[alancalverd]

use a longer lever. It will make the displacement arclength proportionally longer with the same displacement angle as you apply the same force.

What displacement angle? I'm using the lever to prevent the bucket falling down a well.

[Bored chemist]

If the thread is loose enough,

You specifically said "After a force was applied, the bolt didn't turn. "

[hamdani yusuf (OP)]

NONSENSE, as has been explained to you countless times.

Can you point out the best explanation that you know?
Do you agree that all of the equations in the tables are correct?

Despite all the disagreements on setting the standard units, at least it seems that we have all agreed that every equation I included in the tables in reply#941 is correct.

https://www.thenakedscientists.com/forum/index.php?topic=87006.msg745218#msg745218

Those equations place tight constraints on what units are appropriate for each quantity without violating any of those constraints.

[hamdani yusuf (OP)]

You basically ask to determine whether the jammed thread would break before measuring it first.

I did not ask anything of the sort. That's a misunderstanding on your part.

I pointed out that, if I was stood on the floor, the axis of rotation would be different from if I was sat on the bench.

And you still don't understand, but you still assume I'm wrong.

Where is the axis of rotation in each case?
What makes the difference?

[hamdani yusuf (OP)]

use a longer lever. It will make the displacement arclength proportionally longer with the same displacement angle as you apply the same force.

What displacement angle? I'm using the lever to prevent the bucket falling down a well.

The difference in angle of the lever before and after the force is applied.

[alancalverd]

Usual physics: the bucket weighs mg, frictionless pivots, infinitely flexible weightless rope. 

Lever A just prevents the bucket from falling.

Now replace lever A with a stiffer lever B with twice the thickness.

Whatever negligible "displacement angle" θ_A you had with lever A, is now θ_B ≈ 0.5 θ_A.

According to Hamdani, the torque τ = mgr/θ has doubled, so the bucket should rise up the well!

And as it does so, θ decreases and even becomes negative - infinite free energy!

[paul cotter]

Take care Alan, that bucket will obviously race up the well shaft at ever increasing speed and risks decapitating you.

[hamdani yusuf (OP)]

Usual physics: the bucket weighs mg, frictionless pivots, infinitely flexible weightless rope. 

Lever A just prevents the bucket from falling.

Now replace lever A with a stiffer lever B with twice the thickness.

Whatever negligible "displacement angle" θ_A you had with lever A, is now θ_B ≈ 0.5 θ_A.

According to Hamdani, the torque τ = mgr/θ has doubled, so the bucket should rise up the well!

And as it does so, θ decreases and even becomes negative - infinite free energy!

How is the lever connected to the rope?
Where does the counterbalancing force come from?
Can you draw it?
What happens if there was no lever?

[hamdani yusuf (OP)]

In my unit, the meter represents arc length of the rotational displacement.

So what do you call the quantity that a torque meter measures, or a torque wrench delivers?

If I want to use the same force to apply more torque, do I use a longer lever, or keep the lever the same length and move it further?

You seem confused by the fact that torque is dimensionally identical to energy but is intentionally expressed in Nm, not J, to distinguish it. Most people find it amusing, not confusing.

Torque.

use a longer lever. It will make the displacement arclength proportionally longer with the same displacement angle as you apply the same force.

Do you think Nm has the same dimension as Nm/rad?

It seems like you agree that Nm has the same dimension as Nm/rad.

[hamdani yusuf (OP)]

Standard unit for kinetic energy is kg m^2 s^-2
Standard unit for angular velocity is rad s^-1,

No. Dimensions are not units. The unit of energy is the joule.

Don't you think that
1 Joule = 1 kg m^2 s^-2 ?
5 Joule = 5 kg m^2 s^-2 ?
99 Joule = 99 kg m^2 s^-2 ?

It seems like we can agree that
1 Joule = 1 kg m^2 s^-2
5 Joule = 5 kg m^2 s^-2
99 Joule = 99 kg m^2 s^-2

[hamdani yusuf (OP)]

Take care Alan, that bucket will obviously race up the well shaft at ever increasing speed and risks decapitating you.

It seems like you are hallucinating.

[alancalverd]

Usual physics: the bucket weighs mg, frictionless pivots, infinitely flexible weightless rope. 

Lever A just prevents the bucket from falling.

Now replace lever A with a stiffer lever B with twice the thickness.

Whatever negligible "displacement angle" θ_A you had with lever A, is now θ_B ≈ 0.5 θ_A.

According to Hamdani, the torque τ = mgr/θ has doubled, so the bucket should rise up the well!

And as it does so, θ decreases and even becomes negative - infinite free energy!

