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

[Saitama]

Homework Statement

Four similar rods of uniform density are connected with frictionless hinges, and this frame is placed to a horizontal smooth tabletop, such that its shape is a square. Vertex P is acted upon by a horizontal force in the direction of the diagonal, and due to this force it begins to move at an acceleration of a_P. What is the acceleration of the opposite vertex Q at which it begins to move?

Homework Equations

The Attempt at a Solution

I have no idea how to begin with this question. I can't visualize how the body would move when only one one of the vetex is given an acceleration.

Any help is appreciated. Thanks!

 

[SammyS]

Homework Statement

Four similar rods of uniform density are connected with frictionless hinges, and this frame is placed to a horizontal smooth tabletop, such that its shape is a square. Vertex P is acted upon by a horizontal force in the direction of the diagonal, and due to this force it begins to move at an acceleration of a_P. What is the acceleration of the opposite vertex Q at which it begins to move?

Homework Equations

The Attempt at a Solution

I have no idea how to begin with this question. I can't visualize how the body would move when only one one of the vertex is given an acceleration.

Any help is appreciated. Thanks!

Each rod has mass (inertial). Each rod has a moment of inertia (rotational inertia).

Draw a free body diagram for each rod.

 

[Saitama]

Sorry for being so late.

Draw a free body diagram for each rod.

I don't really have a clue what forces are acting on each rod.
The vertex P has acceleration in the left direction. So the force at P acts in the left direction. It rotates about the top most point but that point isn't fixed. After some time it will also move. There are many things going on simultaneously which is confusing me.

 

[haruspex]

By symmetry, you can parameterise the movement by a single angle.
There are three unknown forces. The forces between the rods at Q are clearly in the y direction, likewise at P. The two at the other joints may be in any direction, but are mirror images of each other. In terms of these forces and angles, you can write the equations of motion of the rods. Throw in the fact that two rods must move in the same trajectory where they meet and see what you get.

 

[Saitama]

By symmetry, you can parameterise the movement by a single angle.
There are three unknown forces. The forces between the rods at Q are clearly in the y direction, likewise at P.

How?
The force at P is in the left direction (or x direction).

 

[haruspex]

How?
The force at P is in the left direction (or x direction).

The applied force is in the x direction, but the two rods there will move in the same way in that direction, so the force between the two rods is in the y direction.

 

[Saitama]

The applied force is in the x direction, but the two rods there will move in the same way in that direction, so the force between the two rods is in the y direction.

I still don't get it, can you please show it through a diagram?

 

[haruspex]

Label the other two vertices R, S.
At P, rods PR, PS are each subjected to a force F/2 towards Q. They also exert a force G normally to this on each other. Suppose that's a tension, so it acts towards the line PQ on each. Then there'll be a force parallel to PQ acting between PR and QR at R. Probably a compression - call its magnitude H. Similarly at S. Finally, a force K parallel to RS at Q between QR and QS.
Each of the four points accelerates. R and S will accelerate equally and oppositely. In terms of the accelerations a_P and a_R, you can write the linear and angular accelerations of rod PR. Likewise the other rods.
You can now write out the two linear and one angular force/acceleration equations for each rod. By symmetry, it's only worth doing this for, say, PR and QR. Pls try to to write those out.

 

[Saitama]

Have I mentioned the forces correctly?

Then there'll be a force parallel to PQ acting between PR and QR at R. Probably a compression - call its magnitude H. Similarly at S. Finally, a force K parallel to RS at Q between QR and QS.

How did you figure out these two forces?
Wouldn't there be any force in the y-direction at R? Why there is no horizontal force acting at Q?

Each of the four points accelerates. R and S will accelerate equally and oppositely. In terms of the accelerations a_P and a_R, you can write the linear and angular accelerations of rod PR. Likewise the other rods.

About what point do I need to take moments? R and S are not fixed points so I don't think I can calculate moments about those points.

 

[haruspex]

In terms of your diagram, F acts to the left on the whole system, F/2 on each of PR, PS.
G acts vertically, down on PR, up on PS. This is a force that PR, PS exert on each other. By symmetry, it can only be straight up-and-down.
H acts to the left on QR, QS and to the right on PR, PS.
K acts vertically, down on QR, up on QS. This is a force that QR, QS exert on each other. By symmetry, it can only be straight up-and-down.
(Note: G and K might turn out negative.)

About what point do I need to take moments? R and S are not fixed points so I don't think I can calculate moments about those points.

You're only concerned with instantaneous acceleration, so the points are effectively fixed.

 

[Saitama]

In terms of your diagram, F acts to the left on the whole system, F/2 on each of PR, PS.
G acts vertically, down on PR, up on PS. This is a force that PR, PS exert on each other. By symmetry, it can only be straight up-and-down.
H acts to the left on QR, QS and to the right on PR, PS.
K acts vertically, down on QR, up on QS. This is a force that QR, QS exert on each other. By symmetry, it can only be straight up-and-down.
(Note: G and K might turn out negative.)

