# Where on a rigid body is a resultant force applied?

• zenterix
zenterix
Homework Statement
Three forces are applied to a square plate as shown in the figure below. Find the modulus, direction, and the point of application of the resultant force, if this point is taken on the side ##BC##.
Relevant Equations
##F=ma##

Here is what I came up with

First, let's compute the resultant force.

$$F_{R,x}=F+F\sqrt{2}\cdot\frac{\sqrt{2}}{2}=2F$$

$$F_{R,y}=F-F\sqrt{2}\cdot\frac{\sqrt{2}}{2}=0$$

At this point we already have the magnitude and direction of the resultant force. We need the point of application.

My question is how to find this point. I don't actually know how to do this.

What I thought at first was that the resultant force would have to be applied at a point such that the torque produced by it would be the same as the net torque on the rigid body.

Let ##l## be the length of a side of the square plate.

Net torque about the center is

$$\vec{\tau}=-F\frac{l\sqrt{2}}{2}\cdot 2\hat{k}+F\sqrt{2}\frac{l\sqrt{2}}{2}\hat{k}$$

$$=Fl(1-\sqrt{2})\hat{k}$$

and if ##\vec{r}=x\hat{i}+y\hat{j}## is the position where the resultant force is applied we have

$$\vec{r}\times\vec{F}_R=(x\hat{i}+y\hat{j})\times 2F\hat{i}=-2Fy\hat{k}$$

If we equate this to net torque and solve for ##y## we get

$$y=\frac{l(\sqrt{2}-1)}{2}$$

which seems to be the purple line below

It seems not to matter what the ##x##-coordinate is.

The solution manual says the following:

##F_R=2F## This force is parallel to the diagonal AC and is applied at the midpoint of the side BC.

Which is not in accordance with my attempt at a solution. Which isn't surprising given that I am very unsure about how to find the point of application of a resultant force on a rigid body.

You can choose any point (or line really) of application, but only one choice will result in a net torque of zero. If you pick another line of application it is not sufficient to use a net force only, you would also need to specify a torque.

In the context of this specific problem, is it possible to make a choice that results in a net torque of zero?

There is, after all, a non-zero net torque in this problem.

Is it correct to equate the torque produced by the resultant force to the net torque produced by the individual forces?

zenterix said:
What I thought at first was that the resultant force would have to be applied at a point such that the torque produced by it would be the same as the net torque on the rigid body.
Quite so.
zenterix said:
Net torque about the center is

$$\vec{\tau}=-F\frac{l\sqrt{2}}{2}\cdot 2\hat{k}+F\sqrt{2}\frac{l\sqrt{2}}{2}\hat{k}$$
Check the last term.

zenterix
zenterix said:
In the context of this specific problem, is it possible to make a choice that results in a net torque of zero?

There is, after all, a non-zero net torque in this problem.

Is it correct to equate the torque produced by the resultant force to the net torque produced by the individual forces?
You can make it a bit easier by choosing an axis that lies on the lines of two of the applied forces.

zenterix
Note that the forces at B and D combined produce no torque about the centre. The total torque about the centre is due to the force at A only. The same torque would be provided by a horizontal force of ##2F## at half the diagonal distance above the centre. Hence through the midpoint of BC. No calculations needed!

Last edited:
zenterix, Lnewqban and nasu
haruspex said:
Quite so.
What about the case in which we have a rectangular current loop in a uniform magnetic field?

The net force (ie, the resultant force) is zero.

Yet there is a net torque.

I think you agreed it is correct to say that the resultant force is applied such that the torque due to this force is equal to the net torque on all the infinitesimal pieces of the loop. But the resultant force is zero.

zenterix said:
What about the case in which we have a rectangular current loop in a uniform magnetic field?

The net force (ie, the resultant force) is zero.

Yet there is a net torque.

I think you agreed it is correct to say that the resultant force is applied such that the torque due to this force is equal to the net torque on all the infinitesimal pieces of the loop. But the resultant force is zero.
There are situations in which there is no point with respect to which the net torque is zero. This happens any time the torque has a nom zero component in the force direction.

