Question

1. A meter stick is balanced at its 50.0 cm center of mass. Suppose a 20.0 g mass is now placed at the 40.0 cm position. What torque is being applied by this mass?
2. Given data from ques 1, at what position would a 10.0 g mass have to be placed to achieve equilibrium?
3. Suppose the 10.0 g mass is removed, and an unknown mass is now placed at the 90.0 cm position on the meter stick. Find this unknown mass.
4. Suppose this meter stick is now pivoted at the 40.0 cm position. If an 80.0 g mass is now placed at the 20.0 cm position in order to achieve equilibrium, what is the mass of the meter stick? Draw a sketch to support your answer.

Answer 1

1)Torque applied by 20g mass = = = 0.0196 Nm

2) Given the data in question1 10g mass should have placed at 70cm mark to achieve equilibrium.

This is evident from the above equation, to be in equilibrium both torque should be same.Since the mass is half we have to double the distance

3)Assuming the meter stick in Equilibrium,

4) We have to find the mass of the scale, from equilibrium conditions we get  

m = 160 g

All the question solved. Thank You!