A very simple moments question

[A.T.]

For small angle the large downward force at the wall contact point and the large upward force at the diagonal rod point both go to infinity in your calculation. Is it OK?

OK in what sense? It's not good engineering to make that angle too small.

 

[anuttarasammyak]

The jig would undertake a big torsion and the screw to attach the jig to the wall would break by the large upward force’s torsion, right?

 

[A.T.]

The jig would undertake a big torsion and the screw to attach the jig to the wall would break by the large upward force’s torsion, right?

What would break first depends on the material properties.

 

[anuttarasammyak]

So anyway the screw undertakes larger torsion caused by larger upward force from the jig in the smaller angle case, right?

 

[A.T.]

So anyway the screw undertakes larger torsion caused by larger upward force from the jig in the smaller angle case, right?

Why torsion?

 

[anuttarasammyak]

I consider three cases shown below, in which one of the three points of the jig is either x-free or y-free.

In cases (a) and (b), one of the two contact points is free. This point moves smoothly along the wall and therefore receives only a force perpendicular to the wall. The fixed point receives the y-force W. The two contact points also receive x-forces F, which form a couple that cancels the couple produced by the y-force W.

Case (c) corresponds to the situation in which the contact point between the horizontal bar and the diagonal bar is free to move in the x-direction. The result is similar to that of the “horizontal-bar-only” model. For the total couple to vanish, the value of F must be the same as in cases (a) and (b).
But due to the separation of the jig by x-free point I do not have an idea how to cancel these two x-forces in balance. I assume x-force could be generatated in a manner that infinite elastic constant multiplied by infinitesimal deformation equals finite F

If the diagonal rod is regarded as being outside the system, a force $W(1+L_2/L_1)$ acts on the x-free contact point and produces no couple. In that situation, any pair of x-forces acting along the same line may appear. Since such forces do not generate a couple, their magnitudes could in principle be arbitrary. However, before dividing the system into two parts, if we look at it from a different viewpoint, the force F should still be present. This suggests that something may have been overlooked when we consider the system consisting of only the horizontal bar.　This raises the question of whether case (c) is related to the “horizontal-bar-only” model, which generates the same x-forces as in cases (a) and (b).

In case (c), the y-forces at the wall contacts are larger and diverge as the angle θ approaches zero. This feature is quite interesting, although it could be a drawback in practical applications.
In contrast, in cases (a) and (b) the case θ=0 is not particularly special compared with other angles, although the rigid jig would have to withstand a large stress.

 

[A.T.]

I consider three cases shown below, in which one of the three points of the jig is either x-free or y-free.

In cases (a) and (b), one of the two contact points is free. This point moves smoothly along the wall and therefore receives only a force perpendicular to the wall. The fixed point receives the y-force W. The two contact points also receive x-forces F, which form a couple that cancels the couple produced by the y-force W.

Case (c) corresponds to the situation in which the contact point between the horizontal bar and the diagonal bar is free to move in the x-direction. The result is similar to that of the “horizontal-bar-only” model.

If by “horizontal-bar-only” model you mean the one by @kuruman, then no. None of your 3 cases is similar to that, because you are apparently using torque-transmitting joints (some of your dots filled black), while @kuruman was using only torque-free hinges (his circles filled white ).

Your overall confusion seems, at least partially, to stem from failure to distinguish those two types of joints.

For the total couple to vanish, the value of F must be the same as in cases (a) and (b).

No, if you are using a torque-transmitting joint between the wall and diagonal rod in (c), then F in (c) does not have to be the same as in (a) and (b), in order to have torque and force balance. And that is the relevant condition, not some "total couple to vanish", whatever that means.

 

[anuttarasammyak]

We divide the system into two sub-systems: the horizontal rod (HR) and the diagonal rod (DR), such that the total external force on each sub-system is zero.

In addition, in the simple horizontal-bar model and in case (c) of my previous post, we assume that the torque on each sub-system is zero:
torque(HR) = 0 and torque(DR) = 0.

Is this assumption justified?
Even if the total torque of the whole system is zero, can it be decomposed into sub-systems that individually have non-zero torques which cancel each other?

If it is acceptable, we will be free from $L_1/L_2$- type limit infinite y-forces.

 

[A.T.]

We divide the system into two sub-systems: ...

Please read post #67. You have to be clear on which of your joints are torque-free hinges, and which are torque-transmitting. When cutting at torque-transmitting joints, you have to consider the torques acting through the joints between the parts, not just the forces.

 

[Steve4Physics]

A couple of thoughts...

Even if the total torque of the whole system is zero, can it be decomposed into sub-systems that individually have non-zero torques which cancel each other?

No. If a ’subsystem’ had non-zero torque, then it would have some angular acceleration.

If it is acceptable, we will be free from $L_1/L_2$- type limit infinite y-forces.

