The Displacement of the Platform When the Man and Woman Meet

The Scenario

The man of mass m1 = 82 kg and the woman of mass m2 = 80 kg are standing on opposite ends of the platform of mass m0 = 103 kg which moves with negligible friction and is initially at rest with s = 0. The man and woman begin to approach each other.

The Problem

Find the displacement s of the platform when the two meet if the displacement x1 of the man relative to the platform is 9 m. The length l of the platform is 15.5 m.

The Solution

Answer: Δs = 0.0679 m

Explanation:

  • m₀ = 103 kg
  • m₁ = 82 kg
  • m₂ = 80 kg
  • x₀ = (L/2) = (15.5/2) m = 7.75 m

Initial positions

  • x₁ = 0 m
  • x₂ = L = 15.5 m

Final positions

  • x₁ = 9 m
  • x₂ = L – x₁ = 15.5 m – 9 m = 6.5 m

Xcm = (m₀*X₀+m₁*X₁+m₂*X₂) / (m₀+m₁+m₂)

We get the center of mass in every case

Xcm initial = (103 Kg *7.75 m +82 Kg*0 m+ 80 Kg*15.5 m)/(103+82+80) Kg = 7.6915 m

Xcm final = (103 Kg *7.75 m +82 Kg*9 m+ 80 Kg*6.5 m)/(103+82+80) Kg = 7.7594 m

Variation: Δs = Xcm final - Xcm initial = 7.7594 m - 7.6915 m = 0.0679 m

What are the initial positions of the man and woman on the platform?

The man is at position 0 m and the woman is at position 15.5 m on the platform.

← Designing the smallest concrete section for a simply supported beam Results and discussions about or tambo airport →

More Posts: