# Line of action of resultant force of two parallel forces

We take a rigid body as shown in the figure and apply two parallel forces (which do not have the same lines of action) at the ends of the body. Let us assume that $$P>Q$$. Now as we all know

There will be a transnational motion along with a rotational motion. For the translational motion,we take the resultant as $$P+Q$$ and consider it to be acting at the center of mass of the object. So the equation for the body $$P+Q=ma$$ holds by Newton's second law where $$a$$ is the acceleration of the center of mass.

Now comes the confusing part. We will now derive the point of application of the resultant force. The torque of resultant about the point of application is $$0$$ as the resultant passes through that point. So,the net torques of $$P$$ and $$Q$$ with respect to that point is $$0$$. Such a point $$S$$ will satisfy $$P\times AS=Q\times BS$$ and since $$P$$ is not equal to $$Q$$, $$S$$ can't be the midpoint of $$AB$$(hence center of mass of the body).

So in the translational part,we assumed $$P+Q$$ is acting on the center of mass but in the rotational part,we proved it is acting on some point $$S$$ which satisfies an equation. But a single resultant force can't pass through both the points at the same time. So how is this possible?

If the two forces are not exactly parallel, then the line of action of the resultant will go through the point the two lines of action meet regardless of the magnitude of the forces.

Now slowly imagine the two forces becoming more and more parallel. The meet point goes further and further away, until at the limit of parallelism, the meet point is at infinity.

The location of the combined force line of action follows the lever rule, making be closer to the force with the highest magnitude.

To prove the above, first look at the 2D case where the two forces are parallel and horizontal. Let's call the perpendicular distance of each line of action as $$d_A$$ and $$d_B$$ for the forces $$F_A$$ and $$F_B$$.

Then we are looking for the distance $$d_C$$ of the combined force $$F_C = F_A+ F_B$$. We need to solve the equation of the equipollent torques about the origin

$$d_C F_C = d_A F_A + d_B F_B$$

or

$$d_C = \frac{F_A}{F_A+F_B} d_A + \frac{F_B}{F_A+F_B} d_B$$

which is also known as the lever rule, as the ratio of each force to the total force gives us how close the resulting force is.

In 3D, the story is a bit more complex but still valid. Note that if you have the resultant force vector $$\boldsymbol{F}_C = \boldsymbol{F}_A + \boldsymbol{F}_B$$ and the resultant moment vector

$$\boldsymbol{M}_C = \boldsymbol{r}_A \times \boldsymbol{F}_A + \boldsymbol{r}_B \times \boldsymbol{F}_B$$

you can find the point where the resultant force goes through with

$$\boldsymbol{r}_C = \frac{ \boldsymbol{F}_C \times \boldsymbol{M}_C}{ \| \boldsymbol{F}_C\|^2}$$

where $$\times$$ is the vector cross product, and $$\| \; \|$$ designates the magnitude of a vector. Also $$\boldsymbol{r}_A$$ and $$\boldsymbol{r}_B$$ are the location vectors where the two forces go through.

The above describes the algebra of forces and how they add up together. The result of these forces in the motion of a rigid body is described by newton's 2nd law and the corresponding law of rotation.

• The net force acting on a rigid body describes the motion of the center of mass (point) only
• The net torque about the center of mass describes the motion about the center of mass (body) only.

In the case of two parallel forces of not equal magnitude, if the combined force line of action does not go through the center of mass, then the resulting motion will be a combination of translation and rotation. This situation is entirely similar to a single force applied through the combined line of action.

In summary, the rigid body is agnostic to the fact on how a particular set of the net force on the body and net torque about the center of mass is applied, meaning if it arises from single or multiple forces and in what configuration.

Any combination of forces parallel or not can be described to be equipollent to a single force through the combined line of action*, or a single force at a point and torque about this point, where the choice of point is arbitrary.

*The equipollent system is a single force through the line of action, and a parallel torque about the line of action

• Sir,isn't this approach similar to what i have done in the 2nd part. I assumed the point where the resultant acts to be $S$. Since resultant passes through $S$,torque of resultant about $S$ is $0$. So the net torque due to the components of $S$ which are $P$ and $Q$ are $0$. So we will get $P\times AS-Q\times BS=0$. The point $S$ doesn't come out to be the center of mass since if $COM$ is point $C$ on $AB$,then $AC=BC$ but $P$ is to equal to $Q$,so $P\times AC-Q\times BC$ is non zero. Hence the resultant doesn't pass through the COM. Commented Mar 24, 2023 at 6:17
• Then how can we say that the resultant produces a translational motion through the COM whereas we have just proved that the line of action of resultant doesn't pass through the COM. Commented Mar 24, 2023 at 6:18
• @madness - the resultant does not pass about the COM, and that means there is going to be a net torque about the center of mass for the body. That does not change the fact that Newton's 2nd law applies to the motion of the center of mass only. If there is a net force applied the center of mass will accelerate, and it sais nothing about if the overall motion is going to be a pure translation, pure rotation or combined motion. Commented Mar 24, 2023 at 11:27
• @madness - I made some edits to clarify that point. Commented Mar 24, 2023 at 11:38

Most of the existing answers are correct (and helpful) but I think it may be helpful to underline the central point here. This is:

When using a sum of forces to find the motion of a rigid body, you should NOT assume the line of action of the sum of forces is through the centre of mass (COM).

Instead, to identify a suitable line of action to assign to the resultant force, find the total torque produced by the various forces about some convenient axis (e.g. one running through the COM) and then assign the line of action such that the resultant force produces that same torque about your chosen axis.

