Question

A uniform metre rule is balanced horizontally on a wedge placed under the 40cm mark by a weight of 0.5N hanging from the 20cm mark. a) What is the weight of the metre rule? b) the wedge is moved under the 60cm mark. What weight must now be placed at the end of the rule to keep it horizontal?

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