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Common balance

by krishna mohan
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krishna mohan
#1
Sep4-11, 02:25 AM
P: 118
Confused about a simple thing...

For a common balance, since both arms have equal weight, what causes the pole to choose a horizontal position?
Shouldnt it be stable in any other position as the torque around the centre is zero for any other position too?
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Lobezno
#2
Sep4-11, 04:36 AM
P: 53
The arms of the balance move up and down, leaving the only position for equilibrium as exactly straight, for equal mass arms.
krishna mohan
#3
Sep5-11, 01:48 AM
P: 118
I am a little confused here....

Even if you move the balance to some other position, the potential energy is not going to change..as the increase in PE due to one mass moving up is compensated by the decrease due to the other moving down..

In a slightly tilted position, where does the restoring torque come from? The two masses seem to contribute equal and opposite torque leading to zero torque....

rude man
#4
Sep5-11, 05:22 AM
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Common balance

If the pointer points below the pivot point then if the scale is unbalanced the pointer provides a restoring torque = W*l*sina where W = weight of pointer, l = distance from pivot to pointer's c.g. and a = offset angle. Since scales are always balanced with known weights, the "open-loop" quantity da/dm is of no first-order importance. Scales are always used in a closed-loop manner.

I don't know much about scales but I would think that's how they're designed.


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