9.1 Static fluids

Achievable AP Physics 1

9. Fluids

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Static fluids

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Fluids at rest (liquids and gases) exert pressure on any surface they touch and produce buoyant forces on objects immersed in them.

How the intrinsic property of density leads to hydrostatic pressure
How that pressure transmits via Pascal’s principle
How displaced fluid produces buoyancy via Archimedes’ principle

The goal is to build a clear conceptual picture and then support it with algebra and worked examples.

Density & specific weight

Definition & physical meaning.

The mass density $ρ$ of a substance tells you how much mass is packed into a given volume:

$ρ = \frac{m}{V}, [ρ] = kg/ m^{3} .$

Higher density means more mass per unit volume (for example, lead).
Lower density means less mass per unit volume (for example, air).

Multiplying density by the acceleration of gravity $g$ gives the specific weight:

$γ = ρ g, [γ] = N/ m^{3},$

Specific weight tells you the weight of one unit volume of the material under gravity.

Interpretation. Picture a cube with volume $1 m^{3}$ . If its density is $ρ = 1000$ kg/m³, its weight is $1000 \times 9.8 = 9.8 \times 1 0^{3}$ N. For a foam with $ρ = 50$ kg/m³, the same volume weighs only $490$ N.

Example Problem 1

A block of oak has mass $m = 0.80$ kg and volume $V = 1.0 \times 1 0^{- 3}$ m³. Calculate its density and specific weight.

Compute density:

$ρ = \frac{m}{V} = \frac{0.80}{1.0 \times 1 0 ^{- 3}} = 800 kg/ m^{3} .$
Compute specific weight:

$γ = ρ g = 800 \times 9.8 = 7.84 \times 1 0^{3} N/ m^{3} .$

Discussion.

Oak, with $ρ \approx 800$ kg/m³, is less dense than water ( $1000$ kg/m³), which is why it floats. Its specific weight $7.84 \times 1 0^{3}$ N/m³ means each cubic meter of oak would have that weight under gravity.

Hydrostatic pressure

In a fluid at rest, pressure increases with depth because deeper layers must support the weight of the fluid above them.

Pressure definition:

$p = \frac{F}{A}, [p] = Pa = N/ m^{2} .$

You can feel this effect when you dive deeper in a pool: the pressure on your ears increases because there’s more water above you.

Derivation of the hydrostatic law. Consider a vertical column of fluid with cross-sectional area $A$ and height $h$ .

The volume is $V = A h$ .
The mass is $m = ρ V = ρ A h$ .
The weight is

$W = m g = (ρ V) g = ρ A h g .$

That weight is supported by the pressure force at the bottom, $p A$ :

$p A = W ⟹ p = ρ g h .$

This is the gauge pressure due only to the fluid above the point.

Hydrostatic pressure due to the weight of the fluid

Absolute pressure. If the fluid surface is open to the atmosphere, the total (absolute) pressure at depth $h$ is

$p_{abs} = p_{atm} + ρ g h .$

Example problem 2

What is the pressure 20 m below the surface of freshwater? Use $ρ = 1000$ kg/m³, $g = 9.8$ m/s² and $p_{atm} = 1.01 \times 1 0^{5}$ Pa.

Compute gauge pressure:

$p_{gauge} = ρ g h = 1000 \times 9.8 \times 20 = 1.96 \times 1 0^{5} Pa .$
Add atmospheric pressure:

$p_{abs} = 1.01 \times 1 0^{5} + 1.96 \times 1 0^{5} = 2.97 \times 1 0^{5} Pa .$

Interpretation.

At 20 m depth, the water contributes nearly twice atmospheric pressure on its own. Any diver or submarine must withstand that extra pressure.

Pascal’s principle & hydraulic machines Pascal’s principle statement

Any change in pressure applied to an enclosed incompressible fluid is transmitted equally in all directions throughout the fluid.

Consequence. If two pistons are connected by the same enclosed fluid, the pressure is the same at both pistons. That means a small force applied over a small area can produce a larger force over a larger area.

