Question

a meter stick is free to rotate about its center. A load of 85 gm is hung at 100 cm mark. what load should be placed at 30 cm mark so that the meter stick will remain in a horizontal position?

Answer 1

Answer:

I assume you meant to say the length of the stick is 100 cm.  Set up equilibrium of torques about the pivot point.  Apply the weight of the stick through its midpoint at 50 cm. 

 

M = mass of the stick

 

Mg[(50 - 42.5) cm] = (40.0 grams)g[(42.5 - 20) cm]

 

M = 40.0(42.5 - 20) / (50 - 42.5) grams = 120 grams