Question

A meter stick is found to balance at the 50.0 cm mark when placed on a fulcrum. When a 0.100 kg mass is attached at the 10.0 cm mark, the fulcrum must be move to the 34.0 cm mark for balance.

a) What is the mass of the meter stick?

b) How much torque does the 100 g mass exert on the meter stick around an axis at the new fulcrum (at 34.0 cm)

c) How much torque does the meter stick's weight exert around the same axis?

Answer 1

Xcm = ( (m1*x1) + (m2*x2) )/(m1 + m2)

Xcm = 34 cm

m1 = mass of meter stick

m2 = mass attached to stick = 0.1

x2 = 10 cm

x1 = 50 cm

34 = ( (m1*50) + (0.1*10))/(m1+0.1)

m1 = 0.15 kg

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b)

torque exerted = m2*g*(Xcm - x2)*sin90

torque exerted = 0.1*9.8*(0.34-0.1)*sin90

torque exerted = 0.2352 Nm

(c)

torque exerted = m1*g*(X2 - Xcm)*sin90

torque exerted = 0.15*9.8*(0.5-0.34)*sin90

torque exerted = 0.2352 Nm