# Equilibrium

• Sep 8th 2007, 09:34 AM
fez_329
Equilibrium
I've been diving into dynamics these days as part of my Education in mathematics, but I'm having some problems. One, very particularly biting problem, is that of equilibirum.

Wikipedia states Equilibirum as: "A system is in mechanical equilibrium when the sum of the forces, and torque, on each particle of the system is zero".

This is the definition I have been taught with. I was also taught that when in equilibirum, a system/object has a resultant force of zero in ANY direction/axis. Finding resultant forces was easy when they were all passing through the centre of gravity of an object. But now, as I try to solve scenes of rigid bodies, where there are several forces, with part of them NOT applying on the centre of gravity, I get confused.

See, it's because of this simple example:

A ladder of uniform mass and length (2L) is lying on the floor, horizontally. It has weight, which is being neutralised by the reaction force from the ground. But, there are two forces applying, upwards, at the edge of the ladder, both of magnitude 7N. Thus, their moments cancel, as both their moments are (7L), but in the opposite directions. But how the heck do the sum of forces here result in zero??????????? You can change the question by converting the ladder into a ball, or you can resposition or increase the number of moment-producting forces as long as their moments nullify and yet they remain parallel and in one direction. How do the sum of forces in a direction result in zero in such cases, even when the object is at equilibirum???? If the sum of forces in one direction is not true here, then how can it be true for a case where a ladder has one end on the ground, and the other on a wall, when all the reaction and friction forces which produce torque?
• Sep 8th 2007, 12:28 PM
ticbol
A ladder of uniform mass and length (2L) is lying on the floor, horizontally. It has weight, which is being neutralised by the reaction force from the ground. But, there are two forces applying, upwards, at the edge of the ladder, both of magnitude 7N. Thus, their moments cancel, as both their moments are (7L), but in the opposite directions. But how the heck do the sum of forces here result in zero???????????

Uniform mass? So uniform weight. Say the ladder weighs w N/meter.

Without the two 7-N upward forces, the whole weight of the ladder is supported by the ground. The ground offers w N/meter too. Upwards. To neutralize the w N/m weright of the ladder. The ladder is in equilibrium.

Apply the two 7-N upward forces, one at each end of the L-meter long ladder. The ladder doesn't move.
----Maybe because the total weight of the ladder, wL newtons, is heavier than 14N. Why are the vertical forces equal then? It's because the ground now offers only (wL -14)/L newtons/meter upwards. Not (wL)/L newton/meter anymore.
----Or, the ladder weighs less than 14N but it is tied/anchored to the ground at the middle. So, technically, the ladder is off the ground but won't be lifted anyway because it is pulled back, downwards, by the anchor at its middle. The anchor offers 14N downward to neutralize the two 7N upward forces. [The total force on the anchor is not 14N downwards. It is (14 minus wL) newtons downwards.] So net, or the sum of vertical forces is zero. Then, as you said, the moments of the two upward forces cancel each other. So net moment at the anchor is zero. The net moment anywhere on the ladder is zero. The ladder won't move or tilt whichever way. The ladder is in equilbrium.

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In any of those cases you mentioned, the ball, etc, even if there millions of upward forces apllied on the object and the object still won't be lifted, then something is neutralizing all those millions of upward forces---rendering the object to be in equilibrium verticaly. The weight of the object is greater than all those million upward forces most probably, or the object is anchored vertically.
Apply that to horizontal forces.
To moments.
If the object does not move even after all those billions of random forces are applied on the object, then the object is in equilibrium. You have to investigate why the dang object won't move. :)
• Sep 8th 2007, 11:59 PM
fez_329
Ah, I see... So the ground adjusts it's own reaction force. Thanlks.

But I still don't get it in the horizontal direction. Say, there is a ball of weight 100N on the ground, at rest. It has uniform density, so the centre of mass is it's precise center. The ground offers back an upward reaction force of 100N to neautralize the weight. The ball has radius 3 metres. Now, there are two forces of 10 N acting horizontally, to the left. One is on the topmost curve of the ball, and the other on the bottom. So both their torques are of magnitude 30 Nm, and they cancel each other out. Equilibirum is maintained. But how does the sum of forces in the horizontal direction equal zero, both for a smooth ground surface and a rough ground surface???
• Sep 9th 2007, 01:35 AM
ticbol
But how does the sum of forces in the horizontal direction equal zero, both for a smooth ground surface and a rough ground surface???

On rough ground surface:
If the ball doesn't move, or it is in equilibrium, the two horizontal 10N leftward forces will be neytralized by a friction force equal top 20N rightward.
If that friction force cannot be developed by the ground, the ball will move to the left.
What about then the torque? The friction cannot stop the torque/moment. It is the moment of inertial of the ball that will take care of the external moment. If the ball rotates/rolls, then it is not in equilibrium anymore, because the moment of inertia of the ball is less than the external torque.

On smooth or frictionless ground surface?
Are you sure the two 10N leftward forces will not move the ball to the left?
• Sep 15th 2007, 10:46 PM
fez_329
Ah, yes. I actually forgot that equilibrium isn't possible in every situation. I don't know what the inertial is yet, so I just noted down this example as not having equilibirum possible. Thanks for the help, though.
• Sep 15th 2007, 11:06 PM
ticbol
Quote:

Originally Posted by fez_329
Ah, yes. I actually forgot that equilibrium isn't possible in every situation. I don't know what the inertial is yet, so I just noted down this example as not having equilibirum possible. Thanks for the help, though.

Moment of inertia is the internal "rotational mass" of any object. It the object is at rest, its moment of inertia will try to prevent external torques/moments to rotate the object...until it is overcome by larger external moments.

Research moment of inertia in the internet. Try Google.com or wikipedia.com.