Force Couple System and resultant force

[Northbysouth]

Homework Statement

Replace the force-couple system at point O by a single force. Specify the coordinate yA of the point on the y-axis through which the line of action of this resultant force passes.

Homework Equations

The Attempt at a Solution

I'm not sure how to start this question and I don't understand how I'm supposed to interpret the circular arrow; I think it is a moment.

I think that the 280 N force pointing left would have a moment of zero because it passes through the origin (O), after this I'm not sure what to do with the circular arrow.

Any suggestions?

 

[CWatters]

Imagine the sheet of paper has a pivot at point O.

As drawn there are two forces acting on the paper. One pulling it sideways and one trying to rotate it (the curved arrow). Now remove those two forces and replace them with one equivalent force connected to a point on the Y axis above or below point O. eg one force that also tries to pull it sideways and rotate it.

Lets say you were to try a force pointing to the left connected at Y = +1 meter. That would produce an anti clockwise torque about point O which is obviously wrong.

 

[Northbysouth]

Are you saying that I should select a point A on the y-axis which is a known distance from point O and then try to find the moment of that point?

So for example, if I picked a point 1 meter below point O and labelled it A.

Moment of A = (280 N)(1m) - 80(Nm)
moment of A = 200 Nm

 

[CWatters]

Are you saying that I should select a point A on the y-axis which is a known distance from point O and then try to find the moment of that point?

Yes that bits right but replace "select" by calculate"

So for example, if I picked a point 1 meter below point O and labelled it A.

Moment of A = (280 N)(1m) - 80(Nm)
moment of A = 200 Nm

No, you need to replace the existing force and torque with one force that has the same effect as both.

This isn't the answer (that would make it too easy) but suppose you tried using a force of 300N acting to the left on a point 1.5m below point O. That would be equivalent to a force of 300N acting to the left and a clockwise torque of 300*1.5 = 450NM about point O.

You know what force and torque is required so calculate the distance it needs to be from point O

 

[Northbysouth]

So are you saying:

-80 Nm = 280N*d

I'm sorry but I honestly don't understand. Could you explain again? I just don't see how this works.

 

[Northbysouth]

I was looking over my textbook and I managed to get the correct answer as:

d = -0.285714 m

But I am a little confused about why this is the correct answer. As far as I can tell the system as it is, with the two forces, has a moment about O, M_o, of -80 Nm (negative because it's clockwise), right? Because the 280 N force doesn't impact the moment about O because it passes through O; would this be similar to pushing directly on the narrow side of a door so that the force was directed through the hinge (the pivot point of the door)?

But when we then want to move the force to a point on the y-axis, this is where I get confused. I can understand that the moment would remain as -80Nm because we want the forces constant (if that makes sense) but how do I know that the force is 280N? Couldn't the force be any value, thereby making the distance any value? Why does the force have to be that 280N force?

 

[CWatters]

I was looking over my textbook and I managed to get the correct answer as:

d = -0.285714 m

That's what I make it but for full marks see the footnote on the diagram about tollerance.

Couldn't the force be any value..

No because then the linear force would not be the same same as the original. For example a force of 40N located 2m below point O would produce the right torque (80NM) but the linear force to the left would be 40N not 280N as per the original.

Likewise a 280N force acting vertically at X = -0.285714 would also produce the right torque but the vertical and horizontal linear forces would be wrong.

 

[CWatters]

Here is a more complicated example... Start at the left. It shows two forces and a torque. Calculate the position and magnitude of a single force to replace all three...

Middle diagram shows an intermediate step in the calculation. The green line shows a new force that is equivalent to the 280N and 140N forces BUT not the torque. This is standard vector addition so far.

Force = sqrt(280²+140²)
= 313N

The final diagram shows where that force would have to act to produce the torque as well.

The angles can be calculated from the middle diagram.

To produce the right torque..

R = 80/(sqrt(280²+140²))
=80/313
= 0.356m

Compare the first and last images. To be equivalent the vertical component must be the same, the horizontal component must be the same, and the torque must be the same.