Pascal's principle for equal pressure transmission in a fluid

Since $p_{1} = p_{2}$ , we have

$\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}} ⟹ F_{2} = F_{1} \frac{A _{2}}{A _{1}} .$

Example Problem 3

A hydraulic jack has small piston area $A_{1} = 0.02$ m² and large piston area $A_{2} = 0.80$ m². Find the force $F_{1}$ needed to lift a car of weight $F_{2} = 12000$ N.

Use $F_{2} = F_{1} (A_{2} / A_{1})$ .
Solve:

$F_{1} = F_{2} \frac{A _{1}}{A _{2}} = 12000 \times \frac{0.02}{0.80} = 300 N .$

Sidenote

Practical tip

Hydraulic systems trade force for distance: the small piston must move farther than the large piston rises.

Buoyancy & archimedes’ principle Physical origin of buoyancy

In a fluid, pressure increases with depth. So for an immersed object, the pressure on the bottom surface is greater than the pressure on the top surface. That pressure difference creates a net upward force called the buoyant force.

Here

$p_{top} = p_{atm} + ρ g h_{top}, p_{bot} = p_{atm} + ρ g h_{bot} .$

The net upward force on the block is

$F_{b} = p_{bot} A - p_{top} A = ρ g (h_{bot} - h_{top}) A = ρ g V_{disp},$

where $V_{disp} = A Δ h$ is the volume of fluid displaced.

Archimedes’ principle.

$F_{b} = ρ_{fluid} V_{disp} g .$

This holds whether the object is fully or partially submerged.

Example Problem 4

A cube of side $L = 0.10$ m and density $ρ_{cube} = 600$ kg/m³ floats in water. Sketch the equilibrium and find (a) the submerged depth $h$ , (b) the height above water.

Solution. Weight $W = ρ_{cube} L^{3} g$ , buoyant force $F_{b} = ρ_{water} (L^{2} h) g$ . At equilibrium, set $F_{b} = W$ :

$ρ_{water} L^{2} h g = ρ_{cube} L^{3} g ⟹ h = L \frac{ρ _{cube}}{ρ _{water}} = 0.10 \frac{600}{1000} = 0.060 m .$

Height above water $L - h = 0.10 - 0.06 = 0.04$ m.

Example problem 5

A solid sphere of radius $R = 0.20$ m and density $ρ_{sphere} = 2000$ kg/m³ is attached to a scale and fully submerged in oil ( $ρ_{oil} = 800$ kg/m³). What does the scale read? ( $g = 9.8$ m/s²)

Solution.

Volume of sphere $V = \frac{4}{3} π R^{3} = \frac{4}{3} π (0.2)^{3} = 0.0335$ m³.
Weight of sphere $W = ρ_{sphere} V g = 2000 \cdot 0.0335 \cdot 9.8 = 656$ N.
Buoyant force $F_{b} = ρ_{oil} V g = 800 \cdot 0.0335 \cdot 9.8 = 263$ N.
Net downward force $= W - F_{b} = 656 - 263 = 393$ N. Thus the scale reads $393 N$ .

Reasoning practice

(a) Why is the hydrostatic pressure at a given depth the same in a very wide lake as in a narrow canal?

Explanation: Hydrostatic pressure depends only on depth (and fluid density): $p = ρ g h$ . For any chosen area $A$ , a vertical column of depth $h$ contains mass $ρ A h$ , so the weight per unit area at the bottom is the same. The lake’s width doesn’t appear in $p = ρ g h$ .

(b) A large steel cargo ship and a small wooden canoe both float in the same lake. Using Archimedes’ principle, explain why each vessel displaces exactly its own weight in water, and why the vastly different hull shapes do not change this equilibrium condition.

Explanation: Archimedes’ principle gives the buoyant force:

$F_{b} = ρ_{water} g V_{disp} .$

For a floating object in equilibrium, the net force is zero, so the buoyant force must equal the object’s weight $W = m g$ :

$ρ_{water} g V_{disp} = m g ⟹ V_{disp} = \frac{m}{ρ _{water}} .$

So the displaced volume is set by the object’s mass and the fluid density, not by the hull shape. Shape affects how the object sits in the water (and its stability), but it can’t change the requirement that the displaced water’s weight matches the object’s weight.

Different hull shapes for a cargo ship and canoe

(c) A sealed container full of fluid is shaken vigorously. Will the pressure at the bottom change compared to when it is at rest? Explain.

