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.