Finding Equilibrium in a Weightlifting System: Torque and Statics Explained

[Dave27]

First of all, I will like to say that I'm teaching myself physics because I've never learned it at school. So, please excuse my ignorance and bear with me.

This is not really a homework problem but rather a problem I thought on my own given a situation in real life.

1. Homework Statement
I'm a recreational lifter, so I train and lift weights from time to time. For this problem I would just consider measurements in 2 dimensions or even just one if I consider the other is irrelevant. I have 2 squat stands about 1.5m tall, 6.4cm width, and a barbell about 1.8m long which weights about 6kg. I'll put this bar over the squat stands and put a 20kg plate on one end of the bar as shown in the picture:

Picture 1

As you can see I took some measurements. IRL the base of the stands are 2 buckets filled up with concrete (much wider) each weight about 30kg (I don't think this matters but I'm throwing it out just in case). I've also put point A and B as pivot points. The measurement that says 56.8cm starts exactly at the center of the bar.

I know for a fact that this system will collapse with the weight plate and bar going off to the right side. The question is how much distance I need to move the squat stand at point A for the system to be at equilibrium.

It should look something like this:
Picture 2

For this matter, wouldn't the force of the bar at the left side of point A increase? or is it rather just the torque decreasing because the distance between point A and the Forces on the right side is decreasing. Also, in picture 1, linearly the weight of the bar just cancels out with the stands but what happens with the weight plate? Obviously it doesn't make the bar go through the stands, so, should I view the bar and the plate as whole and just say both objects just exert a normal force on the stands and is really just the torque that would make it fall. I'm pretty confused.

Homework Equations

T = F * d
d being the distance from the pivot point to the force being applied and F of course being the force due to gravity and T for torque.

Linearly also F = m * a

For the system to be at equilibrium the sum of all of the forces must equal 0 same for the torques.

The Attempt at a Solution

Here's the thing, I don't know if I should take both points or just one (any) to solve the problem. I guess I just took point A and see where it leads me.

Let's say m1 = mass of the bar and m2 = mass of the plate

Considering pivot point A on the second picture.

On the right side the forces are applying a negative torque (clockwise)[/B]
T1 = m1g * D1
T2 = m2g * D2
On the left side only the weight of the bar is applying a positve torque but it would be
T3 = m1g * D1

T3 - ( T1 + T2 ) = 0
but T3 = T1 so
-T2 = 0 (?)

That can't be.

Obviously I'm missing an important concept here. Please help.

where D is the distance between each

PD: If you want to know my background on physics (which is non existant) I made a thread here: https://www.physicsforums.com/threads/hello-from-peru.927603/#post-5855350

 

[scottdave]

The way I understand what you are describing, the left end of the 6 kg barbell is rotating up, and the part to the right of B (with the weight) is rotating down.
So what can A do to keep that end of the pole from moving up? I'm assuming that the stands only provide upward force (it has no way to "hold" something down.

 

[Dave27]

The way I understand what you are describing, the left end of the 6 kg barbell is rotating up, and the part to the right of B (with the weight) is rotating down.
So what can A do to keep that end of the pole from moving up? I'm assuming that the stands only provide upward force (it has no way to "hold" something down.

Yeah the stands have no way of holding down anything.

if both forces make the bar rotate clockwise, then somehow the left end of the barbell needs to provide some force going down so there will be a counter clockwise torque that balances it. I'm still clueless to how this would be done, still confused about the other stuff I mentioned.

 

[scottdave]

Since the barbell has mass, it has downward forces from gravity, distributed across its length. I will try to find a good link for you. I am on my phone right now.

 

[Dave27]

Since the barbell has mass, it has downward forces from gravity, distributed across its length. I will try to find a good link for you. I am on my phone right now.

Please do. I don't think it will be as simple as just considering.. say point B and g = 10m/s^2, and then doing something like:
200N * 8.7cm = 60N * Xcm
X = 29cm

 

[haruspex]

Please do. I don't think it will be as simple as just considering.. say point B and g = 10m/s^2, and then doing something like:
200N * 8.7cm = 60N * Xcm
X = 29cm

Not sure where you are getting those numbers from.
The midpoint of the 20kg disk appears to be 0.8+3/2=2.3 cm from the edge of the support. Its moment about that edge is 4.6Nm.
To this, add (26/180)x60x.13=1.13Nm for the section of the bar it is on.
The main part of the bar exerts a torque (1-26/180)x60x(1.8-.26)/2= 39.5 Nm the other way.
In short, the bar weighs a bit under a third of the plate, but its mass centre is over twenty times further from the tipping axis, so how come it collapses?