Explanation: If the fluid remains incompressible and the container has no net acceleration as a whole, the time-averaged pressure difference with depth is still described by $p = ρ g h$ . Shaking can create temporary local pressure fluctuations, but once the fluid settles back to rest, the pressure at depth $h$ returns to $ρ g h$ .

(d) Two pistons of equal area are connected by oil. One piston is pushed down slowly. Describe how Pascal’s principle dictates the motion and why the force on the second piston equals your push.

Explanation:

Piston 1 applies a force $F_{1}$ over area $A$ , creating a pressure increase $Δ p = F_{1} / A$ . Pascal’s principle says that same $Δ p$ is transmitted throughout the fluid, including to piston 2. Since piston 2 has the same area $A$ , the force it experiences is $F_{2} = Δ p A = F_{1}$ . So the second piston responds with the same force as your push.

Example problem 6

A wooden block (density $ρ = 600$ kg/m³) of volume $0.005$ m³ floats in water. Determine the submerged volume and the height floating above the surface.

Weight of block:

$W = ρ_{block} V g = 600 \times 0.005 \times 9.8 = 29.4 N .$
Buoyant force at equilibrium equals weight:

$F_{b} = ρ_{water} V_{disp} g = 29.4.$
Solve for submerged volume:

$V_{disp} = \frac{29.4}{1000 \times 9.8} = 0.0030 m^{3} .$
Fraction submerged $= V_{disp} / V = 0.0030/0.005 = 0.60$ . Thus 60% submerged, 40% above water.

Example problem 7

A solid iron sphere of mass $m = 80$ kg is attached to a string and slowly lowered into mercury ( $ρ = 13600$ kg/m³). When fully submerged and left to stabilize, what tension does the string finally read? Take $g = 9.8$ m/s². Solution :

When the iron sphere is fully submerged in mercury, its buoyant force exceeds its weight:

$F_{b} = ρ_{Hg} V g \approx 1370 N, W = mg = 784 N,$

so $F_{b} > W$ . A string can only pull, not push, so it can’t provide the downward force needed to hold the sphere fully submerged. The string goes slack, and the sphere rises until it reaches floating equilibrium where $F_{b} = W$ . Thus the tension is

$T = 0 (string slack).$

String goes slack due to buoyant force exceeding the weight

Density & Specific Weight\

Density ( $ρ = m / V$ ): mass per unit volume, units kg/m³
Specific weight ( $γ = ρ g$ ): weight per unit volume, units N/m³
Higher density = more mass/weight per volume; lower density = less

Hydrostatic Pressure\

Pressure in fluid at rest increases with depth: $p = ρ g h$
Absolute pressure at depth: $p_{abs} = p_{atm} + ρ g h$
Pressure depends only on depth and fluid density, not container shape/width

Pascal’s Principle & Hydraulic Machines\

Pressure change applied to enclosed fluid transmitted equally in all directions
Hydraulic systems: $F_{2} = F_{1} (A_{2} / A_{1})$ , force multiplied by area ratio
Small input force over small area produces large output force over large area

Buoyancy & Archimedes’ Principle\

Buoyant force arises from pressure difference between bottom and top surfaces
Archimedes’ principle: $F_{b} = ρ_{fluid} V_{disp} g$
Floating equilibrium: object displaces fluid equal to its own weight

Reasoning Practice\

Hydrostatic pressure at a given depth same regardless of container width/shape
Floating objects (regardless of shape) displace fluid equal to their own weight
Shaking a sealed fluid container: average pressure at depth unchanged after settling
Equal-area pistons: force applied to one piston transmitted equally to the other via fluid

Key Example Takeaways\

Fraction submerged = $ρ_{object} / ρ_{fluid}$ for floating objects
Buoyant force reduces apparent weight of submerged objects
If buoyant force exceeds object weight, supporting string goes slack (object floats up)

Next | 9.2 Dynamic fluids

Static fluids

Fluids at rest (liquids and gases) exert pressure on any surface they touch and produce buoyant forces on objects immersed in them.

How the intrinsic property of density leads to hydrostatic pressure
How that pressure transmits via Pascal’s principle
How displaced fluid produces buoyancy via Archimedes’ principle

The goal is to build a clear conceptual picture and then support it with algebra and worked examples.

