Question

A meterstick of uniform density is hung from a string tied at the 23-cm mark. A 0.40-kg object is hung from the zero end of the meterstick, and the meterstick is balanced horizontally. What is the mass of the meterstick? in kg

Answer 1

Using torque balance about the string

Clockwise torque = Counter-clockwise torque

clockwise torque will be due to mass of meterstick, which will be = r2*m*g

r = length = 1 m

r2 = distance of center of mass of meterstick from string = 50 - 23 = 27 cm = 0.27 m

m = mass of meterstick

Clockwise torque = 0.27*m*g

Counter-clockwise torque = r1*F1

r1 = distance of hung object from string = 23 cm = 0.23

F1 = m1*g = 0.40*9.8 = 3.92

So,

0.27*m*9.81 = 0.23*3.92

m = 0.23*3.92/(0.27*9.81)

m = 0.34 kg

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