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Found in: Page 397

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# A twin-sized air mattress used for camping has dimensions of by $100cm$ by $200cm$ when blown up. The weight of the mattress is $2kg$. How heavy a person could the air mattress hold if it is placed in freshwater?

The mass of the person should be less than $298kg$.

See the step by step solution

## Step 1: Conceptual Introduction

Fluid statics, often known as hydrostatics, is a branch of fluid mechanics that investigates the state of balance of a floating and submerged body, as well as the pressure in a fluid, or imposed by a fluid, on an immersed body.

## Step 2: Calculate the volume of the mattress

The ratio of the object's density to the density of the fluid in which it is immersed determines the fraction submerged in the fluid.

Fraction submerged will be ratio of the density of the object divided by the density of the fluid in which the object is.

If the mattress is to be floating than its density should be greater than the density of the mattress.

The volume of the mattress:

${V}_{m}=0.1\text{m}×0.2\text{m}×0.15\text{m}\phantom{\rule{0ex}{0ex}}=0.3 {\text{m}}^{3}\phantom{\rule{0ex}{0ex}}$

The volume of the mattress is ${\text{0.3 m}}^{\text{3}}$

## Step 3: Reasons behind the floatation of the men.

The Density of the mattress will be less than density of the water.

${\rho }_{am}<{\rho }_{w}\phantom{\rule{0ex}{0ex}}{\rho }_{am}<1000\phantom{\rule{0ex}{0ex}}\frac{{m}_{m}+{m}_{P}}{{V}_{m}}<1000\phantom{\rule{0ex}{0ex}}\frac{2+{m}_{P}}{0.3}<1000\phantom{\rule{0ex}{0ex}}2+{m}_{p}<300\phantom{\rule{0ex}{0ex}}{m}_{p}<298\phantom{\rule{0ex}{0ex}}$

Hence, the mass of the person should be less than $298kg$ which is obviously possible in all case.