Conceptual Problem - Static Equilibrium

[daphnelee-mh]

Homework Statement

The truss is hoisted using cable ABC that passes through a very small pulley at B. If the truss is placed in a tipped position, show that it will always return to the horizontal position to maintain equilibrium.

Relevant Equations

ΣFy=0

I am not sure whether I can prove like this, I am trying to explain it with the angle .

 

[Lnewqban]

It seems to me that something is off here.
Perhaps I am wrong, but I believe the pulley, if of little friction, will prevent the truss from naturally returning to a horizontal position.
If the center of mass of the truss can relocate itself to a lower position, it will.

 

[daphnelee-mh]

It seems to me that something is off here.
Perhaps I am wrong, but I believe the pulley, if of little friction, will prevent the truss from naturally returning to a horizontal position.
If the center of mass of the truss can relocate itself to a lower position, it will.

but the question needs us to prove that it will return to equilibrium

 

[Lnewqban]

but the question needs us to prove that it will return to equilibrium

Exactly the problem.
Let's wait for some input from mentors of this forum; they will help you for sure.
Again, I may be wrong.

 

[ProxyMelon]

You are dealing with a rigid body, so you should consider where the forces are acting.

In this problem it looks like the center of mass of your bar is going to stay fixed during the motion, that could help.

 

[haruspex]

It is unclear what your tipped diagram represents. There are two degrees of freedom: the arrangement (including the cable above the pulley) may swing to one side as a unit, with no cable moving through the pulley; and the cable can move through the (stationary) pulley with the bar's mass centre remaining below it. In general, any combination of the two.
Your diagram as drawn seems to represent both aspects, but your equations are only correct for the case of swinging as a unit. E.g. consider the case where so much cable has moved through the pulley that the bar hangs vertically. What do your equations give for that?

Now, I am not entirely certain what the question intends. The careful statement that there is a pulley implies you are to consider cable moving through it. Whether you are also supposed to consider swinging is less clear.
I would argue that the two aspects are largely independent, and the tendency of the swing to keep returning to the centre is obvious and well known, so here we are only concerned with movement of the cable through a stationary pulley.
Hint: think about potential energy.

Trouble is, if I am right about all that, what you are asked to prove is false!

 

[daphnelee-mh]

It is unclear what your tipped diagram represents. There are two degrees of freedom: the arrangement (including the cable above the pulley) may swing to one side as a unit, with no cable moving through the pulley; and the cable can move through the (stationary) pulley with the bar's mass centre remaining below it. In general, any combination of the two.
Your diagram as drawn seems to represent both aspects, but your equations are only correct for the case of swinging as a unit. E.g. consider the case where so much cable has moved through the pulley that the bar hangs vertically. What do your equations give for that?

Now, I am not entirely certain what the question intends. The careful statement that there is a pulley implies you are to consider cable moving through it. Whether you are also supposed to consider swinging is less clear.
I would argue that the two aspects are largely independent, and the tendency of the swing to keep returning to the centre is obvious and well known, so here we are only concerned with movement of the cable through a stationary pulley.
Hint: think about potential energy.

Trouble is, if I am right about all that, what you are asked to prove is false!

Ya, I think I did wrong here.
We are asked to only use
ΣF=0
ΣFx=0
ΣFy=0
but I have no idea about it, whether have to prove it in terms of the angle or anything else

 

[haruspex]

Ya, I think I did wrong here.
We are asked to only use
ΣF=0
ΣFx=0
ΣFy=0
but I have no idea about it, whether have to prove it in terms of the angle or anything else

Then draw your diagram with the beam tilted but its centre directly below the pulley.

 

[daphnelee-mh]

Then draw your diagram with the beam tilted but its centre directly below the pulley.

centre directly below the pulley?

 

[haruspex]

centre directly below the pulley?

Yes, the centre of the beam vertically below the pulley, but with the beam tilted. What is the problem?

 

[daphnelee-mh]

Yes, the centre of the beam vertically below the pulley, but with the beam tilted. What is the problem?

Not sure whether I misunderstood your meaning

how can I use the formula to prove it ?

 

[haruspex]

Not sure whether I misunderstood your meaning
View attachment 266519
how can I use the formula to prove it ?

Write in θ and φ for the two angles, as you did originally, and write expressions for the two torques about the centre of the beam.

 

[daphnelee-mh]

Write in θ and φ for the two angles, as you did originally, and write expressions for the two torques about the centre of the beam.

Is this what you mean?

