a uniform meter stick...

Question

A uniform meter stick has a 40.0 g mass placed at the 20 cm mark as shown in the figure, If a pivot is placed at the 42.5 cm mark and the meter stick remains horizontal in static equilibrium, what is the mass of the meter stick.
a. 45g
b. 72 g
c. 120 g
d. 135 g
(The answer should be c, 120g)

Answer 1

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