Anyway, why not put the supports outside the plates?

 

[Dave27]

Not sure where you are getting those numbers from.

Yeah... I think I got a little ahead of myself. lol.

The midpoint of the 20kg disk appears to be 0.8+3/2=2.3 cm from the edge of the support. Its moment about that edge is 4.6Nm. To this, add (26/180)x60x.13=1.13Nm.

Oohhhh... so you HAVE to take a part of the mass that is on the other side. I actually didn't know that. My bad. And the .13 must come from the center of mass of this part of the bar I suppose, which would be 26cm/2 * 1m/100cm.

The main part of the bar exerts a torque (1-26/180)x60x(1.8-.26)/2= 39.5 Nm the other way.

So here (1.8-.26) would be the length in meters of what's left of the bar and you divide it by 2 to get the distance from the center of mass to the pivot point. And then you get the new weight (or force rather) by multiplying 60 by the distance that's left which would be 154/180 (why did you put 1? maybe you meant 100%?)

In short, the bar weighs a bit under a third of the plate, but its mass centre is over twenty times further from the tipping axis, so how come it collapses?

And then of course we get that the torque on the left side is actually bigger than the torque on the right side... But I swear the bar still falls to the right side... what the heck.. there must be something else I'm not considering. I guess I'll have to repeat the experiment, perhaps my measurements weren't accurate enough.

Anyway, I still don't know if it does it make any difference if the stand at 'A' is pushed a little bit further in the right direction while keeping everything else the same.

Plus, I still fail to see the linear equilibrium. Is the vertical force caused by the weight of the bar + the plate canceled out by the normal force on both stands?

 

[haruspex]

why did you put 1? maybe you meant 100%?

100%=100/100=1

does it make any difference if the stand at 'A' is pushed a little bit further in the right direction while keeping everything else the same.

If it were in equilibrium, sliding A left and right would alter the balance of the two normal forces. If you slide it far enough to the right, you might reduce the normal load on B to zero, whereupon the bar would tip to the left. But if it is tipping right then the normal force on A is zero, and no matter how you slide it around that won't change.

Is the vertical force caused by the weight of the bar + the plate canceled out by the normal force on both stands?

By the sum of the normal forces, yes.

 

[Dave27]

100%=100/100=1

If it were in equilibrium, sliding A left and right would alter the balance of the two normal forces. If you slide it far enough to the right, you might reduce the normal load on B to zero, whereupon the bar would tip to the left. But if it is tipping right then the normal force on A is zero, and no matter how you slide it around that won't change.

I completely fail to understand how moving A left or right would alter the balance of the normal forces.

100%=100/100=1
By the sum of the normal forces, yes.

If Fn = normal force then 2Fn = 260N -> Fn = 130N. Correct?

 

[haruspex]

I completely fail to understand how moving A left or right would alter the balance of the normal forces.

If Fn = normal force then 2Fn = 260N -> Fn = 130N. Correct?

No. The normal forces will only be equal if the supports are equidistant from the mass centre of the load.
E.g. consider a uniform beam length 2m with one support at its middle and one support at one end. The support at the end is redundant. All of the load will be on the central support.
More generally, take moments about the mass centre of the load. Only the normal forces exert a torque about that point, so those two torques must balance. If the distances from the mass centre are x₁ and x₂, on opposite sides of the mass centre, then F₁x₁=F₂x₂.

 

[Dave27]

No. The normal forces will only be equal if the supports are equidistant from the mass centre of the load.
E.g. consider a uniform beam length 2m with one support at its middle and one support at one end. The support at the end is redundant. All of the load will be on the central support.
More generally, take moments about the mass centre of the load. Only the normal forces exert a torque about that point, so those two torques must balance. If the distances from the mass centre are x₁ and x₂, on opposite sides of the mass centre, then F₁x₁=F₂x₂.

Hmmm ok. I think I get it now. I was confusing some concepts here. But the example you just put made it clear. However just to be 100% sure, Let's consider an example like in the picture. Let's suppose that this system is on equilibrium. Let T1 be the torque on the left side and T2 be the torque on the right side and M the mass of the beam. then T1 = Mgx₁ and T2 = mgx₂. Plus, the normal force Fn should be equal to (Mg+mg) correct?