 

[haruspex]

No. The angle formed by the two sections of cable has changed, so the angles they make to the beam will not have the same total as before.
If you want to keep theta for the horizontal arrangement, call the tilted angles φ and ψ.

 

[daphnelee-mh]

No. The angle formed by the two sections of cable has changed, so the angles they make to the beam will not have the same total as before.
If you want to keep theta for the horizontal arrangement, call the tilted angles φ and ψ.

but also the Tension of AB and BC will not longer the same right? I saw another explanation that the Tension BC is assumed as 0 since the cable slacks so the W will have a clockwise moment which cause it to return equilibrium

 

[haruspex]

but also the Tension of AB and BC will not longer the same right?

It is all one cable, passing through a pulley. How could the tension not be the same?

 

[Lnewqban]

but the question needs us to prove that it will return to equilibrium

Daphne,
I have been researching a little about this problem.
As I have told you earlier, that concept of using a small pulley for a cable hoisting a truss is something that I have never seen in many years of working in construction and I believe it can't work.

Any type of sling has eyes at both extremes for connecting it to the hook of the crane or to shackles or to any other rigging hardware.
One of the purposes of those eyes is to avoid accidental tipping of the loads and/or destructive slides of the sling over metal.

Please, see:
https://www.mazzellacompanies.com/R...est-sling-hitch-vertical-basket-choker-bridle

Again, once the load is accidentally placed in a tipped position and that pulley has allowed the cable to slide through it, making leg BA longer than leg BC (as shown in your solution), the load can't return to the horizontal position by itself.

For any tipped position of the truss, its center of mass will be located directly below the hook of the crane, along a vertical line.
If we assume that the CM is located midpoint of the truss, we could call point D to the intersection of that vertical line and side AC of your triangle ABC.

We can consider the distance between vertex B (location of pulley) and point D to be the median of the triangle.
Apollonius's theorem seems to demonstrate that the median of the triangle will increase as lengths of sides BA and BC become different.

$AD=\sqrt{[(AB^2+AC^2)/2]-BD^2}$

That will relocate the CM of the truss lower (respect to its position when truss was horizontal), to a point of less potential energy.
With the hook keeping its spatial position, the extreme situation could be a close to vertical position of the truss and the lowest location of CM.
Because of that, additional external work would be necessary to return the truss to the horizontal position.

Please, see:
https://en.wikipedia.org/wiki/Median_(geometry)

If you have time, also take a look at page 53 (sling configurations) of the following document:
https://www.oshatrain.org/courses/pdf/HoistingRigging_Fundamentals.pdf

 

[haruspex]

Homework Statement:: The truss is hoisted using cable ABC that passes through a very small pulley at B. If the truss is placed in a tipped position, show that it will always return to the horizontal position to maintain equilibrium.
Relevant Equations:: ΣFy=0

View attachment 266159
I am not sure whether I can prove like this, I am trying to explain it with the angle .

This is a duplicate of https://www.physicsforums.com/threads/to-check-prove-that-it-will-return-to-equilibrium-from-the-tipped-position.991402/.
Why have you created a new thread? Do you disbelieve what I have been telling you? I see you are getting the same opinion from @Lnewqban .

Please do not proliferate threads.

 

[PeterDonis]

Moderator's note: Posts from a duplicate thread on the same question have been merged into this thread. The duplicate thread has been deleted.

 

[Lnewqban]

This is a duplicate of https://www.physicsforums.com/threads/to-check-prove-that-it-will-return-to-equilibrium-from-the-tipped-position.991402/.
Why have you created a new thread? Do you disbelieve what I have been telling you?
...

I don't believe that the OP has not been appreciative of your excellent advice, haruspex.
Please, note the date of the first OP (old thread): the thread that contains your posts was created more recently.

Please, excuse me for any confusion that my last response to the old thread may have caused.
It is my fault, I should have posted in the more recently created thread instead, but I had initiated a draft in the old one several days ago, and finished it today, not paying proper attention to avoiding duplication.

Again, my sincere apologies to moderators, to OP and to you.

 

[haruspex]

I don't believe that the OP has not been appreciative of your excellent advice, haruspex.
Please, note the date of the first OP (old thread): the thread that contains your posts was created more recently.

Please, excuse me for any confusion that my last response to the old thread may have caused.
It is my fault, I should have posted in the more recently created thread instead, but I had initiated a draft in the old one several days ago, and finished it today, not paying proper attention to avoiding duplication.

Again, my sincere apologies to moderators, to OP and to you.

Ok, thanks for explaining that. I just assumed the thread I saw initially was the original.
Anyway, they are merged now.