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Can you give some examples of stable, neutral and unstable equilibriums in reality?

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**bob bundy****Moderator**- Registered: 2010-06-20
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hi

There's one solid that can demonstrate all three. A cone.

When it is standing on its base it is stable. If you push the vertex sideways it will lift up but the C of G rises, so when you let go, it drops back down.

When it is on its side it will roll about when you push it, but the C of G stays at the same height, so it is neutral.

If you stand it on its point, it will readily topple over because the slightest movement will cause the C of G to drop after which it continues to drop.

Bob

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**anonimnystefy****Real Member**- From: Harlan's World
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Hi Agnishom

I like the example of a ball on a curved surface.

When the surface is straight, the equilibrium is neutral, because you can move the ball around and it will stay in the place you move it.

When the surface is curved down, the equilibrium is unstable on the highest point of the surface, because if you push the ball just a little bit, it will fall down.

When the surface is curved up, the ball will fall back to the lowest point when you push it, so it's a stable equilibrium.

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Isn't that a metaphor rather than an example? How does it have got anything to do with the sum of resulting forces?

How about this one:

You are lying on your bed: Neutral Equilibrium

You throw a stone upwards, the point when the stone is at its maximum height, it is in unstable equilibrium because it has max potential energy and min kinetic energy

When the bob (not bundy) of a pendulum is halfway through the path of the pendulum, it is in stable equilibrium with max kinetic energy and least potential energy.

Could you please list some examples like this and/or explain these ones in a better way?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bob bundy****Moderator**- Registered: 2010-06-20
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But stones moving through the air are not in equilibrium. Neither is a moving pendulum. What you should be asking is "If I make a small displacement to the stationary object, will it (i) just stay in its new position; or (ii) try to return to its previous position; or (iii) move to a new lower potential position ?

Here's a practical, and important (if you want to avoid pain), example.

I lean a ladder against a wall at about 60 to the horizontal and climb up. I'm in equilibrium. I push against the wall. The ladder moves away from the wall but drops back towards it. This is because the C of G of the ladder + me rises and hence gains potential energy. So when I stop pushing the C of G returns to its previous position. My equilibrium is stable.

A foolish person leans the ladder against the wall at 75 to the horizontal. At that angle the C of G is just over the point where the foot of the ladder touches the ground. He is in equilibrium. He pushes against the wall. The C of G moves downwards and is now outside the base so it can continue to go downwards. The ladder topples over backwards and the fool is badly injured. This equilibrium is unstable.

moral: angles are important!

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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Define a mechanical equilibrium, precisely, you will, hmm? yess, please?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bob bundy****Moderator**- Registered: 2010-06-20
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The external forces acting on the object have a resultant of zero, so there is no tendency to change the state of uniform motion of the object

AND

there is no resultant couple, and hence, no tendency for the object to rotate.

Regret, I do, that talk like Yoda, Newton, did not, hmm.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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Second derivative < 0: The potential energy is at a local maximum, which means that the system is in an unstable equilibrium state. If the system is displaced an arbitrarily small distance from the equilibrium state, the forces of the system cause it to move even farther away.

Stable equilibria

Second derivative > 0: The potential energy is at a local minimum. This is a stable equilibrium. The response to a small perturbation is forces that tend to restore the equilibrium. If more than one stable equilibrium state is possible for a system, any equilibria whose potential energy is higher than the absolute minimum represent metastable states.

Neutral equilibria

Second derivative = 0 or does not exist: The second derivative test fails, and one must typically resort to using the first derivative test. Both of the previous results are still possible, as is a third: this could be a region in which the energy does not vary, in which case the equilibrium is called neutral or indifferent or marginally stable. To lowest order, if the system is displaced a small amount, it will stay in the new state.

Whose derivative, asking to take, they are, yesss?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

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**bob bundy****Moderator**- Registered: 2010-06-20
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answer in prep.

Bob

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**bob bundy****Moderator**- Registered: 2010-06-20
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As an object is moved about its potential energy may vary. If you could plot a 4-D graph, then the independent variables would be space coordinates and the dependent variable would be the PE. I think that's what you are supposed to understand from the Wiki article. Let's simplify so we can at least picture what is going on.

Imagine a ball that is able to roll along a linear, ball-sized roller coaster. I'll assume that the ball can only travel in a single, direction measured by x, the distance along the track. But my track has hills and valleys even though there's only a 1-D measurement for where the ball is. I'll calculate the PE of the ball by imagining it's in a gravitational field acting 'down'. As the ball goes uphill it loses KE and gains PE. On downward parts of the track it gains speed, so KE goes up and PE goes down.

Let's say there is a hill coming up and I halt the ball exactly at the top of the hill. Its weight acts exactly down onto the track and is balanced by the reaction of the track on the ball. So the resultant force is zero and it's in equilibrium.

If you give it a slight sideways push, it will roll off the hill and speed up losing PE and gaining KE. If you knew the PE as a function of x, you could differentiate to get dP/dx. At the hilltop this would be zero, a turning point. Differentiating again, d2P/dx^2, would be negative, indicating a local maximum, and so this gives a way to determine unstable equilibrium.

When the ball is at the bottom of a valley, dP/dx is again zero, but this time d2P/dx^2 is positive, indicating a local minimum, and this gives a way to determine stable equilibrium.

Neutral equilibrium would be when the track is completely flat; dP/dx is again zero, but the second derivative doesn't show a max or min.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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Wow, thanks

Appreciate it, I must. Hmmmmm

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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