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.

Question

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)

Answer 1

Now, we will solve for x:

0.27*32 = 0.28*(x-50)
8.64 = 0.28x - 14
8.64 + 14 = 0.28x
22.64 = 0.28x

Divide both sides by 0.28:

x ≈ 80.86 cm

So the 0.28 kg mass should be placed approximately 80.86 cm from the left end of the meter stick to balance the 0.27 kg mass.