 

[Dave27]

Hmmm ok. I think I get it now. I was confusing some concepts here. But the example you just put made it clear. However just to be 100% sure, Let's consider an example like in the picture. Let's suppose that this system is on equilibrium. Let T1 be the torque on the left side and T2 be the torque on the right side and M the mass of the beam. then T1 = Mgx₁ and T2 = mgx₂. Plus, the normal force Fn should be equal to (Mg+mg) correct?
View attachment 212469

Actually, this is wrong I have to consider the distance from B to each end of the object, plus the change in center of mass when evaluating each side. Suppose the distance from B to the end of the right side was L/4 (L being the total length). Also to make it easy for myself let's suppose the distance from m to b was L/8 then:
T2 = m x g x (L/8) + (L/4) x M x g x (L/8)
and T1 = (3L/4) x M x g x (3L/8)

I think this is fine now. But I still think (M+m) x g = Fn

 

[haruspex]

T2 = m x g x (L/8) + (L/4) x M x g x (L/8)
and T1 = (3L/4) x M x g x (3L/8)

Not quite, you have some extra Ls in there.
The mass M is one quarter to the right of B, so its mass contributing to T2 is M/4. That mass is centred L/8 from B so we get (M/4)g(L/8).

If we look at how M contributes to T1 and subtract its contribution to T2 we get MgL((3/4)(3/8)-(1/4)(1/8))=MgL/4. Note this is the same as just regarding M as a single mass acting at its mass centre, which is L/4 left of B. Indeed, this is a general result and allows us to avoid having to consider beams in two parts in a question like this. But when it comes to beam stress analysis the two part approach is needed.

(M+m) x g = Fn

Yes.

 

[Dave27]

Not quite, you have some extra Ls in there.
The mass M is one quarter to the right of B, so its mass contributing to T2 is M/4. That mass is centred L/8 from B so we get (M/4)g(L/8).

If we look at how M contributes to T1 and subtract its contribution to T2 we get MgL((3/4)(3/8)-(1/4)(1/8))=MgL/4. Note this is the same as just regarding M as a single mass acting at its mass centre, which is L/4 left of B. Indeed, this is a general result and allows us to avoid having to consider beams in two parts in a question like this. But when it comes to beam stress analysis the two part approach is needed.

Yes.

Thank you!

One last thing, what resources / books would you recommend me to brush up on these topics?

 

[haruspex]

Thank you!

One last thing, what resources / books would you recommend me to brush up on these topics?

Sorry, can't help with that. There is a lot online these days.

 

[scottdave]

Thank you!

One last thing, what resources / books would you recommend me to brush up on these topics?

You may want to have a look in the textbook section here. https://www.physicsforums.com/forums/science-and-math-textbooks.21/
There are some free books available online. Personally, i also like the Hyperphysics site hosted by Georgia State University. Here is the link. http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

 

[scottdave]

If you like videos, you may want to look at Khan Academy for instructional videos. https://www.khanacademy.org/science/physics
Also, MIT has several courses freely available at https://ocw.mit.edu/index.htm

 

[Dave27]

Sorry, can't help with that. There is a lot online these days.

Oh well...

Today I performed the experiment, and you were right! The bar actually slides off to the left side. What happens is as the bar goes to the left side, it tryes to pull the 20kg plate with it and it starts rolling in the perpendicular direction (if you look at the drawing that would be towards you) and since the pins of the stand aren't very tall (that's my design fault) to prevent that from happening the bar slips in that direction. So, previously I may have thought that it slips to the right side as my ignorant intuition told me. But now I know that this is not the case.

With your help I calculated I have to put about 27kg on the other side for the system to be static (at the same point of the center of mass of the part of the bar). Of course I took into account that each time I put a plate on the bar each plate exerts a different torque because it is at a slightly different distance from the pivot point. Also some plates I have don't have the same thickness and it has nothing to do with how much they weight, that's another thing I would have to considerate if I want a more exact result I guess.

Anyways, with a little help of my brother I was able to verify this was more or less correct. In reality we ended up putting around 30kg on one end to balance everything out.

If you like videos, you may want to look at Khan Academy for instructional videos. https://www.khanacademy.org/science/physics
Also, MIT has several courses freely available at https://ocw.mit.edu/index.htm

Thank you very much.