@anuttarasammyak, you don’t have to worry about infinite forces here!

Consider a simple case: a horizontal bar of negligible weight, supported at one end by a hinge and carrying a load $\vec F$ at the other end. Vertical force, $\vec A$, is applied to balance moments about the hinge. (Hinge reaction is $\vec B$.):

Balancing moments about the hinge gives $A = \frac {L_2}{L_1} F$.
This means as $L_1 \to 0$ then $A \to \infty$.

Note that the limiting case ($L_1 = 0$) is not physically meaningful. An infinite force acting with a zero lever-arm is unphysical.

Edited.

 

[kuruman]

Note that the limiting case $(L1=0)$ is not physically meaningful. An infinite force acting with a zero lever-arm is unphysical.

Not quite. The equation says that you need an infinite force to open a door if you push at the hinge. When the location of the hinge is not obvious, to prevent people from pushing at the wrong place (and to avoid unnecessary cussing), doors in public places often have PUSH signs to ensure that a finite force will do the job.

 

[anuttarasammyak]

If by “horizontal-bar-only” model you mean the one by @kuruman, then no. None of your 3 cases is similar to that, because you are apparently using torque-transmitting joints (some of your dots filled black), while @kuruman was using only torque-free hinges (his circles filled white ).

Yes, my black point means hard fixed for an example by welding. I am not familiar with torque-free hinges. By replacing welding with these hinges can we change forces and torques in the static configuration ?

 

[Steve4Physics]

Not quite. The equation says that you need an infinite force to open a door if you push at the hinge. When the location of the hinge is not obvious, to prevent people from pushing at the wrong place (and to avoid unnecessary cussing), doors in public places often have PUSH signs to ensure that a finite force will do the job.

Here in the UK, building insurance often excludes cover for damage caused by infinite forces. For this reason, all our builders and carpenters now receive mandatory physics training in order to reduce the risk of incorrectly positioned PUSH (and PULL) signs.

 

[anuttarasammyak]

No. If a ’subsystem’ had non-zero torque, then it would have some angular acceleration.

I agree that they will turn if we separate them free. When two opposite motors give opposite torques to the connected rod, the rod does not turn. If we cut the rod in half, these halved rods turn with one plus the other minus way.

 

[A.T.]

I am not familiar with torque-free hinges.

You don't know what a hinge is?

By replacing welding with these hinges can we change forces and torques in the static configuration ?

Yes, of course.

 

[anuttarasammyak]

Yes, of course.

I should appreciate it if you would show it in any of my examples. I really have no idea. Torque free hinge in statics.

 

[anuttarasammyak]

@anuttarasammyak, you don’t have to worry about infinite forces here!

I thought it would be a headache for a carpenter fixing a shelf on the wall. What size of bolt should I use to fix the shelf to stop it goes up, upper side, down, downer side.

 

[Steve4Physics]

I agree that they will turn if we separate them free. When two opposite motors give opposite torques to the connected rod, the rod does not turn. If we cut the rod in half, these halved rods turn with one plus the other minus way.

You are comparing two different systems – not two versions of the same system. So your comparison isn’t valid. I'll try to explain why.

In the initial system, you have two motors $M_1$ (trying to make rod rotate clockwise (CW)) and $M_2$ (trying to make rod rotate anticlockwise (ACW)). The motors exert equal magnitude, opposite sense, moments (torques).

The torques will deform the rod, tending to make a dip in the middle. In response, the rod resists by developing internal forces: the bottom of the rod is in a state of tension and the top of the rod is in a state of compression.

We can consider the left and right sides of the rod (before the cut) and mark the forces acting on each side to show the forces due to the tension and compression:

Note that the net moment on each half of the rod is zero. That's because P and Q exert an ACW moment on the left half of the rod. And P' and Q' exert a CW moment on the right half of the rod.

If the rod is cut at along AA’, the two sections of the rod are free to move.
Forces P, Q, P' and Q' become zero. The net moment on each half of the rod is no longer zero. The system has changed:

Edited several times.

 

[A.T.]

I should appreciate it if you would show it in any of my examples.

In your example (c) from post #66 F is zero (assuming welded wall connections and no friction between the rods,).

If you replace all 3 connections with hinges then F is not zero (see solution by @kuruman).

 

[A.T.]

If we cut the rod in half, these halved rods turn with one plus the other minus way.

"Cutting" in the context of free body diagrams doesn't mean "analyzing would would happen if we actually physically cut the system", because that would change the system we are trying to analyze.

It rather means defining virtual boundaries, and analyzing the forces and moments that act across these boundaries, in the physical system as it is.

 

[anuttarasammyak]

You are comparing two different systems – not two versions of the same system. So your comparison isn’t valid. I'll try to explain why.

Thanks for the nice figures.

In this connected situation which is a correct saying about torque of subsystems?

