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Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
This is probably a basic physics question, but it suddenly started to get to me:
Why a balance scale balances? Now I now most of this. For a classic pan-balance scale not only does the weight (mg) times the moment arm (d) have to equal both sides, but the center of gravity comes into play . Otherwise, a scale with the fulcrum at the center of mass that has perfectly matched weights on both sides will stay where ever it is moved. My question is what weight is used for the center of gravity moment arm? The trig value for the masses in the pans cancel out so I won't enter these...
What value is used for Mbar? Is it the mass of the whole bar?
Another way this question could be asked is is if I know the length of the scales arms, say 1 meter each, and the center of gravity is 1 cm below the fulcrum, what will the deflection be on the scale from the vertical for masses m1 and m2?
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
For a rigid body, its entire mass can be thought to act through the COG. When the body turns, all particles that move nearer to or further away from the rotation axis act to change the moments.
However for a balance, for small angles around the balance point, the scale pans do not move like this, so they get left out of the COG/rotation summation, and contribute only Mleft*g*d moment as you have identified. However the beam is rigid, so its entire mass is used for Mbar. Just make sure that when you calculate the position of the COG for the balance bar, you include exactly all of the mass of the rigid bar in the calculation, not the pans.
<edit> damnit, you've got me getting it wrong now.
A little thought experiment will show that if the suspension point for the pans is significantly higher than the fulcrum of the bar, then the balance will be unstable, it will flop over to one side and stay, much lower than the fulcrum and it will extra stable.
This means that the mass of the pans does affect the position of the COG of the whole balance, we must include the mass of the pans. BUT, as the pans are free to rotate independantly of the bar, the mass of each pan acts through its suspension point, and not as a rigid extended object. This means the effective COG position and thus offset sensitivity will vary with pan load, unless the fulcrum and the two pan suspension points are co-linear. In the case that they are co-linear, then the COG position is the same as that calculated from the mass distribution of the bar alone. I suspect that most balances will be made with this geometry.
Regardless of whether the three fulcra are in a straight line (got fed up with typing co-linear), if you want to calculate the frequency of oscillation of the balance, then the potential energy term is the COG * sin(x)* yada , the kinetic term contains the moment of inertia of the rigid bar + only mass*d^2 for each pan, there is no MOI term to add for the rotation of each pan+load as they do not rotate with the bar.
Registered Member #193
Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
Most balances have the pivot point slightly higher than the COG. The beam of some balances is "T" shaped, this lowers the COG and gives an easy way to read how far from "balanced" the weights are. There's a nice picture here. The pointer reaching down from the centre of the beam serves to indicate how far out of balance the 2 weights are, and also gives a COG below the pivot points. The other thing to look out for are the 2 threaded weights attached to the beam that adjust the sensitivity slightly.
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Balancing a scale
