# Equilibrium Force

1. Nov 3, 2005

### Augusto1987

What is the procedure I must make to obtain the point of application of a equilibirum force in a system of forces? Thanks in advance.

2. Nov 3, 2005

### Pyrrhus

Could you be more clear?, Well if you want the point of application of a Force the resultant moment is the null vector will be the condition you'll need.

3. Nov 3, 2005

### Augusto1987

When solving a system of forces, if the system has no equilibrium, one has to find the force that gives the equilibrium to that system, which is opposite to the resultant of the system. What I want to know is where to put the equilibrium force, that is its point of application. Setting the equilibrium force in that point will give the force the opposite torque to the resultant of the system, but I don't know how to calcule the point. Am I clear? Thanks

4. Nov 3, 2005

### Pyrrhus

Yes it's clear now, on a system of two forces, both forces will have the same magnitude, will be colineal, both opposite directions. In other words, the point of application of the force you have to find will be an arbitrary point along the line of action of the resultant force, both forces will have the same slope.

5. Nov 4, 2005

### Augusto1987

In a system of forces which forces "cross" or "start" (dunno how to explain properly in english, sorry) in a point is easy to find where to place the equilibrium force because that place is the point the forces cross or start.
Now, in a system of forces that do not cross or start in a point, where do I place the equilibrium force? I know the Fy of the equilibrium force will be the negative of the Fy of the resultant system force, same with Fx and the its torque will be the negative of the resultant force torque. So what I want to know is how to calcule that point, in which placing the force there will cause that force to have the negative torque of the resultant, and set the system in equilibirum. Can you explain me that?
Thanks for helping (and bearing me! :p)

6. Nov 4, 2005

### Pyrrhus

Augusto, explicamelo en español, porqué parece que no te estoy entendiendo bien, ¿vale?

7. Nov 4, 2005

### Pyrrhus

Well after you do a statically equivalent system of a resultant force and resultant couple, for non concurrent forces. If the couple it's not perpendicular to the resultant force, then its effect can't dissappear completely, but there are some special cases where this is possible. For these cases you'll need to include also a couple to cancel the other couple, but i doubt that's your case. I believe you are asking for a resultant force that will eliminate the resultant couple (the special case, no?) for perpendicular couple to the force?

8. Nov 5, 2005

### Augusto1987

Yes. I want to find the point of application of the equilibrium force. In that point, the equilibirum force will have the opposite torque or couple of the resultant force. That will give total equilibirum (Fy, Fx, and Torque).
I want to know how to calcule that point of application.

9. Nov 5, 2005

### Pyrrhus

On system of parallel forces, (coplanar or no coplanar) you can use the transport couple procedure to get a Resultant Force and a Resultant couple which will be perpendicular to the resultant force, then you can find a position vector which will give the coordinates where to put the Resultant force which will generate that same resultant couple (thus only having a resultant force). Now you'll simply put the equilibrium force at the coordinates of the resultant force.
To find the point of application use
$$\sum \vec{M}_{o} = \vec{r} \times \vec{R}$$

where o is any arbitrary point, you'll need to find the position vector r, which will give you the coordinates you want.

10. Nov 5, 2005

### vaishakh

i think you are referring ti the free body diagram. but your question is not at lal specific

11. Nov 6, 2005

### Augusto1987

Last edited by a moderator: May 2, 2017
12. Nov 6, 2005

### Pyrrhus

Ok first reduce the system to a Resultant Couple - Resultant Force system. It could be about the origin. Notice this is the coplanar case, hence the resultant force will be perpendicular to the resultant couple, so you can find the position vector for the resultant force, the opposite direction of the resultant force will be the equilibrium force you want.

Also, you might want to read this Augusto, https://www.physicsforums.com/showthread.php?t=4825

Last edited: Nov 6, 2005
13. Nov 7, 2005

### Augusto1987

Well I'll try to resolve it here...

Reduction point: A (2,4)

Fx = 8N cos330º + 10N cos45º + 20N cos270º = 6.92N + 7.07N + 0N = 14N

Fy = 8N sin330º + 10N sin45º + 20N sin270º = -4N + 7.07N - 20N = -16,93N

T = |F2x||y2-yA| - |F2y||x2-xA| - |F3y||x3-xA| = 21.21Nm - 21.21Nm - 80Nm = 80Nm

Resultant Force: Fr(14N, -16.93N) Angle= arctan Fy/Fx = 130º

Resultant Torque : 80Nm

Equilibirum Torque: -80Nm

Equilibirum Force: Fe( ? , ? )=(-14N, 16.93)

How do I find the point of application of this force, so that its torque is -80Nm?

14. Nov 7, 2005

### Pyrrhus

By using

$$\sum M_{o} = R_{y}d$$

where d is the lever arm for the Ry component of the resultant, which will give you the same application point for the equilibrium force.

15. Nov 8, 2005

### Augusto1987

So:
xe-xA = -80Nm/16.93N = -4.72
xe = -4.72 + 2
xe = -2.72

Is this ok? And for ye I use Rx?

16. Nov 8, 2005

### Pyrrhus

What you're basicly finding is a point where the line of action of the Equilibrium force goes throught. This is because of the transmissibility principle which states a force can move along its line of action without altering its effects on the rigid solid.

What is xe-xA? and also yes for Rx.

Remember

$$yR_{x} = \sum M_{o}$$

$$xR_{y} = \sum M_{o}$$

What is Rx will give you the y coordinate which the force intersects on the y axis, and Ry will give you x coordinate the force intersects on the x axis.

17. Nov 9, 2005

### Augusto1987

xe minus xa is: xe is the x coordinate of the equilibirum force and xa the x coordinate of the reduction point.

I thought that ecuation wasn't useful because I had seen this formule:

Mo= [+/-] Ey*x [+/-] Ex*y

18. Nov 10, 2005

### Pyrrhus

Well that's the general approach, but if you look at it, you have two unknowns. Nevertheless, if you use the above equations, you eliminate one of the components moment by calculating the moment where that component has none, those moments will be 0 on the intersection on the x-axis or y-axis.

Also xa coordinate of reduction point? what?