__DENSITY
AND PRESSURE__

The **density** of a
material is defined as the **mass per unit volume**.

density = **r** = mass/Volume (the Greek letter rho)

The
units are = kilograms/meter^{3} =
kg/m^{3}

*It’s a measure of how tightly
the atoms of a material are packed*. It has nothing to do with the hardness of
the material.

Examples:

Material |
Density (kg/m |

air |
1.29 |

ice |
917 |

water |
1000 |

aluminum |
2700 |

lead |
11300 |

gold |
19300 |

The **specific
gravity** of a material is the ratio of its density to that of water. For example, the specific gravity of
aluminum would be 2.7. This number is
dimensionless.

The **pressure**,
**P**, is defined as the ratio of force to area:

The units
of pressure are: Newtons/meter^{2} = N/m^{2} = Pascals = Pa

__Example:__

__ __

A
hammer supplies a force of 700 N. The
hammer head has an area of 7.1 x 10^{-4} m^{2}. What is the pressure?

P =
F/A

P =
(700 N/7.1 x 10^{-4} m^{2})

P =
9.86 x 10^{5} N/m^{2}

P = 9.86 x 10^{5} Pa

__ __

__VARIATION OF PRESSURE WITH DEPTH__

If a
fluid is at rest, then all points at the same depth must be at the same
pressure (otherwise it would be moving!).
However the pressure WILL vary with depth: it will have the weight of the
fluid on top of it.

P = pressure,
P_{o} = pressure due to the air (atmospheric pressure)

Because
the fluid is at rest, the net force will be
_________ ?

So we can write:

Fnet = 0

PA – mg – P_{o}A = 0

However we know that:

Mass =
density x volume = density x (area x height)

M
= rV = rAh

So we can substitute:

PA – (rAh)g - P_{o}A = 0

cancel out the Area, A:

P
– (rh)g - P_{o} = 0

and the pressure at any depth will be:

P = P_{o}
+ rgh

In words this says: the pressure at a depth ‘h’
will be the atmospheric pressure(14.7 lbs/in^{2}) + (rgh)

__Example 1__: Calculate the pressure at a depth of 3.0 m
in a swimming pool.

r_{water} =
(1000 kg/m^{3})

P = P_{o} + rgh

P = (1.013 x 10^{5} Pa) +
(1000 kg/m^{3})(9.8 m/s^{2})(3.0 m)

P = (1.013 x 10^{5} Pa) +
2.94 x 10^{4} Pa

P = 1.31 x 10^{5} Pa =
131,000 Pa = 19 lb/in^{2}.

at a depth of 30 m (100 ft) the pressure would
be: 57
lb/in^{2}.

at a depth of 300 m (1000 ft) the pressure would
be: 440 lb/in^{2}.

__Example 2: __

You are a CIA operative in the jungles of Burma
escaping from a rebel force. You see a
stream and reeds and you cut off a reed, jump into the stream and lie about
half a meter under water breathing thru the reed. Can you breathe?

Answer:
let’s determine the pressure on your chest…

P = rgh

P = (1000 kg/m^{3})(9.8 m/s^{2})(0.5
m)

P = 4900
Pa = 0.71 psi

If your chest is 12 in. by 12 in. (approximately) then the area = 144 sq. in.

Force = Pressure x Area = 0.7 psi x
144 sq.in. = 100 pounds

So, yeh, probably…but don’t go much deeper!

**PASCAL’S PRINCIPLE: **

** **

Pressure applied to an
enclosed fluid is transmitted equally to all points of the fluid (and therefore
to all walls of the vessel containing the fluid)

What good
is this?

We
can turn this phenomenon to our advantage if we alter the areas exposed to
equal pressures: as in an hydraulic lift:

Since the pressure must be the same everywhere:

P_{at
1} = P_{at }_{2}

But Pressure = Force/Area, so

F_{1}/A_{1} = F_{2}/A_{2}

F_{2} = (F_{1}/A_{1})A_{2}

This says that the Force at the outlet (2) is
augmented by the size of the area of the outlet!

So if we make the area 1000 times
larger, we can lift 1000 times the force we apply at F_{1}!!!