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Old 09-09-2011, 10:57 PM   #3 (permalink)
Patrick
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It's been a long time, but I think this is how you solve it algebraically, IIRC.

W1 = initial weight on each side.
W2 = weight to add on light side.
D1 = initial distance to weight on each side.
D2 = distance to weight on light side.

Summation of torques must equal zero.

W1(D1) = W1(D1)

Weight is moved X distance.

X = distance weight was moved.

The equation then becomes W1(D1+X) = (W1 + W2)(D1-X). Call this Equation 1.

Solve for W2: W1(D1+X)/(D1-X) = (W1 + W2), W1(D1+X)/(D1-X) - W1 = W2. Call this equation 2.

Test Equation 2: Let W1 = 5 lbs, D1 = 4 feet, X = 2 feet.

W2 = 5(4+2)/(4-2) - 5 = 5(6)/2 - 5 = 30/2 - 5 = 15 - 5 = 10. W2 = 10 lbs.

Plugging those numbers into Equation 1: 5(4+2) = (5+10)(4-2), 5(6) = (15)(2), 30 = 30. Check.
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