Simple question.

[nicbordeaux]

Considering Newton's 3rd law, in the drawing below, where a rotating wheel with a weight on extension arm is mounted on a pendulum arm, if the arm is locked into place, and the wheel+weight is rotated 90° from bdc 180° and locked, is total F greater than in the "all vertical" scenario ?

[agor95]

Considering Newton's 3rd law

'Force has an equal and opposite force'

The mass of the wheel+weight are the same so the 'F' force is the
same in the down direction.

There is torque on the pendulum arm mounting in the second case.

That is because the center of gravity on the wheel+weight is slightly to the right.

The pendulum searches to place the COG directly below the mounting.

P.S. all this is in a static case. When you go dynamic it gets complex.

Regards

[Fletcher]

Hi nic .. long time.

What about a thought experiment that you could actually real world test to get your answers.

Imagine you are standing on your very accurate bathroom scale (you know, the one you hate coz it lies).

You stand tall and straight (and suck in the six-pack) and at chest level you let the pendulum hang vertically downwards in your hands without touching anything. The pendulum is in the configuration where the red weight is at 6 o'cl. You get your wife to read the scales. It of course measures only weight force. There is no rotational forces on the end of the pendulum you are holding so that doesn't matter.

Next you reconfigure the pendulum to the second state with the red weight at 3 o'cl and hop back on again (you had to expend some energy to lift that red weight from 6 to 3 o'cl but that is incidental).

Once again you hold the end of the pendulum shaft again at chest level. Dang, it want's to move to the side and find its level of lowest GPE. That means its CoG is vertical etc. Your wife reads off the same weight force.

Now you move the pendulum shaft to vertical by force. It's a bit of a strain on the wrists to hold it there coz of the torque its offset CoM/CoG causes. It wants to rotate and find the position of lowest GPE where you are both comfortable. But being strong and fearless you hold out. Your wife observes that the scales still read the same.

But you think how can that be. I can feel the torque force in my wrists and that must be acting at my feet. Yes, but the scales don't measure the torque your feet feel.

I reserve the right to be wrong - plz don't supply photos ;7)

[nicbordeaux]

Thx. Yep, it's been a while, got a bit involved in politics and stuff. Complete waste of time, energy, money.

I agree w/ you 100 % Fletch. The weight registering on the scales is identical whether the OB weight is at 6, 3 or 12.

At 6 , it takes a given effort to rotate that wheel to 12 and gain GPE. You move it to 12 because that's where you'd usually start a OB wheel rotating.

Moving it to 3 requires 1/2 of the energy or F to get 50% of the GPE available at 12 . The offset weight seeking to reach a new position translates as F (or torque) which I strain to resist. Same weight, as agreed.

If the pendulum arm is constrained during all this rotating to different poisitions, that offset at 3 , even if it doesn't register on scales as "vertical weight", and why should it, still exists as F pushing against the restraint (if the wheel is restrained too) .

I read that as 3 = 50% input requirement for 100 % of vertical (downwards) F compared to 12 , but w/ extra torque, strain etc which is available until such time as everything is released and dynamic, in which cases things do get complicated and the Force/"torque" get in each others way, the lateral "F" doesn't naturally help the rotation of the wheel. Unless you have some cunning setup of fishing line over almost frictionless rollers feeding that torque elsewhere etc. Plus if you do use that torque, the Com of the system drops.

IOW, if at 3 you had a weighless rod extending left from the arm to a digital scale which acted as the restraint for the pendulum arm, you's get a positive reading.

I don't need to reserve the right to be wrong, because I'm always wrong. I do however leave open a infinitesimal possibility of being at least partially right by mistake ;-)

It's just the static situation at 3 which tickles me. Identical vertical force, + lateral or rotational force as the system seeks without success to keel.

[nicbordeaux]

PS: Like me, the Wife's eyesight isn't what it once was, so she'd need to be bent down pretty close to those scales I'm standing on, and I can't see how the experiment could possibly be extended long enough in those sort of circumstances.

[agor95]

It's the really simple, in plain sight observations, that are surprisingly tricky.

Regards

[nicbordeaux]

OK, I am going to do something right now: stand on the bathroom scales, read weight. Then lean sideways as much as I can with arm extended. That should be equivalent of an OB offset. The weight reading should be the same, even if the scales aren't Roberval.

Then I move the scales to near a wall and repeat, but when I lean laterally, I will be putting weight on the wall by way of my arm resting against the wall (the equivalent of the pendulum arm restraint).

If things go as I think, the scales will show less weight. That "loss" is what the wall is getting. The total of the two should be equivalent to standing vertical with no wall ?

[agor95]

You might want to take the kitchen scales with you.

So you can see how much pressure you put on the wall.

Also note the arm a 3 0'clock is not contacting a wall so you need
to keep away from friction forces agained the wall.

Good look

[nicbordeaux]

Yep, 71kgs vertical, 71 leaning sideways give or take some miniscule movement of the scale needle due to the way the scales are sensitive to CoM. And 67 with the scales nearer the wall , same amount of lean but hand against wall supporting some or all of the offset body weight. You'd really need scales on the wall to measure exact amount of weight or "F" transferred to wall.

Does all this zero sum adding and subtracting conform with the know Laws of physics ? It should do :-)

[nicbordeaux]

You might want to take the kitchen scales with you.

So you can see how much pressure you put on the wall.

Also note the arm a 3 0'clock is not contacting a wall so you need
to keep away from friction forces agained the wall.

Good look

Oops, just saw this, we agree on the two scales needed. The pendulum arm is restrained by a hook to a "wall" (well, the table) and the wheel at "3" is held in position by way of a fishing line to the pendulum arm.

[agor95]

If you think about it you could have placed one foot on the scales and the other of the floor.

Make sure you have a book to raise your floor foot to the same high as the scales.

Then move your weight from one foot to the other.

What proportion of you weight is on each foot is related to your center of mass location.

The total mass should be the same.

Regards

[Georg Künstler]

Hi Nic,
You have two contact Points, your foot on the ground and the hand on the wall.
If you stand vertical then you can't get any force to the Wall.
If you do so you will fail.
You will fail Back.

Bessler did It different, He used a moving Wall, when you do If now, then you will fail Forward. You win If you loose.

[WaltzCee]

You might be able to eliminate one of the scales if you had a pulley attached to the ceiling,
and a little rope and a 50 pound weight, some duct tape . . .

Nah, don't try that at home. If you want to lose weight, it would be best to move to Mars.