How is the lever connected to the rope?
Where does the counterbalancing force come from?
Can you draw it?
What happens if there was no lever?

Surely you've seen a well with a windlass and bucket?

https://thumbs.dreamstime.com/z/old-well-yard-picture-180628993.jpg  but imagine it is designed by a physicist, not built by an engineer.

Let the bucket fall down the well, but not quite to the bottom of the chain. Now pin or tie the handle to the upright so it can't move. According to your analysis, if I just make the handle twice as long, the bucket will rise all by itself!

[Bored chemist]

You basically ask to determine whether the jammed thread would break before measuring it first.

I did not ask anything of the sort. That's a misunderstanding on your part.

I pointed out that, if I was stood on the floor, the axis of rotation would be different from if I was sat on the bench.

And you still don't understand, but you still assume I'm wrong.

Where is the axis of rotation in each case?
What makes the difference?

Re "Where is the axis of rotation in each case? "
You can work that out for yourself.

Re "What makes the difference?".

As I already said...
"if I was stood on the floor, the axis of rotation would be different from if I was sat on the bench."

[hamdani yusuf (OP)]

Usual physics: the bucket weighs mg, frictionless pivots, infinitely flexible weightless rope. 

Lever A just prevents the bucket from falling.

Now replace lever A with a stiffer lever B with twice the thickness.

Whatever negligible "displacement angle" θ_A you had with lever A, is now θ_B ≈ 0.5 θ_A.

According to Hamdani, the torque τ = mgr/θ has doubled, so the bucket should rise up the well!

And as it does so, θ decreases and even becomes negative - infinite free energy!

How is the lever connected to the rope?
Where does the counterbalancing force come from?
Can you draw it?
What happens if there was no lever?

Surely you've seen a well with a windlass and bucket?

https://thumbs.dreamstime.com/z/old-well-yard-picture-180628993.jpg  but imagine it is designed by a physicist, not built by an engineer.

Let the bucket fall down the well, but not quite to the bottom of the chain. Now pin or tie the handle to the upright so it can't move. According to your analysis, if I just make the handle twice as long, the bucket will rise all by itself!

The bucket will be accelerated down by gravity when the windlass is free to rotate. When it's pinned, the bucket will decelerate until it stop. If the rope is elastic, it might bounce up and down for a while until the mechanical energy is dissipated.
Thus the force at the pin depends on the velocity of the bucket and the elasticity of the rope.

[hamdani yusuf (OP)]

You basically ask to determine whether the jammed thread would break before measuring it first.

I did not ask anything of the sort. That's a misunderstanding on your part.

I pointed out that, if I was stood on the floor, the axis of rotation would be different from if I was sat on the bench.

And you still don't understand, but you still assume I'm wrong.

Where is the axis of rotation in each case?
What makes the difference?

Re "Where is the axis of rotation in each case? "
You can work that out for yourself.

Re "What makes the difference?".

As I already said...
"if I was stood on the floor, the axis of rotation would be different from if I was sat on the bench."

If  you stood on the floor, the axis is the foot of the bench.
If you sit on the bench, the axis is the center of the bolt.
Is it correct?

[paul cotter]

In #971 Alan is talking about a static situation with the bucket near the bottom of it's travel. You continually confuse statics with dynamics and this is the root cause of your errors. Because you have included 1/rad in your torque expression the torque will increase without bound as the angle is reduced- hence the bucket will fly up the well shaft at a speed without bound.

[alancalverd]

Same problem with brake pads. Just like the guys who bolt the doors on to Boeing aircraft, Hamdani has no concept of static torque,

[Bored chemist]

If  you stood on the floor, the axis is the foot of the bench.
If you sit on the bench, the axis is the center of the bolt.
Is it correct?

Or the other foot of the bench if I choose to push.
or the middle of the wrench if that's where it happens to break.


My point is that you can not say you know what the torque is unless you state what point you are measuring it about.

[hamdani yusuf (OP)]

In #971 Alan is talking about a static situation with the bucket near the bottom of it's travel. You continually confuse statics with dynamics and this is the root cause of your errors. Because you have included 1/rad in your torque expression the torque will increase without bound as the angle is reduced- hence the bucket will fly up the well shaft at a speed without bound.

If you imagine unphysical situation, don't be surprised if you get unphysical numerical results.

[hamdani yusuf (OP)]

Same problem with brake pads. Just like the guys who bolt the doors on to Boeing aircraft, Hamdani has no concept of static torque,

It's just a torque that's counterbalanced.
What do you think is its difference from canceled torque?