Paradox of balance scales

[Mike-from-the-Bronx]

 

The pivot point is at the center of mass of the beam in my example.

Oh. Then I with draw my criticism.

[Janus]

edit to add:

after seeing your model, I noticed that you did not make the lever forces equal. i.e. no equilibrium.

That's because, with the pivot point of the beam being inline with the points from which the pans hang,and unequal weights in the pans there is no equilibrium unless the beam is allowed to reach a vertical position.

That would imply the balance arrm is spinning and never stops spinning.

No, as the arm approaches vertical with the heavier weight being the lower, F1 and f2 extend until they equal Fg1 and Fg2 respectively and Fl1 and Fl2 go to zero. If the scale as I drew it were released from the shown position, the beam will rotate clockwise until it reaches the vertical position. Momentum will carry it past vertical, Fl1 and Fl2 will grow again, but now the net torque will be counter-clockwise. It will slow, stop, reverse and swing back to vertical. It will keep swing back and forth in smaller arcs until friction brings it to rest in the vertical position. This is the equilibrium point for this set up.

 

In order to get the standard beam balance effect where the equilibrium when the weights are equal is with the beam horizontal (or at some less than vertical angle if the weights are unequal), the pivot point of the beam and "hang points" of the pans cannot be on the same line.

[swansont]

The real force is always pointed straight down. However the force that acts on the balance is the lever force which is always perpendicular to the cross bar. if the cross bar is crooked or its not perpendicular to the gravity force then the net force and the gravity force are seperated by a delta force

When the masses are equal then the cross bar is already horizontal and there is no need for a returning force.

I was showing what happens the instant after the unequal masses become equal when the cross bar is not horizontal. then the delta force is the returning force so as the cross bar approaches the horrizontal the delta force approaches zero. when the delta force is zero, then the gravity force is equal to the lever force and they both point straight down but that only happens when the bar is horrizontal.

Last I checked gravity points down. There's a force at the pivot point. What is the source of this additional force?

Oh. Then I with draw my criticism.

 

Thank you.

[TakenItSeriously]

The real force is always pointed straight down. However the force that acts on the balance is the lever force which is always perpendicular to the cross bar. If the cross bar is crooked or its not perpendicular to the gravity force then the lever force and the gravity force are seperated by a delta force

When the masses are equal then the cross bar is already horizontal and there is no need for a returning force.

 

I was showing what happens the instant after the unequal masses become equal when the cross bar is not horizontal. then the delta force is the returning force so as the cross bar approaches the horrizontal the delta force approaches zero. when the delta force is zero, then the gravity force is equal to the lever force and they both point straight down but that only happens when the bar is horizontal.

Last I checked gravity points down.

I highlighted every reference to the gravity force, I made one typo where I said net force instead of lever force which I corrected in red, but I'm not sure hoy how you interpreted anything that is in disagreement with gravity forces pointing down.

 

There's a force at the pivot point. What is the source of this additional force?

This is simple Newtonian physics.

The force the pivot has on the cross bar is the offsetting force that points straight up and offsets the weight of the system which includes the weight of the cross bar two pans the chain or coord that suspends the pans and any extra forces or masses acting on those pans.

Edited July 22, 2017 by TakenItSeriously

[J.C.MacSwell]

 

 

Yes. There's no restoring torque before or after. So moving the CoM is not the impetus.

None of the examples where there is no restoring torque has any movement of the CoM. Laterally or vertically.

I higlighted every reference to the gravity force, I'm not sure hoy how you interpreted anything that is in disagreement with gravity forces pointing down.

 

This is simple Newtonian physics.

The force the pivot has on the cross bar is the offsetting force that points straight up and offsets the weight of the system which includes the weight of the cross bar two pans the chain or coord that suspends the pans and any extra forces or masses acting on those pans.

When the CoM does move (accelerate) laterally, there is an offsetting lateral force as well.

[swansont]

I highlighted every reference to the gravity force, I made one typo where I said net force instead of lever force which I corrected in red, but I'm not sure hoy how you interpreted anything that is in disagreement with gravity forces pointing down.This is simple Newtonian physics.

The force the pivot has on the cross bar is the offsetting force that points straight up and offsets the weight of the system which includes the weight of the cross bar two pans the chain or coord that suspends the pans and any extra forces or masses acting on those pans.

Simple Newtonian physics is that the net force on an object is the sum of the forces acting on it.

 

We have gravity, and a force on the pivot. What is this other force acting ON the balance?

None of the examples where there is no restoring torque has any movement of the CoM. Laterally or vertically.

 

Mine did. Simultaneously add or remove equal masses to/from both pans.

[TakenItSeriously]

None of the examples where there is no restoring torque has any movement of the CoM. Laterally or vertically.

When the CoM does move (accelerate) laterally, there is an offsetting lateral force as well.

It depends on if your adding or removing mass or force. But in the scenario where we remove our hand or remove mass from one pan in order to return the two pans to an equal and symmetrical state, then the CoM has not accelerated it has simply changed location in accordance to the new distribution of mass to somewhere along the vertical center line. Any acceleration is done off system such as removing the mass.

 

However their is a reaction to this change which is the rotation of the cross bar back to horrizontal.

 

Its more useful to compare the the states along points in time.

with Unbalanced forces

The bar is in a slanted state half above and half below the pivot and the tension on the coords are vertical but no longer in allignment with the lever forces that have rotated with the levers. Therefore their is a delta force that fits in the gap head to toe between the lever and gravity forces on both sides of the balance, but it is important to note that the delta forces are pointed in opposite directions. Therefore their is no lateral displacement or acceleration force.

 

With balanced forces at its initial unbalanced, assymetrical state, and rotated state

The instant you unload one side to have equal masses, the delta forces are no longer resisting the unballenced forces and the frame starts rotating back to horizontal to reduce those delta forces.

 

Why the delta forces reduce is because they are now only their due to frame rotation and not to offset the asymetrical loading. The balance wants to return to the state of lowest energy where delta forces are zero the frame rotation is zero and the system is symmetrical.

 

When returning we see the classical overshoot and the two pans will oscilate up and down in a decaying fournier series, just like would happen with a spring.

 

So loading one side of a balance rotates the balance to a greater energy state like a wind up toy. unloading the balance allows the balance to unwind back to its lowest energy state..

Edited July 22, 2017 by TakenItSeriously

[J.C.MacSwell]

 

Mine did. Simultaneously add or remove equal masses to/from both pans.

Right...was one a... duck?