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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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Consider the problem in terms of work/energy/power. You know about gravitational potential, and you know about the power dissipated by a current through a resistor. What can you do with that? –  dmckee Mar 11 '11 at 17:31
    
@Kyle: is the motor lifting the weight? Is it allowing the weight to accelerate? Allowing it to fall at a constant speed, and using the energy as a generator for the circuit? –  Jerry Schirmer Mar 11 '11 at 17:44
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@ Kyle: What is the relation of the power plant in Your town to the wattage of the lamp on Your desk? –  Georg Mar 11 '11 at 18:15
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@Kyle: if this is not a homework problem (or something of that nature), let us know and perhaps we can approach it differently. –  David Z Mar 11 '11 at 19:04
    
hmmm. Not a homework problem. The original question comes from a hammer mill. The motor lifts the weight and drops it. So your telling me there isn't a direct relation in Ohm's law? –  Kyle Parisi Mar 16 '11 at 2:02

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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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