You're only concerned with instantaneous acceleration, so the points are effectively fixed.

Please confirm that I have mentioned the forces correctly.
After your confirmation, I will begin making equations.

Thank you for your help so far. :)

 

[haruspex]

Please confirm that I have mentioned the forces correctly.
After your confirmation, I will begin making equations.

That is indeed what I wrote, but I realize now I did make one mistake. At junctions R and S there will be a vertical force as well. It will be upwards on RQ, SP and downwards on RP, SQ. Sorry about that.

 

[wrobel]

I tell the students little bit another version of this problem. It is not very nice that a problem drifts in the internet for years but a solution can not be found anywhere
see the attachment

 

[haruspex]

Fwiw, here is a solution using elementary force and torque methods:

Let the points be PRQS, starting where the force F is applied on the right and working anticlockwise.
p the acceleration of P, q the acceleration of Q, positive left.
Each rod is length 2√2, for convenience, mass 1.
Fx is X axis force at R and S (positive right on PR, PS), similarly Fy (positive down on PR, etc.)
Fp is Y axis force between PR and PS. Similarly Fq.
The accelerations of R are (p+q)/2 in X, (p-q)/2 up in Y. The midpoint of PR accelerates (3p+q)/4 left, (p-q)/4 up, etc.
If the magnitude of the angular acceleration of each rod is α=(p-q)/4. (Consider vertical acceleration of R in relation to P and the rotation.)
MoI of a rod is I=2/3.

Rod PR:
F-Fx=(3p+q)4 ... [1]
Fp-Fy=(p-q)/4 ... [2]
Iα=F+Fx-Fy-Fp ...[3]
[2+3] Iα+(p-q)/4=F+Fx-2Fy ... [4]
[1+4] = 2Fx-2Fy+(3p+q)/4 ... [5]
So Iα = 2Fx-2Fy+(2p+2q)/4 ... [6]

Rod RQ:
Fx=(p+3q)/4 ... [7]
Fy-Fq=(p-q)/4 ... [8]
Iα=Fx+Fy+Fq ... [9]
[8+9] = Fx+2Fy-(p-q)/4 ... [10]
[7+10] =2Fy+q ... [11]

[6+7] Iα = -2Fy+(p+2q) ... [12]
[11+12] 2Iα = p+3q ... [13]
Substituting for I and α
[13] (p-q)/3=p+3q ... [14]
q = -p/5

 

[wrobel]

Fwiw, elementary solution is presented at the end of the paper

 

[Adesh]

At junctions R and S there will be a vertical force as well. It will be upwards on RQ, SP and downwards on RP, SQ.

Can you please tell me how those vertical forces are caused? I understood the origin of the force $H$ (it’s the resultant of the normal force) but how the vertical forces are caused? Please help.

 

[haruspex]

Can you please tell me how those vertical forces are caused? I understood the origin of the force $H$ (it’s the resultant of the normal force) but how the vertical forces are caused? Please help.

I should have worded that differently. There could be a vertical component to the reactions at R and S. We don't need to worry about which way they act, we just need to allow for them.
What I wrote was a bit of a guess, and I could well be wrong.

 

[haruspex]

... or can we apply the Principle of Least Action? I.e. argue that the system moves so as to minimise the rate of gain of KE?
In time Δt, PR gains KE in the X direction of ½m((3p+q)Δt/4)².
Meanwhile, RQ gains KE in the X direction of ½m((p+3q)Δt/4)².
Both gain ½m((p-q)Δt/4)² in the Y direction and rotationally ½I(αΔt)²=⅓m(LαΔt)².
As noted in post #14, Lα=(p-q)/4.
Adding these up:
$\Delta KE=\frac 12m(\frac 14\Delta t)^2((3p+q)^2+(p+3q)^2+2(p-q)^2+\frac 43(p-q)^2)$
$=\frac 16m(\frac 14\Delta t)^2(40p^2+16pq+40q^2)$
Minimising wrt q gives q=-p/5, as before.

 

[Adesh]

@haruspex The direction of normal forces has quite confused me (I’m in one of those states of mind, which I mentioned earlier, where I began to object every clause). Consider this diagram of our actual question:

In this wikipedia article, it is written that normal force always acts in a direction so as to restrict one object to move into the other.

If we follow the above definition for the direction of normal force, then let’s analyse the normal force on the junction $R$. The rod PR won’t allow the rod QR to move into it, therefore, one component of normal force will be along $\vec{RQ}$, similarly RQ won’t allow PR to move into it, therefore, the other component of the normal force will be along $\vec{RP}$. Adding these two components will give us the result normal force at R in the direction of $\vec{RS}$.

