read that when a force doesn't go through the center of mass, then you have to split it up into two components, one parallel to the offset and one perpendicular to it. The parallel part of that force obviously now goes through the center of mass and thus "a=F/m" applies whereas the perpendicular force applies a torque, allowing me to use "alpha = torque/moment of inertia".

In my tests, this has worked perfectly, for as long as only one force acts on the body at a time.

However how (if at all) do I need to resolve multiple forces that influence the same body? At first I thought I can simply resolve each force in the pattern I described above and thus I calculated the sum of the linear forces on the CoM as well as the sum of the torque components. However this doesn't work for a number of examples, such as one where f1 and f2 act on either "side" of the circle in the same direction and both have the same length. For each individual force, there is no parallel component and thus no force is acted on the body. The sum of the torque is 0 as well, thus the body is not accelerated at all. However from playing KSP I realize that the net force applied on the body should be f1+f2.

This problem should have been solved numerous times, however I am unable to find a proper numerical solution (or don't use the right search terms). Can you guys help me out?

  • \$\begingroup\$ in your single force point if you put a single force tangential (torque only) on the circle it should have some acceleration \$\endgroup\$ Commented Sep 18, 2014 at 11:06
  • \$\begingroup\$ Do you mean linear or angular acceleration? From my understanding it should only cause an angular acceleration, shouldn't it? \$\endgroup\$
    – Simon
    Commented Sep 18, 2014 at 12:19
  • \$\begingroup\$ not on a free body, in KSP what happens if you put a single RCS on the side of the craft and try to rotate using it? you get translated as well \$\endgroup\$ Commented Sep 18, 2014 at 12:23
  • \$\begingroup\$ Perhaps related are these two questions about forces acting on different points of a modular spaceship. \$\endgroup\$
    – Anko
    Commented Sep 18, 2014 at 13:45
  • \$\begingroup\$ @ratchetfreak It took me a while to understand what you were saying. My intuition was totally off on this one. I have changed my simulation to behave as you described and it looks great now! \$\endgroup\$
    – Simon
    Commented Sep 19, 2014 at 8:15

1 Answer 1


You were mistaken about how to separate the forces.

The torque applied is the cross product between a_t and r where r is the vector from center of mass to the point the force is applied to. (or length(a_t)*length(r)* sin(theta) with theta the angle between the two)

The translation force is just a_t.

These results can be manipulate independently which means that your symmetric forces example (where theta1==-theta2) produces no torque (because length(a)*length(r)* sin(theta)+length(a)*length(r)* sin(-theta)=0 and all acceleration.

On Physic.SE there are several questions explaining it.

  • \$\begingroup\$ Thanks for your answer! To clarify, does this mean that the force applied to the center of mass is the same, regardless of the angle and force arm? \$\endgroup\$
    – Simon
    Commented Sep 18, 2014 at 14:23
  • \$\begingroup\$ @simon yes it does \$\endgroup\$ Commented Sep 18, 2014 at 14:23

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