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Suppose that a meter stick is balanced at its center. A 0.27-kg is positioned at 18 cm from the left end of the meter stick. Where should a 0.28 kg mass be placed to balance the 0.27 kg mass? Express your answer in terms of the position (in cm) of the 0.28-kg mass as measured from the left end of the meter stick.

Compute the total torque about the center due to the two forces. Mass m1 is 50-18 = 32 cm from the center. If x is the distance of mass m2 from the left end, then it is x-50 to the right of the center pivot. Set the total torque equal to zero and solve for x.
-0.27g*(50-18) + 0.28g(x-50) = 0
You can factor out the g.
0.27*32 = 0.28*(x-50)

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