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How can I relate $mg$ to $R$?

Just to give you a scenario: If a free hanging weight was tied to a motor... I know the voltage ($V$) and current ($I$) so I can use $V = IR$ to find $R$. What if I reduce the weight by say half? Will that be the same thing as $.5R$?

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If you want to see how much energy is saved, you can get it in 2 ways, either mechanical, or electric.

1) If you measure U and I to the motor (directly, using voltmeter and ampermeter, before and afer change), then energy saved is t*U1*i1-t*U2*I2, where t is time, in seconds.

2) Mechanically. You need to know either total distance traved by the weight, L, or weight speed, v. Evergy saved is (m1-m2)gL, or (m1-m2)gtv.

V=IR is not the right formula for your purpose.

You need to use formula to connect the electrical power of the motor (P1) to mechanical energy of raising weight (P2), P1=P2.

Power P1 (in watts) to/from vertically moving weight is proportional to its speed v and mass m: P1 = vmg

Electric power P2=$IU$ for direct current, P2=$IU0.77$ for AC, IIRC that constant.
From P1=P2 you get

    vmg = IU0.77

Depending on how the motor is controlled, if weight reduces by half, then either I or U need to be reduced by half.

If you want to reduce I or U by connecting resistor R2 in sequence to the motor, then you need, at the very least, to know I and U, and calculate the dissipation power. You might be surprised.

Small resistors are between 0.1W to 1W only. What is power of your motor. Motor can be kilowatts.

Resistors are good for several watts only. Above that, resistor will be huge size, it might overheat, and it will waste energy.

Resisters are not how bigger motors are controlled!

Better methods of controlling big AC motor are transformer or semiconductor circuit, not resistors.

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Awesome. This is what I was looking for. I'll give it another 24 hours to hear anyone else's input. You did draw an interesting conclusion on my intentions. Ultimately I was looking to see how much energy would be saved by reducing the weight. –  Kyle Parisi Mar 16 '11 at 18:24
see my added remarks above, at the top of the answer. –  Andrei Mar 16 '11 at 21:30

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