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

[Adesh]

In your reasoning, you added the normal force you expected PR to exert on QR to the normal force you expected QR to exert on PR. Those are not two forces acting on the same object, nor exerted by the same object, so you cannot add them to get anything meaningful.
To get two addable forces you have to turn one of them around into its equal and opposite reaction. When you do that, you discover the resultant is parallel to PQ, not RS.

But if I follow that rule, I’m getting the wrong direction for the force $\mathbf G$.

Considering our usual frame : the force SP will exert on PR to keep it out is in the direction SP, therefore the force PR will exert is in the direction PS. The force that PR will exert on PS to keep it out will be in direction RP, adding those two forces (as they are exerted by the same object) we have $$\hat{PS} + \hat{RP}= \hat{RS}$$ But according to the diagram in post #9 (to which you have certified) my conclusion about G is wrong.

 

[etotheipi]

Here are the two rods, hinge not shown. The two pairs of contact forces are as shown in red/orange, and their resultants in green. Note that the red and orange components are not necessarily of equal magnitude! Any combination of magnitudes is initially possible. Though in the case that both components happen to have the same magnitudes, the resultants are indeed horizontal if the rods are initially at 45 degrees.

 

[Adesh]

Here are the two rods, hinge not shown. The two pairs of contact forces are as shown in red/orange, and their resultants in green. Note that the red and orange components are not necessarily of equal magnitude! Any combination of magnitudes is initially possible. Though in the case that both components happen to have the same magnitudes, the resultants are indeed horizontal if the rods are initially at 45 degrees.

View attachment 261808

I really admire your artistic work :)

I was talking about the rods PR and SP. Please draw the diagrams for them, it’s a request.

 

[etotheipi]

Here's what I would do. On each I've shown the external forces of F/2, and the internal forces as smaller arrows.

The red $x$ components have to be zero because of top/bottom symmetry. I.e. we can't favour any particular side to have the forward force and the other to have the backward force, so they both have to be zero.

Similarly, we know that the blue and purple forces on the top rod will be the same as the blue and purple forces on the bottom rod, again but symmetry. But the magnitude of the blue doesn't necessarily equal that of the purple.

From then on, you can start writing down equations to try and figure out what some of them are!

 

[wrobel]

By the way, if we keep pulling the frame with constant force F then it will behave like the mathematical pendulum. The angle $\alpha$ from the paper satisfies the mathematical pendulum equation. Even asymptotic solutions will be present.

 

[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.