Fluid Pressure

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Fluid: A substance that has definite mass but not definite shape.
Examples: Gases and Liquids

Aim 31: Representing Force & Mass for a Fluid

What weighs more... a pound of feathers or a pound of bricks?

Density ($\rho$) = $\frac{m}{V}$

$m$ = Mass $V$ = Volume

Units: $1 \text{ g/cm}^3$, $1000 \text{ kg/m}^3$, $1 \text{ kg/L}$

The same mass of feathers and bricks would have very different volumes. You would need a much larger pile of feathers to weigh the same as a pile of bricks.

Weight of a Fluid: The force gravity exerts on a fluid = $mg$
Since $m = \rho V$, then $F_g = \rho V g$

Specific Gravity

DEFINITION: Specific Gravity is the density of a fluid compared to the density of water.

$\rho_{water} = 1000 \text{ kg/m}^3 = 1 \text{ kg/L} = 1 \text{ g/cm}^3$

Example Problem: A liquid has a specific gravity of 0.357. What is its density?
A) $357 \text{ kg/m}^3$ B) $643 \text{ kg/m}^3$ C) $1000 \text{ kg/m}^3$ D) $3570 \text{ kg/m}^3$

Answer: A) 357 kg/m$^3$

Aim 32: Buoyancy Force (Archimedes' Principle)

An object floats when its density is LESS THAN the fluid.

$F_B = \rho_{fluid} \cdot V_{displaced} \cdot g$

Archimedes' Principle: A fluid exerts a net upward force on any object it surrounds. The Buoyancy force ($F_B$) is equal to the force gravity would exert on the displaced liquid.

The buoyant force is always present regardless of whether the object sinks or floats in the fluid.

Floating and Density:

Percent Submerged = $\left( \frac{\rho_{object}}{\rho_{liquid}} \right) \times 100$
Any floating object displaces its own weight of fluid.

Ex 1) Ice has a density of $934 \text{ kg/m}^3$. What percent of the ice cube is above water level in freshwater?

Ex 2) A 280 g mass hangs from a scale. When placed in water, it displaces $8.2 \text{ cm}^3$ of water. What does the scale read in the water?

Aim 32: CER Key

EX 1 - FLOATING ICE CUBE: 1. Percent Submerged = $(\rho_{ice} / \rho_{water}) \times 100 = (934 / 1000) \times 100 = \mathbf{93.4\%}$ Submerged.
2. Percent Above Water = $100\% - 93.4\% = \mathbf{6.6\%}$ Above Water.
Note: In salt water ($\rho = 1030 \text{ kg/m}^3$), it would be $(934/1030)\times 100 = 90.6\%$ submerged.

EX 2 - APPARENT WEIGHT: 1. $F_{net} = 0 \implies F_{scale} + F_B = mg$
2. $F_{scale} = mg - F_B$
3. The scale reads the apparent weight, which is the actual weight minus the buoyant force of the displaced fluid.

Aim 33: Solving Buoyancy Problems

Flask on a Scale: A flask of water rests on a scale. If you dip your finger into the water, without touching the flask, does the reading on the scale (a) increase, (b) decrease, or (c) remain the same?

The Crown: The crown weighs 6 N in air but only 4 N in water. What is the volume of the crown? ($\rho_{water} = 1000 \text{ kg/m}^3$)

Melting Ice: A cup is filled to the brim with water and a floating ice cube. When the ice melts, which of the following occurs?
1. Water overflows the cup
2. The water level decreases
3. The water level remains the same

Raft Problem: A large rectangular raft (density $650 \text{ kg/m}^3$) is floating on a lake. The surface area of the top of the raft is $8.2 \text{ m}^2$ and its volume is $1.80 \text{ m}^3$. The density of lake water is $1000 \text{ kg/m}^3$. Calculate the height $h$ of the portion of the raft that is above the surrounding water.

Aim 33: CER Key

FLASK ON SCALE (Q1): (a) Increase. Even though your finger is not touching the flask, the buoyant force acting on your finger (which is the weight of the displaced water) is transmitted to the water and, consequently, to the scale. This results in an increased reading.

THE CROWN (Q2): 1. Buoyant Force ($F_B$) = Weight in air - Weight in water = $6\text{N} - 4\text{N} = 2\text{N}$.
2. $F_B = \rho_{water} V g \implies 2 = (1000)(V)(10)$.
3. $V = 2 / 10000 = \mathbf{2 \times 10^{-4} \text{ m}^3}$.

