Static Fluids: Pressure, Pascal’s Laws, and Buoyancy

Static Fluids and Pressure

Definition of Pressure in Fluids

In a static fluid, the particles are in constant random motion and collide with any surface they contact. The collective force from these collisions creates pressure on the surfaces.

• Pressure (P): The force exerted per unit area by a fluid on a surface.

• Mathematically, pressure is given by:

$P = \frac{F_{\perp}}{A}$

• Where $F_{\perp}$ is the force perpendicular to the surface, and $A$ is the area.

• SI Unit: Pascal (Pa), where $1\ \text{Pa} = 1\ \text{N}/\text{m}^2$.

Key Effects:

• Fluids push outward on container walls.

• Fluids push upward on objects submerged in them.

Pascal’s Laws

Pascal’s First Law: Transmission of Pressure

Pascal’s First Law states that a pressure change at one point in an enclosed fluid is transmitted undiminished to all points in the fluid.

• This principle allows for mechanical advantage in hydraulic systems.

$P_{\text{in}} = P_{\text{out}}$

• For a hydraulic lift:

$\frac{F_{\text{in}}}{A_{\text{in}}} = \frac{F_{\text{out}}}{A_{\text{out}}}$

• If $A_{\text{out}} > A_{\text{in}}$, then $F_{\text{out}} > F_{\text{in}}$ (mechanical advantage).

Example 1: Hydraulic Lift

• Problem: A 150 N force is applied to a circular piston (radius 5 cm) to lift a 2000 kg car. What is the area of the output piston?

• Solution Steps:

  • Calculate input area: $A_{\text{in}} = \pi r^2 = \pi (0.05\ \text{m})^2 = 0.00785\ \text{m}^2$

  • Input force: $F_{\text{in}} = 150\ \text{N}$

  • Output force: $F_{\text{out}} = mg = (2000\ \text{kg})(9.8\ \text{m}/\text{s}^2) = 19,600\ \text{N}$

  • Apply Pascal’s Law:

$A_{\text{out}} = \frac{F_{\text{out}}}{F_{\text{in}}} A_{\text{in}} = \frac{19,600\ \text{N}}{150\ \text{N}} (0.00785\ \text{m}^2) = 1.026\ \text{m}^2$

Pascal’s Second Law: Pressure Variation with Depth

Pascal’s Second Law states that the pressure at a point in a uniformly dense fluid depends only on its depth below the fluid’s surface.

• For a static fluid column:

$P_1 = P_2 + \rho g (y_2 - y_1)$

• If point 2 is at the top of the fluid (open to atmosphere):

$P = P_{\text{atm}} + \rho g h$

• Where $\rho$ is fluid density, $g$ is acceleration due to gravity, and $h$ is depth below the surface.

Concept Check: Pressure and Fluid Level

• For containers with the same fluid and depth, the pressure at the bottom is the same, regardless of shape or volume.

• In a connected pipe, points at the same vertical level in the same fluid have equal pressure.

Example 2: Pressure at Depth

• Problem: The top of an open-air water tank is 40 m above ground. A plug with area $0.025\ \text{m}^2$ is at ground level. How much force does the water apply to the plug?

• Solution Steps:

  • Find pressure at the plug:

$P = P_{\text{atm}} + \rho g h = (1 \times 10^5\ \text{Pa}) + (1000\ \text{kg}/\text{m}^3)(9.8\ \text{m}/\text{s}^2)(40\ \text{m}) = 4.92 \times 10^5\ \text{Pa}$

• Find force on the plug:

$F = PA = (4.92 \times 10^5\ \text{Pa})(0.025\ \text{m}^2) = 12,300\ \text{N}$

Buoyant Force and Archimedes’ Principle

Buoyant Force

A fluid exerts an upward buoyant force on an object immersed in it. This force increases as the object displaces more fluid while sinking.

• The magnitude of the buoyant force is equal to the weight of the displaced fluid.

$F_B = \rho_{\text{fluid}} V_{\text{disp}} g$

• Where $V_{\text{disp}}$ is the volume of fluid displaced.

Archimedes’ Principle

Archimedes’ Principle: The buoyant force on an object is equal to the weight of the fluid displaced by the object.

• If $F_B = W_{\text{object}}$, the object floats completely submerged.

• If $F_B < W_{\text{object}}$, the object sinks.

• If $F_B > W_{\text{object}}$, the object rises until partially submerged.

Floating vs. Sinking

• Floating (partially submerged): $\rho_{\text{fluid}} > \rho_{\text{object}}$

• Completely submerged (neutral buoyancy): $\rho_{\text{fluid}} = \rho_{\text{object}}$

• Sinking: $\rho_{\text{fluid}} < \rho_{\text{object}}$

Example 3: Floating Log

• Problem: A wooden log with mass 8 kg floats in fresh water. How much of the log is beneath the water line?

• Solution Steps:

  • Set buoyant force equal to log’s weight:

$F_B = W_{\text{log}}$

$\rho_{\text{water}} V_{\text{disp}} g = m_{\text{log}} g$

$V_{\text{disp}} = \frac{m_{\text{log}}}{\rho_{\text{water}}} = \frac{8\ \text{kg}}{1000\ \text{kg}/\text{m}^3} = 0.008\ \text{m}^3$

Demonstrations: Buoyant Force and Weight

• Removing air from around an object (in a vacuum) does not change its weight, but removes the buoyant force due to air.

• An object weighs less in a fluid than in air due to the upward buoyant force.

When submerged:

$W' = W - F_B$

• Where $W'$ is the apparent weight in fluid, $W$ is the true weight, and $F_B$ is the buoyant force.

Summary Table: Key Fluid Statics Equations

Concept	Equation	Description
Pressure	$P = \frac{F}{A}$	Force per unit area
Pascal’s First Law	$P_{\text{in}} = P_{\text{out}}$	Pressure transmitted undiminished in enclosed fluid
Hydraulic Lift	$\frac{F_{\text{in}}}{A_{\text{in}}} = \frac{F_{\text{out}}}{A_{\text{out}}}$	Mechanical advantage in fluids
Pressure with Depth	$P = P_{\text{atm}} + \rho g h$	Pressure increases with depth in a fluid
Buoyant Force	$F_B = \rho_{\text{fluid}} V_{\text{disp}} g$	Upward force equals weight of displaced fluid

Additional info: These notes cover the core concepts of static fluids, including pressure, Pascal’s laws, and buoyancy, as outlined in Chapter 13 of a typical college physics curriculum.