## Density and pressure

The building blocks of ordinary matter are atoms and molecules.  Molecules are two or more atoms held together by a chemical bond.  Atoms themselves are made up of more fundamental particles, i.e. protons, neutrons, and electrons.  In a solid, the atoms and molecules are densely packed and held in place by intermolecular forces.  The atoms in a solid can be modeled as being held together by tiny springs that permit them to vibrate back and forth about their equilibrium position, but not to exchange positions with other atoms.  Solids are nearly incompressible.  In a liquid the atoms and molecules are also densely packed.  They cannot easily escape from one another, but they are free to move with respect to each other.  Liquids are nearly incompressible.  In a gas, intermolecular forces are weak and short ranged, and the atoms and molecules can move about nearly independently.  Gases are compressible.  Gases and liquids are fluids, i.e. collections of atoms or molecules that are free to move with respect to each other.

### Density

Assume a substance in a volume V has mass M and is made up of N particles.
We define the particle density ρparticle as the number of particles per unit volume, ρparticle = N/V.
We define the density ρ of the substance as its mass per unit volume, ρ = M/V.

#### Problem:

A king orders a gold crown having a mass of 0.5 kg.  When it arrives from the metal smith, the volume of the crown is found to be 185 cm3.  Is the crown made of solid gold?  The density of gold is 1.9*103 kg/m3.

Solution:
The density of the crown is (0.5 kg)/(185 cm3) * (106 cm3)/(1 m3) = 2.7*103 kg/m3.  The crown is not made of solid gold.

### Pressure

We define the pressure P as the magnitude of the normal force F exerted over an area A, divided by the area A.  Pressure equals force per unit area.

P = F/A

In SI units, the units of pressure are N/m2 = Pa (pascal).

#### Examples:

A brick is at rest on a table.  The force with which the brick pushes on the table is its weight, F = mg.  The pressure it exerts on the contact area depends on the brick's orientation.  If the contact area between brick and table is larger, the brick exerts less pressure on the contact area.

Grab a pencil with one hand and press the flat end against your other hand.  Press hard.  Does it hurt?
Now turn the pencil around and press the tip of the pencil against your other hand with the same force.  Does it hurt now?

A 50 kg woman balances on one heel of a high-heel shoe.  If the heel is circular with radius 0.5 cm, the pressure she exert on the floor is
P = F/A = (50 kg 9.8 m/s2)/(π(0.005 m)2) = 6.2*106 N/m2 = 6.2 MPa.
(1 MPa = 1000000 Pa)
If the woman wears a flat-heel shoe with a radius of 3 cm then the pressure she exerts on the floor is
P = F/A = (50 kg 9.8 m/s2)/(π(0.03 m)2) = 1.7*105N/m2 = 0.17 MPa.
The pressure she exerts is reduced by a factor of 36.  Be careful around women with high-heel shoes!

### Hydrostatics

The pressure at a point below the surface of a liquid in a constant gravitational field depends only on the depth of that point and the pressure at the surface.

Any change in the pressure at the surface is therefore transmitted to every point in the liquid.

This is called Pascal's law.

Consider a large pool of water on the surface of the earth and a box-shaped volume of water at some depth in the pool.  Imagine it enclosed by some weightless container.

The volume of water is in equilibrium and stays in place.  It does not rise and it does not fall.  The net force on it must be zero.  The vertical component of the net force is

Fnet = PbottomA - PtopA - Mg = 0.
Fnet = PbottomA - PtopA - ρhAg = 0.
Pbottom - Ptop = ρhg.

The pressure in the pool increases with depth.  If we let h denote the vertical distance of a point below the surface of the water, then we can write the pressure at this point as

Pbelow = Ptop + ρhg.

P is the pressure at depth h and P0 is the pressure at the surface.  Very often this pressure is atmospheric pressure.  The atmospheric pressure at sea level at room temperature is approximately

1 atmosphere = 101 kPa = 14.7 pounds per square inch (psi).

Gauge pressure is the pressure relative to atmospheric pressure. Gauge pressure is positive for pressures above atmospheric pressure, and negative for pressures below it.  Absolute pressure is the sum of gauge pressure and atmospheric pressure.

#### Problem:

Determine the absolute pressure at the bottom of a lake that is 30 m deep.

Solution:
P = P0 + ρgh.  P0 = 101 kPa = 1 atm.
ρwater = 1000 kg/m3.
ρgh = 1000*9.8*30 N/m2 = 294 kPa ≈ 3 atm.
P ≈ 4 atm.

The water pressure increases by approximately 1 atm for every 10 m increase in depth.

#### Problem:

Intravenous infusions are usually made with the help of the gravitational force.  Assume that the density of the fluid being administered is very close to that of water (1 g/cm3).  At what height should the IV bag be placed above the entry point so that the fluid just enters the vein if the blood pressure in the vein is 18 torr = 2392 Pa above atmospheric pressure?   Assume that the IV bag is collapsible.

Solution:
For the fluid to enter the vein, its pressure at entry must just exceed the blood pressure in the vein.
Pbelow = Ptop + ρhg.  The gauge pressure ρhg must be just greater than 2392 Pa.
Solving for  h = (2392 Pa)/(ρg) = (2392 N/m2)/(1000 kg/m3 *9.8 m/s2) = 0.24 m = 24 cm.
The IV bag be placed a little over 24 cm above the entry point for the fluid to just enters the vein.

#### Problem:

What is the hydrostatic force on the back of Grand Coulee Dam if the water in the reservoir is 150 m deep and the width of the dam is 1200 m.

Solution:
Let us denote the distance below the surface by y.  The pressure on the side of the dam facing the water at depth y is P0 + ρgy.  The force on a surface element of width w and height dy at depth y is
dF = (P0 + ρgy)dA = (P0 + ρgy)wdy.
The total force on this side of the dam is
0150dF = wP00150dy + wρg∫0150ydy =wP0150 + wρg 1502/2,
outward, if all quantities are measured in SI units.
The total force on the side of the dam facing air is wP0150, inward.
The net force on the dam is F = wρg1502/2, outward.  With w = 1200 m and ρ = 1000 kg/m3, F = 1.32*1011 N.

#### Problem:

The figure on the right shows an aerial view from directly above two dams.  Both dams are equally long and equally deep.  The dam on the left holds back a very large lake and the dam on the right holds back a narrow river.  Which dam has to be build more strongly?

Solution:
The net force on both dams is the same, the strength of both dams must be the same.

#### Problem:

A hydraulic system is used to lift a 2000 kg vehicle in an auto garage. If the vehicle sits on a piston of area 1 m2, and a force is applied to a piston of area 0.03 m2, what is the minimum force that must be applied to lift the vehicle?

Solution:
Pascal's law:
When a force is applied to a contained, incompressible fluid, the pressure increases by the same amount throughout the fluid.  This property of incompressible fluids can be used to amplify forces, as is done in hydraulic lifts.

P1 = P2,  F1/A1 = F2/A2,  F1 = (A1/A2)*F2 = (0.03/0.5)*(2000 kg*9.8 m/s2) = 1180 N.

Conservation of energy requires that the work done on the system must balance the work done by the system.  In the hydraulic lift the distance over which F1 is applied is be greater than the distance over which F2 acts by the exact same ratio as the force multiplier.
F1/F2 = d2/d1.