Magnitude & Direction of Resultant Force & Equilibrant Force

[JMxBelfast]

Okay, I'm back, yes I know so soon but I brought answers with me nonetheless, I came here to check them with your great knowledge or if it happens to be, someone else's great knowledge. So where did we leave off.. ah yes, the moments. Question 'iv' asked, "Find the position of the resultant and equilibrant force by taking moments about point "A"". I manage to conjure some sort of answer for a force, only one though, couldn't get my head round (what a surprise) the other force. I got an answer for what I believe to be the postion of the 'resultant force', whether it's correct or not remains a mystery, I do hope by postion though it means in my case, 'm' for meters, I don't think it is Nm for this case but wouldn't surprise me if I were wrong.

These are the calculations that I done:

M = (-8.66x2)-(-10.6x4)-(10.8xd) = 0

25.08 - 10.8d = 0

d = 25.08 / 10.8

d = 2.32m

Update
Had a bit of a brain fart and thought that since the equilibrant force had to be in the opposite direction of the resultant force, would I just take 2.32 from 4 (4 being the whole length of the tool) and then that would give me the position of the equilibrant force? Honestly at this point I'm clinging onto anything that seems to be some sort of way of obtaining an answer, whether it's right or wrong.

 

[gneill]

These are the calculations that I done:

M = (-8.66x2)-(-10.6x4)-(10.8xd) = 0

A good start, however...

It looks like you've used the y-components of F1 and F2 and the magnitude of the equilibrant force to write a moment equation. But there are a few problems here. First, they're not all the forces acting to produce a moment, and second, since the moment is to be taken about point A, all distances should be measured from that point. So the distance for first term will not be 2 meters, but zero meters (So F1 won't contribute to the moment about A since it passes through A). Third, you don't want to use the magnitudes of any of the forces involved, only their y-components. Only their y-components act perpendicularly to the tool to cause a torque.

Write your equation again using the y-components of all the forces acting.

 

[JMxBelfast]

Okay thanks for the help again, I re-wrote the equation to try and fit what you said:

M = (-8.66x0)-(-10.6x4)-(-5x2)-(15x2)
= 22.4

Should I include F1 even though it has no effect on the moment? I removed it from the equation and done it again and got a different answer (-62.4) although I tried again with just a '0' first as that is what that single sum (-8.66x0) equals and I got 22.4 again.

 

[gneill]

That looks okay (although you should put units on the result of the sum). You can include F1 or not so long as the distance for it is correct. It should not change the sum.

So that sum represents the moment about point A due to the existing forces. Can you now place the equilibrant force to cancel it out?

 

[JMxBelfast]

That looks okay (although you should put units on the result of the sum). You can include F1 or not so long as the distance for it is correct. It should not change the sum.

So that sum represents the moment about point A due to the existing forces. Can you now place the equilibrant force to cancel it out?

Honestly, nope hahaha, do I divide it by the equilibrium magnitude? I don't have my notes with me at the moment so I can't remember what I wrote down

 

[gneill]

You had an excellent approach in post #31. You just left out a few forces and slipped up on the distances. Now you have a good sum for the moment due to all the applied forces but the equilibrant. You want to apply the equilibrant at a location such that its y-component balances the previous sum (makes the net moment zero). Or just start again and write the full moment sum out including the equilibrant's contribution.

When forces are at various angles you almost never want to use the magnitude of the force for anything but computing the component of the force that's of interest.

 

[JMxBelfast]

Ah okay, so can I add the equilibrant force to the sum like this:

M = (-8.66x0) - (-10.6x4) - (-5x2) - (15x2) - (10.8xd) = 0

And then,

(-8.66x0) - (-10.6x4) - (-5x2) - (15x2) = 22.4 (I'm not sure if this is to be kN or Nm)

22.4 - 10.8d

d = 22.4 / 10.8

d = 2 (Nm?)

Sorry for not getting back sooner.

 

[gneill]

Ah okay, so can I add the equilibrant force to the sum like this:

M = (-8.66x0) - (-10.6x4) - (-5x2) - (15x2) - (10.8xd) = 0

A couple of things. First, I'm not sure why you choose negative terms to sum (the minus signs between terms). Why not simply add all the torques? You can multiply the equation through by -1 and it remains true. Second, you're employing the magnitude of the equilibrant force again. You want to use only its component that will result in torque -- it's y-component, which is perpendicular to the tool.

And then,

(-8.66x0) - (-10.6x4) - (-5x2) - (15x2) = 22.4 (I'm not sure if this is to be kN or Nm)

It would be kNm. kN times meters.

22.4 - 10.8d

d = 22.4 / 10.8

d = 2 (Nm?)

