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Q68PE

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Found in: Page 398

### College Physics (Urone)

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

# During forced exhalation, such as when blowing up a balloon, the diaphragm and chest muscles create a pressure of ${\rm{60}}{\rm{.0 mm Hg}}$ between the lungs and chest wall. What force in newton’s does this pressure create on the ${\rm{600 c}}{{\rm{m}}^{\rm{2}}}$ surface area of the diaphragm?

The force acting into the surface per area is obtained as: ${\rm{480 N}}$.

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: Given data.

The force per unit perpendicular area across which the force is exerted is known as pressure. Pressure is defined as follows in equation form:

$$P = \frac{F}{A}$$

The pressure given is: ${\rm{60 mm Hg}}$.

The surface area creating the pressure is ${\rm{600 c}}{{\rm{m}}^{\rm{2}}}$.

## Step 3: Force acting on the body

The force is given by,

$$F = P \times A$$

By putting all the value into the equation, we get:

$\begin{array}{c}F = P \times A\\F = h\rho g \times A\\ = \left( {0.06\;{\rm{m}}} \right)\left( {13600\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) \times 0.006\;{{\rm{m}}^2}\\ = 480\,\;{\rm{N}}\end{array}$

Therefore, force acting into the surface per area is: ${\rm{480 N}}$.