If I follow that same definition for the direction of normal forces, I would get this diagram:

Which is quite against what you wrote here:

At P, rods PR, PS are each subjected to a force F/2 towards Q. They also exert a force G normally to this on each other. Suppose that's a tension, so it acts towards the line PQ on each. Then there'll be a force parallel to PQ acting between PR and QR at R. Probably a compression - call its magnitude H. Similarly at S. Finally, a force K parallel to RS at Q between QR and QS.

Please resolve this discrepancy.

 

[haruspex]

it is written that normal force always acts in a direction so as to restrict one object to move into the other.

That assumes there is no linkage. In the present problem, R, as one end of PR, cannot move in any direction relative to R as one end of QR.
It may be clearer if you model the joint as a pin (normal to the page) on PR inserted in a ring on QR. the normal force between pin and ring can be in any direction.

Besides, even with the logic you gave you reached the wrong conclusion. Wrt keeping PR out of QR, the force QR exerts on PR would be in the direction QR, so the force PR exerts on QR would be in the direction RQ. Wrt keeping QR out of PR, the force PR exerts on QR would be in the direction PR. If these are of the same magnitude then the net force PR exerts on QR would be in the direction PQ, not RS.

 

[wrobel]

I.e. argue that the system moves so as to minimise the rate of gain of KE?

I thought that the principle of least action is about $\int T-V$

 

[etotheipi]

I thought that the principle of least action is about $\int T-V$

But in this case isn't $V=0$? Since we neglect gravitational or other conservative interactions between the bodies?

 

[wrobel]

But in this case isn't V=0V=0? Since we neglect gravitational or other conservative interactions between the bodies.

how can we neglect the force F?

 

[etotheipi]

how can we neglect force F?

Ah, you got me. Though I thought in your proof you introduced it as a generalised force and not a potential energy? I know very little about the calculus of variations, so I take your word to be true!

 

[wrobel]

Though I thought in your proof you introduced it as a generalised force and not a potential energy?

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

 

[etotheipi]

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 Larrange equations with $L=T-V$ and the right-hand side vanished

Never knew you could do this! Thanks! I'll try and read up about it.

 

[etotheipi]

If we follow the above definition for the direction of normal force, then let’s analyse the normal force on the junction $R$. The rod PR won’t allow the rod QR to move into it, therefore, one component of normal force will be along $\vec{RQ}$, similarly RQ won’t allow PR to move into it, therefore, the other component of the normal force will be along $\vec{RP}$. Adding these two components will give us the result normal force at R in the direction of $\vec{RS}$.

@Adesh there are two ways I like to think about joints/hinges. The first is that it's just a means of the two rods exerting a pair of contact forces on each other, which are necessarily equal and opposite. This doesn't constrain the possible directions, and I quite like @haruspex's idea of a pin inside a ring for this. Sometimes by symmetry considerations you can simplify things (i.e. eliminate the possibility of certain components), but a priori you can make no assumptions.

The second is in an engineering-style "method of joints" approach, where so long as the structure is a truss (only axial forces) in static equilibrium you can resolve forces on the joint to be zero. If the joint is accelerating, this approach no longer works since we have no idea what the "mass" of the joint is. You need to be much more careful with this method, since we have to assume the struts are massless (so that it is a two force member) for the contact forces to be parallel to the struts.

So in this problem only the first method applies, and we just end up with essentially 16 internal contact forces, 4 on each rod.

 

[Adesh]

Besides, even with the logic you gave you reached the wrong conclusion. Wrt keeping PR out of QR, the force QR exerts on PR would be in the direction QR, so the force PR exerts on QR would be in the direction RQ. Wrt keeping QR out of PR, the force PR exerts on QR would be in the direction PR. If these are of the same magnitude then the net force PR exerts on QR would be in the direction PQ, not RS.

Would you please explain the above quote? I don’t think I quite got you.

 

[etotheipi]

I like this set of Purdue notes; trusses and frames are treated separately. Frames are much more general and there is the possibility of shear stress in addition to axial stress on each member. Truss structures are special cases where every member is is a two force member with no shear stress.

So for any hinge in a frame, the force pairs can point in potentially any direction since there is no restriction on shear components of the contact force. The method of joints is no longer very helpful for frames, then, since its power relies on the directions of the contact forces already being known. The joint is still in equilibrium once you evaluate the forces, though.

I would say try not to think about 'preventing the two rods moving into each other', since it's impossible for me to visualise the orientation of the interface at the joint without doing some work first. All we can say with certainty is that the rods exert equal and opposite forces on each other, which could potentially be in any direction. @haruspex is an expert with infinitely more insight than I can offer, however, so I'll wait to see if he has a hot take on it!

 

[haruspex]

Would you please explain the above quote? I don’t think I quite got you.

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.

 

[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