# Fluid Statics

Density (ρ) | mass / volume (kg/m ^{3}) | Characteristic property of a pure substance. Below air density values (ρ
| ||||||
---|---|---|---|---|---|---|---|---|

Specific Gravity (SG) | unitless | ( ρ of substance ) / ( ρ of water at 4° C) | ||||||

Pressure | Pressure = ( Force on fluid ) / ( Surface area of fluid ) | |||||||

Standard Pressure (P_{0}) | atm | Standard pressure (atmospheric pressure) is the pressure the atmosphere exerts on a surface.
1 atm = 1.013 · 10^{5} Pa = 14.69 psi
| ||||||

Water Pressure at Depth | P = F / A = ρ A h g / A = ρ g h The pressure at a given depth is not dependent on area, just on the depth; generally, the pressure at a given depth of a liquid will be uniform. Note that this is just the water pressure. If the container is open to the atmosphere, then total pressure will be: P = P_{0} + ρ g h | |||||||

Pressure on Surface | ½ ρ g l ( y_{2}^{2} - y_{1}^{2} )The pressure a vertical surface such as an aquarium viewing window is determined by the above equation, where y_{2} is the upper edge's depth below water and y_{1} is the lower edge's depth below water. Note that because the surface is vertical, atmospheric pressure does not play a role. | |||||||

Pressure at Elevation | P = P | |||||||

Gauge Pressure | Gauges measure pressure relative to atmospheric pressure. For example, a tire gauge may read off 29 psi as the tire pressure. In fact, the absolute tire pressure is atmospheric pressure P_{0} (14.69 psi) plus the gauge reading of pressure P_{G (29 psi) for a total absolute tire pressure of 43.69 psi.}P = P |

Specific Gravity = ( ρ of substance ) / ( ρ of water at 4° C) = ( mass of substance / volume of bottle ) / ( mass of water / volume of bottle ) = ( mass of substance ) / ( mass of water )

Mass of substance = 80 g - 20 g = 60 grams Mass of water = 100 g - 20 g = 80 grams 60 grams / 80 grams = 0.75 is the specific gravity.

P_{head} = P_{feet} + ρ h g
h = 1.5 m
ρ = density of blood ≈ 1050 kg / m^{3}

P_{head} - P_{feet} = ρ h g = 1050 · 1.5 · 9.8 = 15435 Pa
15435 Pa · 760 / 101300 = 116 mm-Hg

Pascal's Principle | P | |
---|---|---|

Archimedes' Principle | F F | |

Buoyant Force (F_{B}) | Archimedes' Principle addresses buoyant force. This is the phenomenon whereby an object weighs less underwater because the pressure at the bottom of the object is greater than at the top of the object. Thus, the water exerts a net force upward against gravity. |

Pascal's Principle: P_{in} = F_{out} / A_{out}

The problem provides pressure and diameter, so it is apparent that Pascal's Principle will provide an answer.

A_{out} = π r^{2} = π · ( ½ · .20 )^{2} = 0.0314159265 m^{2}
F_{out} = m_{car} · g = m_{car} · 9.8 m s^{-1}
P_{in} = 15.0 atm · 101325 Pascals / atm = 1519875 Pascals

1519875 = m_{car} · 9.8 / 0.0314159265
m_{car} = 4872.2736 kg
4870 kg is the maximum weight that the hydraulic can lift.

F_{B} = ρ_{fluid} V g

F_{B} = ( 10.00 - 8.00 ) · g = 2.00 · g
ρ_{fluid} = 1000 kg m^{-3}
ρ_{object} = ?
V = ?

2.00 · g = 1000 · V · g
2.00 = 1000 · V
V = 2 · 10^{-3} m^{3} → volume of water displaced, and of object.

Density of object → 10.00 kg / ( 2 · 10^{-3} m^{3} ) = 5.00 · 10^{3} kg m^{-3} = 5000 kg m^{-3}

^{-3}. If the cube floats so that it is 80& in the water and 20& in the oil, what is the mass of the cube? What is the buoyant force on the cube?

F_{B} = ρ_{fluid} V g

ρ_{oil} = 900 kg m^{-3}
ρ_{water} = 1000 kg m^{-3}
V_{cube} = .100^{3} m^{3} = 10^{-3} m^{3}
V_{cube} in oil = 10^{-3} m^{3} · 20% = 10^{-3} m^{3} · .20 = .0002
V_{cube} in water = 10^{-3} m^{3} · 80% = 10^{-3} m^{3} · .80 = .0008

F_{B} in water = 1000 · .0008 · 9.8 = 7.84
F_{B} in oil = 900 · .0002 · 9.8 = 1.764
F_{B} total on cube = 7.84 + 1.764 = 9.604 Newtons is buoyant force on the cube

Mass of cube = 9.604 N / 9.8 m s^{-2} = 0.98 kg is mass of the cube

F_{B} = ρ_{fluid} V g

ρ_{fluid} = 1000 kg m^{-3}
F_{B} = ?
V = ?
SG_{wood} = 0.50
ρ_{wood} = SG_{wood} · 1000 kg m^{-3} = 500 kg m^{-3}

Notice that the tribemembers are just getting their feet wet.
Therefore, the raft must be entirely submerged.
Volume of log ≈ Volume for a cylinder = π r_{2} l = π · .25^{2} · 4 = 3.14159265 m^{3}

Recall that the upward (buoyant) force must break even with the raft's *and* the peoples' mass.
F_{B} = ( m_{raft} + m_{troops} ) · g = ( V_{raft} · ρ_{wood} + N · 40 ) · g = ( 1570.79633 + N · 40 ) g
Where N is the number of people on the raft.

( 1570.79633 + N · 40 ) g = 1000 · 3.14159265 · g 1570.79633 + N · 40 = 3141.59265 N · 55 = 1570.79633 N = 28.5599333 → The raft can support 28 people.

This problem is worded in a complicated manner, and with a variety of different units. The first step is to take the material and make it simple, in SI units.

Atmospheric pressure = 1013.25 mbar = 101325 Pa
Tank pressure reading = 30.0 cm → In mm-Hg, this would simply be 30.0 cm · 10 mm cm^{-1} = 300 mm-Hg → Converting Torr (mm-Hg) to Pascals = 39996.7105 Pa

The problem posed is to find the absolute pressure in the tank. Absolute pressure is gauge (or relative) pressure + atmospheric pressure.

101325 Pa + 39996.7105 Pa = 141321.71 Pa → 141000 Pa

This problem is simple, but the presentation is complicated due to wordiness and varying units. Take the material and simplify it all into SI units.

Atmospheric pressure = 1013.25 mbar = 101325 Pa
Tank pressure reading = -7.0 cm → In mm-Hg, this would simply be -7.0 cm · 10 mm cm^{-1} = -70 mm-Hg → Converting Torr (mm-Hg) to Pascals = -9332.56579 Pa

With the above information, it is straightforward to figure out the absolute pressure in the tank. Absolute pressure is gauge (or relative) pressure + atmospheric pressure.

101325 Pa + ( -9332.56579 Pa ) = 91992.4342 Pa → 92000 Pa