Question

A meter stick has the pivot placed at the 28 cm mark. A mass of 156 g is placed at the 3 cm mark, and another mass of 94 g is placed at the 61 cm mark. If these masses create equilibrium what is the mass of the meter stick in grams? Round your answer such that it has no decimal.

Answer 1

Gravitational acceleration = g

Mass of the meter stick = M

Mass of the first mass placed on the meter stick = m₁ = 156 g

Mass of the second mass placed on the meter stick = m₂ = 94 g

Position of the pivot = X = 28 cm = 0.28 m

Position of the first mass on the meter stick = X₁ = 3 cm = 0.03 m

Position of the second mass on the meter stick = X₂ = 61 cm = 0.61 m

The center of gravity of the meter stick will lie at the midpoint of the length of the meter stick.

Position of the center of gravity of the meter stick = X₃ = 50 cm = 0.5 m

Distance of the first mass from the pivot = D₁

D₁ = X - X₁

D₁ = 0.28 - 0.03

D₁ = 0.25 m

Distance of the second mass from the pivot = D₂

D₂ = X₂ - X

D₂ = 0.61 - 0.28

D₂ = 0.33 m

Distance of the center of gravity of the meter stick from the pivot = D₃

D₃ = X₃ - X

D₃ = 0.5 - 0.28

D₃ = 0.22 m

Taking moment balance about the pivot,

m₁gD₁ = m₂gD₂ + MgD₃

m₁D₁ = m₂D₂ + MD₃

(156)(0.25) = (94)(0.33) + M(0.22)

M = 36 g

Mass of the meter stick = 36 g