# Applying a force on a rigid body which is used to create torque

In the above picture, I have a rigid body, in turquoise, which is connected to a point, the red circle. The dotted lines are used to divide objects or lengths, they are not part of the rigid body. The distance from the red circle to the application point of Force 1 is X+1. The length of the lower arm of the rigid body has a length of X + 1.5, and the distance of the red circle to the perpendicular line of the lower arm is X wide. The yellow arrow around the red circle is the direction of torque that we want to measure.

If I produce a Force, Force 1 at the point shown, is has a certain magnitude and angle. Now the turquoise body is a rigid body that does not flex or bend. Am I right to believe that because this body is rigid and not rotating, that it is in Translation? If it is in Translation, should all particles of the upper arm be feeling the same force and angle of Force 1? And if so, am I safe to assume that the Force 2 traveling down the rigid body is the cosine of the angles between Force 1 and Force 2, thus decreasing the magnitude of Force 2 relative to Force 1 because of this force deflection?

The final answer that I am grappling with is how do I correctly calculate torque at the red circle. I now that torque is Force x Distance. Which is the proper way to calculate torque?

Is it:

Force 1 x (X+1) - regardless of the deflections of Force 1 to Force 2?

or

Force 2 x (X+1.5) - which would be this decreased force multiplied by the lower arm length?

I hope this wasn't too convoluted, but I was told that regardless of the elbow type fashion of this rigid body, the only force that matters when calculating the Torque is Force 1 multiplied by its distance to the red circle, X+1.

If anybody can help answer this question and give me the physical proof as to why either method is correct, I would truly appreciative it.

Thanks.

• Done correctly, both methods should give the same answer. That being said, I know of no easy way to compute $F_2$ without using the concept of torque (and therefore the first method). Commented Aug 8, 2014 at 21:58
• You need to provide the direction of $F_1$ to correctly calculate torque. Commented Aug 9, 2014 at 4:23

## 2 Answers

You can translate a force F along its vector (line of action) without changing the torque. For example, you can slide $F_1$ all the way to the vertical member - the one that force $F_2$ acts along.

Now if you look at the diagram I drew:

you can see two different ways to compute the torque.

Method 1: Extend $F_1$ along its direction, draw the perpendicular line to the origin (in turquoise), multiply. The answer is $F_1\cdot d$.

Method 2: Extend $F_1$ until it intersects the line of action of $F_2$. Now decompose it along perpendicular directions - one orthogonal to the radius vector to the origin, and one along it. I labeled the perpendicular component $F_3$ (red). Now the vector $F_2$ that you need to compute is given by looking for the reverse projection of $F_3$ along the line of action of $F_2$. I hope it is clear from the diagram. This is obviously a roundabout way to construct it. It would be quicker to compute torque according to the above, then just set $F_2 \cdot (x_1 + 1) = F_1 \cdot d$.

Note - it may be that you intended to show $d = x$, but that is not clear from the diagram. And if that is indeed not the case, then you need to know the angle of $F_1$ in order to solve this problem - you imply that you do know that. Even so - as presented your problem cannot be solved unless we can compute $d$ for which we need the height of your turquoise shape, or the angle of the dotted line $x$.

Torque is calculated with $$\sum \vec{\tau} = \sum_i( \vec{r}_i \times \vec{F}_i)$$ where $\times$ is the vector cross product, $\vec{r}_i$ the location of each force $\vec{F}_i$.

So based on your diagram you have

$$\vec{\tau} = \begin{pmatrix} x \\ \sqrt{2 x +1} \\ 0 \end{pmatrix} \times \begin{pmatrix} F_{1}x \\ F_{1y} \\ 0 \end{pmatrix} + \begin{pmatrix} x+1.5 \\ 0 \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ F_{2y} \\ 0 \end{pmatrix}$$

$$\vec{\tau} = \begin{pmatrix} 0 \\ 0 \\ F_{1y} x - F_{1x} \sqrt{2 x+1} \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ F_{2y} (x+1.5) \end{pmatrix}$$

Here the force components are positive when directed to the right and up. So for example, if in your diagram you have Force 2 being $1000\, \rm N$ downwards, then $F_{2y}=-1000$.

The deflections are not generally considered in statics because their are orders of magnitude smaller than the overall size of the structure. Of course if deflections need to be considered the location of the force point going to change so an iterative approach is needed.

I did not quite follow the logic about the force felt being reduced, and no a rigid body is not in translation in general. The general motion is a rotation about a point either on the body or away (the center of rotation). The body is translating only when the net torques about the center of mass are zero, and the rotation center moves to infinity.

Please read my answer here (https://physics.stackexchange.com/a/130058/392) for a better discussion of how force and torque interact.

Appendix

If the sum of the forces is zero (as in statics) then the point by which torques are calculated does not matter. On the other hand if there are accelerations (dynamics) then only net torque by the center of mass should be considered.

See accepted answer here (https://physics.stackexchange.com/a/111348/392) for a more elaborate explanation.

• Nice answer. Small elaboration: in a "rigid body" (as posited in the question) you set deflection = 0 by definition of "rigid", not because "it is orders of magnitude smaller". Also I think he's asking about the $F_2$ that gives the equivalent torque to $F_1$, not the torque due to their combination (although by setting their combination == 0 you could solve for $-F_2$, I suppose). But it's not unambiguous from the question. Commented Aug 9, 2014 at 7:50
• Yes to find the static equilibrium you set $\sum \vec{F} =0$ and $\sum \vec{\tau}=0$ and solve for three (in 2D problems) components like $F_{2y}$ and the two reactions at the pivot point (red circle). Commented Aug 9, 2014 at 18:18