MELTING ICE CUBE (Q3): (c) Remains the same. The ice cube displaces an amount of water equal to its weight. When the ice melts, it turns into water, and the volume of the resulting water will be exactly equal to the volume of water it displaced when it was ice.

RAFT PROBLEM (Q4): 1. Percent Submerged = $650 / 1000 = 65\%$. The raft is $35\%$ above water.
2. Total Height of Raft: $Volume = Area \times h \implies 1.80 = 8.2 \times h \implies h_{total} = 0.2195 \text{ m}$.
3. Height Above Water = $0.2195 \times 0.35 = \mathbf{0.077 \text{ m}}$.

Aim 34: Pressure in Static Fluids

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

Units: Pascal (Pa) = $\text{N/m}^2$

Concept: Pressure is the same in all directions, and acts at RIGHT ANGLES to any surface. A difference in pressure will create a FORCE. The direction of the force will be from HIGH pressure to LOW pressure.

Atmospheric Pressure ($P_0$): At sea level, atmospheric pressure is about $1.01 \times 10^5 \text{ Pa}$ or $1 \text{ atm}$ (or $760 \text{ mm Hg}$). Surface air pressure = weight of air in column above unit area.

Dependence on Depth: $P = P_0 + \rho g h$

Pressure in a liquid does not depend on the shape of the container (for incompressible fluids).

Glass Comparison: Two drinking glasses, A and B, are filled with water to the same depth. Glass A has twice the diameter of glass B. Compare the weight of the water and the pressure at the bottom.

U-tube Manometer: Calculate the pressure in the box ($P_2$) if the height difference $h$ is $12 \text{ cm}$. The tube is filled with Mercury ($\rho = 13.6 \times 10^3 \text{ kg/m}^3$).

Aim 34: CER Key

GLASS COMPARISON (Q1): 1. Twice the radius means 4 times the Area. Larger glass has 4x the weight of water.
2. Pressure at the bottom is the SAME. Depth ($h$) is the only factor determining fluid pressure ($P = \rho g h$), not container shape or width.

U-TUBE MANOMETER (Q2): Pressure at the same height is equal. $P_2 = P_0 + \rho g h$
$P_2 = 1.01 \times 10^5 \text{ Pa} + (13.6 \times 10^3 \text{ kg/m}^3)(10 \text{ m/s}^2)(0.12 \text{ m})$
$P_2 = 1.01 \times 10^5 + 16,320 = \mathbf{1.17 \times 10^5 \text{ Pa}}$.

Aim 35: Pascal's Principle

Ideal Fluids:

Will NOT compress.
Transmit applied pressure in all directions.
Have no viscosity (friction) or turbulent flow.

Pascal's Principle: Any change in the pressure applied to a completely closed fluid is transmitted undiminished to all parts of the fluid and its container.

$P_1 = P_2 \implies \frac{F_1}{A_1} = \frac{F_2}{A_2}$

Conservation Laws in Hydraulics:

Conservation of Mass/Volume: Volume in = Volume out $\implies A_1 \Delta d_1 = A_2 \Delta d_2$
Conservation of Energy: Work in = Work out $\implies F_1 d_1 = F_2 d_2$

A reservoir of incompressible fluid is topped by two pistons. The larger piston has four times the surface area of the smaller piston ($4A$). If a force $F$ pushes the smaller piston down by distance $d$, what is the upward force and distance traveled by the larger piston?

Hydraulic Car Lift: Calculate the force $F_1$ that can lift a $20,500 \text{ N}$ car. The bottom surface of the plunger and the lift are at the same level. The radius $r_1$ is $0.012 \text{ m}$ and the radius $r_2$ is $0.15 \text{ m}$.

Aim 35: CER Key

PISTON PROPORTIONS (Q1): 1. Pressure is constant. $\frac{F}{A} = \frac{F_2}{4A} \implies F_2 = \mathbf{4F}$.
2. Volume is constant. $A \cdot d = 4A \cdot d_2 \implies d_2 = \mathbf{d/4}$.

HYDRAULIC LIFT (Q2): 1. Areas: $A_1 = \pi(0.012)^2$ and $A_2 = \pi(0.15)^2$.
2. Pascal's Principle: $\frac{F_1}{r_1^2} = \frac{F_2}{r_2^2}$ (Pi cancels out).
3. $\frac{F_1}{0.000144} = \frac{20500}{0.0225} \implies F_1 = \mathbf{131.2 \text{ N}}$.