Two students sit on either side of a teeter-totter that is 2.8 m in length.
The teeter-totter balances when the student on the left side is 1.1 m
from the center and the student on the right is 1.4 m from the center.
The total mass of the two students is 78 kg . Assume that the teeter-
totter itself pivots at the center and produces zero torque.

What is the mass of the student on the left side of the teeter totter

Since the teeter totter is in equilibrium, the sum of moments about its center of mass is zero. That is, the clockwise moment, M = anticlockwise moments, M'

Taking moments of the students weight about the center, we have

m₁gx₁ = m₂gx₂ where

m₁ = mass of student on left,
x₁ = distance of student on left from center = 1.1 m,
m₂ = mass of student on right,
x₂ = distance of student on right from center = 1.4 m and
g = acceleration due to gravity.

m₁x₁ = m₂x₂

Given that their total mass is 78 kg, we have that m₁ + m₂ = 78.

So, m₂ = 78 - m₁

Substituting m₂ into the equation, we have

m₁x₁ = m₂x₂

m₁x₁ = (78 - m₁)x₂

Making m₁ subject of the formula, we have

m₁x₁ = 78x₂ - m₁x₂

m₁x₁ + m₁x₂ = 78x₂

m₁(x₁ + x₂) = 78x₂

m₁ = 78x₂/(x₁ + x₂)

Mass of student of left side of teeter totter

Substituting the values of the variables into the equation, we have

m₁ = 78x₂/(x₁ + x₂)

m₁ = 78 × 1.4 m/(1.1 m + 1.4 m)

m₁ = 109.2 kgm/2.5 m

m₁ = 43.68 kg

So, the mass of the student on the left side of the teeter totter is 43.68 kg

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