# The Displacement of the Platform When the Man and Woman Meet

## The Scenario

**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.