Question

A uniform meterstick in static rotational equilibrium when a mass of 220 g is suspended from the 5.0 cm mark, a mass of 120 g is suspended from the 90 cm mark, and the support stand is placed at the 40 cm mark. What is the mass of the meterstick in grams?

Answer 1

Answer:

m = 170 g

Explanation:

Meter stick is suspended at 40 cm mark

So here the torque due to additional mass and torque due to weight of the spring must be counter balanced

given that

1) 220 g is suspended at x = 5 cm

2) 120 g is suspended at x = 90 cm

3) mass of the scale is acting at its mid point i.e. x = 50 cm

now with respect to the suspension point the torque must be balanced

so we have

$220(40 - 5) = m(50 - 40) + 120(90 - 40)$

$220(35) = 10 m + 6000$

by solving above equation we have

m = 170 g