1 Torque (S1)=Torque(S2)=0

2 Torque(S1) = - Torque(S2) $\neq$ 0

 

[anuttarasammyak]

In your example (c) from post #66 F is zero (assuming welded wall connections and no friction between the rods,).

If you replace all 3 connections with hinges then F is not zero (see solution by @kuruman).

Thank you. I had found faults in my c) study explained in my posts after. So Let me confirm of your teaching in another approach.

We have three joints and the 8 ways to choose welding or torqe-free joint. Do we have 8 (maybe some of them are same, OK) patterns of forces and torque according to the choice ?

[EDIT] Further variation : No joints or X(Y) binded Y(X) free joints as in #66 some of which matter physics.

 

[Steve4Physics]

Thanks for the nice figures.
View attachment 370225
In this connected situation which is a correct saying about torque of subsystems?

1 Torque (S1)=Torque(S2)=0

2 Torque(S1) = - Torque(S2) $\neq$ 0

In the connected situation (S1 and S2 joined) condition 1 applies:
Torque (S1) = Torque(S2) = 0

If any object - even if it is part of a larger object - has zero angular acceleration ($\alpha$), e.g. if the object is stationary, then the net torque ($\tau$) on the object MUST be zero. $\tau \propto \alpha$.

Your condition 2 can only apply after the rod is cut, i.e. when S1 and S2 can move independently.

 

[anuttarasammyak]

Thank you, By your advice I will restate #68 refrain from using the word subsystem.

Red forces act on HR. They produce minus torque. Blue forces act on DR. They produce plus torque. These torques cancel each other.

I hope it is correct and we can alalyse the system by these forces.

X=0 for case (a), X=W for case (b) of #66. Share of load at the two wall contact points decide the parameter X. Usually X, W-X > 0. The DR angle $\theta$ does not matter. No infinite forces come out in usual settings. I hope to share this result with the "only horizontal bar model".

 

[A.T.]

We have three joints and the 8 ways to choose welding or torqe-free joint. Do we have 8 (maybe some of them are same, OK) patterns of forces and torque according to the choice ?

No, because some joint combinations will be underdetermined, yielding infinitely many solutions for forces and torques.

[EDIT] Further variation : No joints or X(Y) binded Y(X) free joints as in #66 some of which matter physics.

When you combine sliding joints with the two types above, some combinations will have unique solutions, some will be underdetermined (infinitely many solutions), and some will collapse (no solution).

Note that even if you don't have a unique solution for all the internal forces in the construct, you still might have a unique solution for the single external force component F, that the original question asks about. That's why it's OK for the original question not to be specific about the joint types used in the construct, because it only asks about F, not about all the forces in the construct.

 

[A.T.]

I hope it is correct and we can alalyse the system by these forces.

Choosing X for the vertical component is not optimal here, as it was used for the horizontal direction throughout the thread.

X=0 for case (a), X=W for case (b) of #66.

Yes, and those joint combinations give unique solutions

Share of load at the two wall contact points decide the parameter X. Usually X, W-X > 0.

The share of vertical load at the two wall contact points can be either determined or indeterminate, depending on the type joints you use.

The DR angle $\theta$ does not matter.

It depends. For example, if all 3 joints are hinges, the diagonal rod angle $\theta$ does matter.

 

[kuruman]

I hope it is correct and we can alalyse the system by these forces.View attachment 370230

Whether it is correct or incorrect, it cannot help analyze the system because you have not defined what the system is. Defining the system sets the boundary between the system and the external world thereby determining the forces that you put in the diagram. Any entity outside this boundary may exert a force on the system and any entity inside may not. Specifically, action-reaction pairs in a free body diagram are superfluous and a source of confusion because when you add them to find the net force on the system, they give zero. Your drawing seems to have included such pairs, but who's to know since you have not defined your system.

Thus, the first three steps for drawing FBDs are
1. Define the system and its boundary.
2. List the physical entities that interact with the system through the boundary and count them.
3. Draw a single arrow representing the force exerted by each entity on the system. Any force may be resolved into component when you write equations later.

For example, in post #13
1. The system is the horizontal rod.
2. There are 3 physical entities interacting with with it (a) the weight placed on it; (b) the wall glued to its left end; (c) the diagonal piece glued between its two ends so that it forms angle $\theta$ with the vertical.
3. Three arrows exerted by each entity are drawn, labeled and color-coded.

At his point one can write the equilibrium equations and find what is being asked, as I did in post #13.

Note that the rod as shown is in equilibrium. If you were to put it in free space at rest and then magically applied the 3 forces with magnitudes and directions as shown, it will remain in equilibrium. (The drawing is to scale using the given values in post #1.)

If you were to choose the horizontal bar and the diagonal piece as your system, then the entities exerting forces on the system are the (a) the wall at two places instead of one and (b) the weight on the horizontal bar. The cross of forces labeled F and X in post #84 do not belong.