(As other answers have mentioned, an alternative, and useful, way to proceed is to replace the total force with three forces: one through the COM and the others acting as an equal-and-opposite pair producing the correct amount of torque.)

So where did the false idea that the resultant always passes through the COM come from? I expect it came from the gravitational case, as follows.

In a gravitational field there are gravitational forces on all the parts of the body, each acting locally on that part. If the field is uniform then these forces are parallel and each is proportional to the local mass in each small region of the body. The net effect here is that there will be no torque about the COM, and therefore for the total gravitational force in a uniform field one should assign the net gravitational force as going through the COM.

Note, all this is only useful for a rigid body. For a non-rigid body (a jelly, a fluid, etc.) you can still get the magnitude and direction of the total force (and therefore the linear acceleration) by using a resultant, but now there is no single place where you can imagine the force acting in order to find the overall response of the non-rigid body.

• Sir,as for replacing a force with the three forces,one passing through the center of mass,and the other a couple. About which axis will the couple rotate? Since one force of the couple is acting at the end and the other one through the center of mass. (Please refer to the diagram 1 of Farcher) Commented Mar 24, 2023 at 13:40

Start off with force $$Q$$ as in diagram $$1$$.

Diagram $$2$$
Add two equal magnitude opposite direction, line of action parallel to force $$P$$ forces , $$Q'$$ and $$Q''$$, acting at centre of mass $$C$$.
You now have a force $$Q''$$ acting through the centre of mass which produces a translational acceleration of the centre of mass and a couple $$Q'Q$$ which produces a rotational acceleration of the body.

Diagrams $$3$$ and $$4$$
The same process can be done for force $$P$$.

Diagram $$5$$
The body under the action of force $$P''+Q''$$ through the centre of mass producing transnational acceleration and two couples, $$Q'Q$$ and $$P'P$$, which produce the rotational acceleration.

• Sir your one seems perfect to me too but i don't understand what is wrong with my logic in finding point $S$. The resultant of the forces $P$ and $Q$ passes through $S$. Since the resultant passes through $S$,torque of resultant about $S$ is $0$. So net torque about $S$ due to $P$ and $Q$ should also be $0$. Hence our point $S$ is uniquely determined by $P\times AS=Q\times BS$. But in your answer,you showed that the resultant passes through center of mass which doesn't satisy $C\times AC=Q\times BC$. I am confused since the logic regarding $S$ seems fine to me. Commented Mar 23, 2023 at 16:19
• youtu.be/mfQYzFpWusI Commented Mar 24, 2023 at 6:11
• Sir,actually in your answer,you have showed that the forces which are two couples and the horizontal forces. But in the video given,we were asked to find a single resultant force which again brings to the premise of my question. Commented Mar 24, 2023 at 6:12

We will now derive the point of application of the resultant force. The torque of resultant about the point of application is $$0$$ as the resultant passes through that point. So,the net torques of $$P$$ and $$Q$$ with respect to that point is $$0$$.

The net torque about the center of mass (COM) is the sum of the individual torque contributions produced by $$P$$ and $$Q$$ about the COM prior to moving the forces to act at the COM.

Let $$d_P$$ be the perpendicular distance between the line of action of $$P$$ and the COM and $$d_Q$$ be the perpendicular distance between the line of action of $$Q$$ and the COM. The net torque is then, taking counterclockwise as positive,

$$\tau_{net}=Qd_{Q}-Pd_P$$

Only if $$Qd_{Q}=Pd_P$$ will the net torque be zero in which case there will only be translational acceleration of the COM.

Hope this helps.

• Sir actually i was not taking moment of the forces wrt center of mass. Rather i defined the point through which the resultant acts to be $S$. Since resultant passes through $S$,torque about $S$ due to resultant is $0$. So net torque about $S$ due to $P$ and $Q$ is $0$ meaning $Qd_Q-Pd_P=0$. Here $d_Q$ and $d_P$ are perpendicular distances from $S$ which we still don't know if it is center of mass. So,$Qd_Q=Pd_P$. Surely the center of mass doesn't satisfy this since $P>Q$. So,won't the object rotate about the point $S$ instead of center of mass? Commented Mar 23, 2023 at 16:25
• @madness I don't follow you. The point of application of the resultant force, which you call $S$, is the center of mass since the acceleration of the body due to the net force is the acceleration of the COM. Commented Mar 23, 2023 at 16:30
• Sir,we know that if a force passes through a point,then torque of the force about that point is $0$. As you said,the resultant passes through the center of mass,so the torque of resultant about center of mass is $0$. Hence sum of the torques of its components about center of mass is $0$. If we assume the forces are being applied at the end points of diameters,then perpendicular distance from center of mass to the forces are the same which is the length of radius. In that case $d_P=d_Q=d$. But then $Pd-Qd$ is not equal to $0$ whereas in the first line we said torque of the resultant about Commented Mar 23, 2023 at 18:17
• The center of mass is $0$. Isn't this a contradiction,Sir? Commented Mar 23, 2023 at 18:21
• @madness I'm not saying the resulting force acts through the center of mass. That would only be the case if the lines of action of both P and Q actually pass through the COM. I'm saying in my answer you can move each force so that they act through the COM provided you add a moment about the COM to account for the torque they produce at the actual location. It's called an equivalent force system. Certainly IF the two forces are the same distance from the COM AND both forces are equal in magnitude, then there will be no rotation about the COM. Commented Mar 23, 2023 at 18:33