The above doesn't make sense as you are using the magnitude of the equilibrant again.

Fix the first equation for M and use it to solve for d.

Sorry for not getting back sooner.

No worries. I don't have any deadlines for homework

 

[JMxBelfast]

A couple of things. First, I'm not sure why you choose negative terms to sum (the minus signs between terms). Why not simply add all the torques? You can multiply the equation through by -1 and it remains true. Second, you're employing the magnitude of the equilibrant force again. You want to use only its component that will result in torque -- it's y-component, which is perpendicular to the tool.

The equilibrium's y-component? The resultant's y-component was -9.26kN so should the equilibrium's y-component be 9.26kN, since it's in the opposite direction?

M = (-8.66x0) + (-10.6x4) + (-5x2) + (15x2) + (9.26xd) = 0

And then,

(-8.66x0) + (-10.6x4) + (-5x2) + (15x2) = -22.4kNm

-22.4 x -1 = 22.4kNm

22.4 - 9.26d

d = 22.4 / 9.26

d = 2.41

Is this the correct way or is it a shambles again?

 

[gneill]

No shambles; Looks fine. Just be sure to add the units to the final result.

 

[JMxBelfast]

No shambles; Looks fine. Just be sure to add the units to the final result.

Okay thanks, and the units being..? Newton Meters?

 

[gneill]

Okay thanks, and the units being..? Newton Meters?

d is a distance.

 

[JMxBelfast]

d is a distance.

Ah yes fair enough, and is d the position of both the resultant and equilibrant force?

 

[gneill]

Ah yes fair enough, and is d the position of both the resultant and equilibrant force?

It's the position of the equilibrant force. Presumably the resultant force would be positioned (if it had a position) symmetrically opposite on the other side of the center of mass of the tool (I can guess this because of the inherent symmetry in the setup. You'd have to check where the actual center of rotation would occur for the applied forces).

The resultant force isn't something you can point a finger at and say "Here! This is what is applying the resultant and at this location!". It's a mathematical sum of forces scattered over different locations.

 

[JMxBelfast]

It's the position of the equilibrant force. Presumably the resultant force would be positioned (if it had a position) symmetrically opposite on the other side of the center of mass of the tool (I can guess this because of the inherent symmetry in the setup. You'd have to check where the actual center of rotation would occur for the applied forces).

The resultant force isn't something you can point a finger at and say "Here! This is what is applying the resultant and at this location!". It's a mathematical sum of forces scattered over different locations.

Do you think you could use you're great knowledge once again, to help me find the position of the resultant force, it asks in the question to find it and I'm not sure what to do. It also says in the question that "the weight of the tool can be assumed to act through its center" if that is relevant.

 

[gneill]

Ask yourself this question: If the tool was floating by itself in space with no forces acting on it except for the equilibrant and one we'll call the resultant, with the equilibrant being located where you previously calculated it to be, where would you place the resultant so that no motion (translation or rotation) would occur?

 

[JMxBelfast]

Ask yourself this question: If the tool was floating by itself in space with no forces acting on it except for the equilibrant and one we'll call the resultant, with the equilibrant being located where you previously calculated it to be, where would you place the resultant so that no motion (translation or rotation) would occur?

That's a good way to put it to be honest, I'm guessing 2.41m on the opposite side of the equilibrant?

So basically what you said but dumbed-down for myself..

 

[gneill]

That's a good way to put it to be honest, I'm guessing 2.41m on the opposite side of the equilibrant?

Yes, that's where I'd put it. Since we weren't given any information about the 3D dimensions of the tool we have to treat it like a slender rod, and all forces effectively act on the axis of the rod. So conceptually you have a situation like this:

 

[JMxBelfast]

Yes, that's where I'd put it. Since we weren't given any information about the 3D dimensions of the tool we have to treat it like a slender rod, and all forces effectively act on the axis of the rod. So conceptually you have a situation like this:

View attachment 109203

Ah okay thanks a bunch, that's basically me done that whole question now, wouldn't have made it without you to be honest so thank you very much, and did you draw that diagram just to show me? If so I applaud your dedication to helping someone understand something :)

 

[gneill]

Ah okay thanks a bunch, that's basically me done that whole question now, wouldn't have made it without you to be honest so thank you very much, and did you draw that diagram just to show me? If so I applaud your dedication to helping someone understand something :)

You're welcome.

Yes, I made the drawing. Took a couple of seconds using Visio, so very much worth the effort if it helps makes things clear.

 

[JMxBelfast]

You're welcome.

Yes, I made the drawing. Took a couple of seconds using Visio, so very much worth the effort if it helps makes things clear.

I'll have a wee look into it, thanks again bud