In a two dimensional setting this occurs only when the net force is zero because the torque is then always perpendicular to the two dimensional plane. It is then rather moot as it really doesn’t matter where a force of zero is applied.

In a 3D setting the best you can do is to make force and torque parallel.

Orodruin said:
There are situations in which there is no point with respect to which the net torque is zero. This happens any time the torque has a nom zero component in the force direction.

In a two dimensional setting this occurs only when the net force is zero because the torque is then always perpendicular to the two dimensional plane. It is then rather moot as it really doesn’t matter where a force of zero is applied.

In a 3D setting the best you can do is to make force and torque parallel.
"It doesn't matter where the force of zero is applied" because the torque due to this force will always be zero. Yet the net torque isn't zero.

So again I have this question, which actually I haven't really understood yet from previous questions of mine: it need not be the case then that we can find a single force that produces the same torque as the individual forces acting on a rigid body?

In the case of zero translation acceleration, ie zero resultant force, the answer seems to be "no".

Yet, in the problem in the OP above, in which there was a non-zero resultant force, we were able to find a point of application such that a single force produced the net translation and the net rotation. In this case, the answer seems to be "yes".

zenterix said:
I think you agreed it is correct to say that the resultant force is applied such that the torque due to this force is equal to the net torque on all the infinitesimal pieces of the loop. But the resultant force is zero.
For clarity, I will distinguish between a force vector, having no particular line of action, and a force instance. I don't know if there is a standard terminology.

For any system of force instances you can sum their force vectors to find a net force vector.
If you also choose an axis line or axis point, you can sum their torques about that line or point to find a net torque.

In the 2D case, the axis is necessarily normal to the plane. If there is a nonzero net force vector then you can represent the system completely by a single force instance producing both the net force vector and the net torque.
If the net force vector is zero but the torque is nonzero, the system is a pure torque and cannot be represented by a force instance.

In the 3D case, if the axis is a line then you can first reduce it to 2D by projecting onto the plane normal to the axis. Solve this 2D case to find a net force instance, then add to its force vector the sum of the force vectors' components parallel to the axis. Since these have no torque about the axis, the resulting force instance represents the system.

But in 3D, the axis might be a point. Although each force vector is normal to its force instance's torque about the point, the net force vector need not be normal to the net torque. In that case there is no way to represent the system by a single force instance.
I would guess it can be represented by the sum of a force instance and a parallel pure torque.

zenterix said:
"It doesn't matter where the force of zero is applied" because the torque due to this force will always be zero.
Yes. This is talking about the resultant force. You can have a non-zero torque even if the net force is zero. The easiest example is taking any equal and opposite forces acting at points whose separation is not in the force direction.

The torque is not produced by the net force.

zenterix said:
Yet the net torque isn't zero.
So what? This is perfectly possible.

zenterix said:
So again I have this question, which actually I haven't really understood yet from previous questions of mine: it need not be the case then that we can find a single force that produces the same torque as the individual forces acting on a rigid body?
Again, there can be a net torque without a net force.

haruspex said:
For clarity, I will distinguish between a force vector, having no particular line of action, and a force instance. I don't know if there is a standard terminology.
Once one digs a little deeper into analysis on manifolds, it becomes natural to associate vectors in physics such as forces, velocities, etc, to a point as they are defined to be elements of a points tangent space. Only in the case of a flat base space, such as the one we usually consider, does it make any sense to start adding vectors from different points.

Of course, all of this is above the thread level. Just a curiosity based on the comment.

All inroads into physics have always pointed towards analysis on manifolds to answer many of my questions about the math. I have to learn that.

• Introductory Physics Homework Help
Replies
3
Views
273
• Introductory Physics Homework Help
Replies
25
Views
323
• Introductory Physics Homework Help
Replies
3
Views
276
• Introductory Physics Homework Help
Replies
12
Views
261
• Introductory Physics Homework Help
Replies
10
Views
319
• Introductory Physics Homework Help
Replies
1
Views
949
• Introductory Physics Homework Help
Replies
4
Views
769
• Introductory Physics Homework Help
Replies
16
Views
752
• Introductory Physics Homework Help
Replies
2
Views
322
• Introductory Physics Homework Help
Replies
25
Views
1K