How Does Force Applied to One Vertex Affect Opposite Vertex in a Square Frame?

[Adesh]

@etotheipi But you haven’t drawn normal forces at the ends. What I’m saying is if we consider this frame:

Forces on PR are:

1. The normal force going in the direction $\vec{SP}$.

2. The normal force going in the direction $\vec{PR}$.

Oho! I got it now! The resulatant would be in PQ direction.

 

[etotheipi]

@etotheipi But you haven’t drawn normal forces at the ends. What I’m saying is if we

I'm not sure what you mean. The orange, blue, red, purple arrows are the normal forces, or if you will, the components of the normal force. The resultant normal force between any two rods can be in any direction.

 

[Adesh]

But @haruspex for each point we have a choice, the opposite direction of force $\mathbf K$ is also possible. It’s more about on which rod we are analysing the normal forces. Will it affect our further analysis?

 

[Adesh]

The orange, blue, red, purple arrows are the normal forces.

Normal forces must be normal to the rods.

 

[etotheipi]

Normal forces must be normal to the rods.

They don't have to be.

Sure, at any interface between two materials the normal force is normal to the interface. That's the definition.

But here we don't know what the interface looks like. As @haruspex said, consider the hinge to be a pin inside a circular ring. The normal force between the rods can then act in any direction.

 

[Adesh]

They don't have to be.

Sure, at any interface between two materials the normal force is normal to the interface. That's the definition.

But here we don't know what the interface looks like. As @haruspex said, consider the hinge to be a pin inside a circular ring. The normal force between the rods can then act in any direction.

But in post #32 you have drawn them perpendicular to the rods.

 

[etotheipi]

But in post #32 you have drawn them perpendicular to the rods.

You can set up your orthogonal basis however you like. I chose to decompose the reaction force in those directions just because it was what you and @haruspex were discussing.

Don't think of the force between the rods as a normal force, just think of it as a reaction force. Two equal and opposite forces on each rod, no prior constraints on the direction. You can split these into components just as you can any other force.

 

[etotheipi]

It's perhaps unintuitive. I think the confusion stems mainly from the fact that it is hard to visualise the hinge. So long as you accept that the hinge is capable of transmitting reaction forces in any possible direction, then there's only a few things you can say. Primarily, Newton's third law must be obeyed. Take a look at this:

 

[haruspex]

yes
I used a version of the Lagrange equations that has no relation to variation principles. But if you wish you can treat F as a constant vector then it is a potential force and you can find corresponding potential energy and obtain variational Lagrange equations with $L=T-V$ and the right-hand side vanished

Are you saying that it is valid to use the Principle of Least Action here? I really don't know, which is why I posted it as a question. It does give the right answer.

Edit: looking at a simpler case, it seems to be that having the acceleration of the point where the force is applied as a given makes this least KE method work.

 

[Adesh]

@haruspex I would like to apologise for asking too much, but please be with me because I’m not a good student .

Can you please tell me if the force $\mathbf G$ is drawn correctly in this post ?

 

[haruspex]

@haruspex I would like to apologise for asking too much, but please be with me because I’m not a good student .

Can you please tell me if the force $\mathbf G$ is drawn correctly in this post ?

There are several things wrong with that diagram. It is unclear what any of the forces is acting on. It needs to be broken into separate FBDs for PR and QR.
G, as I defined it, acts vertically.
But my own post preceding it also has errors.

 

[Adesh]

There are several things wrong with that diagram. It is unclear what any of the forces is acting on. It needs to be broken into separate FBDs for PR and QR.
G, as I defined it, acts vertically.
But my own post preceding it also has errors.

Sir can we begin from beginning? Please guide me the way you were guiding OP.

As far as I contemplated on your advice, the force $\mathbf G$ would act vertically up on the rod PR. This is because we got two normal forces on PR, one in the direction of SP and the other in the direction of PR and hence their sum will be in the direction of SR.

 

[wrobel]

Are you saying that it is valid to use the Principle of Least Action here

yes but the potential energy of F must be employed

 

[haruspex]

yes but the potential energy of F must be employed

So, why does ignoring that give the right answer?
I also tried the trivial case of a force applied normally to one end of a rod. Provided the acceleration of the point where the force is applied is taken as the given (the force being unspecified) , minimising the KE again gives the right answer.

 

[wrobel]

So, why does ignoring that give the right answer?

because that what you use is not the Least Action Principle

 

[etotheipi]

because that what you use is not the Least Action Principle

But in that case, what is being used instead that is causing the right answer to be produced with this method?

 

[haruspex]

because that what you use is not the Least Action Principle

Umm.. that doesn’t explain why it gives the right answer. I could have used any number of things that are not the least action principle. I find it hard to believe it is just coincidence. Have I chanced upon some other (named) principle?

 

[wrobel]

I do not think that argument like "right answer" can prove anything by itself

 

[haruspex]

I do not think that argument like "right answer" can prove anything by itself

No, but it is strongly suggestive after it worked in two cases.