Density & specific weight

Definition & physical meaning.

The mass density $ρ$ of a substance tells you how much mass is packed into a given volume:

$ρ = \frac{m}{V}, [ρ] = kg/ m^{3} .$

Higher density means more mass per unit volume (for example, lead).
Lower density means less mass per unit volume (for example, air).

Multiplying density by the acceleration of gravity $g$ gives the specific weight:

$γ = ρ g, [γ] = N/ m^{3},$

Specific weight tells you the weight of one unit volume of the material under gravity.

Example Problem 1

A block of oak has mass $m = 0.80$ kg and volume $V = 1.0 \times 1 0^{- 3}$ m³. Calculate its density and specific weight.

Compute density:

$ρ = \frac{m}{V} = \frac{0.80}{1.0 \times 1 0 ^{- 3}} = 800 kg/ m^{3} .$
Compute specific weight:

$γ = ρ g = 800 \times 9.8 = 7.84 \times 1 0^{3} N/ m^{3} .$

Discussion.

Hydrostatic pressure

In a fluid at rest, pressure increases with depth because deeper layers must support the weight of the fluid above them.

Pressure definition:

$p = \frac{F}{A}, [p] = Pa = N/ m^{2} .$

You can feel this effect when you dive deeper in a pool: the pressure on your ears increases because there’s more water above you.

Derivation of the hydrostatic law. Consider a vertical column of fluid with cross-sectional area $A$ and height $h$ .

The volume is $V = A h$ .
The mass is $m = ρ V = ρ A h$ .
The weight is

$W = m g = (ρ V) g = ρ A h g .$

That weight is supported by the pressure force at the bottom, $p A$ :

$p A = W ⟹ p = ρ g h .$

This is the gauge pressure due only to the fluid above the point.

Absolute pressure. If the fluid surface is open to the atmosphere, the total (absolute) pressure at depth $h$ is

$p_{abs} = p_{atm} + ρ g h .$

Example problem 2

What is the pressure 20 m below the surface of freshwater? Use $ρ = 1000$ kg/m³, $g = 9.8$ m/s² and $p_{atm} = 1.01 \times 1 0^{5}$ Pa.

Compute gauge pressure:

$p_{gauge} = ρ g h = 1000 \times 9.8 \times 20 = 1.96 \times 1 0^{5} Pa .$
Add atmospheric pressure:

$p_{abs} = 1.01 \times 1 0^{5} + 1.96 \times 1 0^{5} = 2.97 \times 1 0^{5} Pa .$

Interpretation.

At 20 m depth, the water contributes nearly twice atmospheric pressure on its own. Any diver or submarine must withstand that extra pressure.

Pascal’s principle & hydraulic machines Pascal’s principle statement

Any change in pressure applied to an enclosed incompressible fluid is transmitted equally in all directions throughout the fluid.

Since $p_{1} = p_{2}$ , we have

$\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}} ⟹ F_{2} = F_{1} \frac{A _{2}}{A _{1}} .$

Example Problem 3

A hydraulic jack has small piston area $A_{1} = 0.02$ m² and large piston area $A_{2} = 0.80$ m². Find the force $F_{1}$ needed to lift a car of weight $F_{2} = 12000$ N.

Use $F_{2} = F_{1} (A_{2} / A_{1})$ .
Solve:

$F_{1} = F_{2} \frac{A _{1}}{A _{2}} = 12000 \times \frac{0.02}{0.80} = 300 N .$

Sidenote

Practical tip

Hydraulic systems trade force for distance: the small piston must move farther than the large piston rises.

Buoyancy & archimedes’ principle Physical origin of buoyancy

Here

$p_{top} = p_{atm} + ρ g h_{top}, p_{bot} = p_{atm} + ρ g h_{bot} .$

The net upward force on the block is

$F_{b} = p_{bot} A - p_{top} A = ρ g (h_{bot} - h_{top}) A = ρ g V_{disp},$

where $V_{disp} = A Δ h$ is the volume of fluid displaced.

Archimedes’ principle.

$F_{b} = ρ_{fluid} V_{disp} g .$

This holds whether the object is fully or partially submerged.

Example Problem 4

A cube of side $L = 0.10$ m and density $ρ_{cube} = 600$ kg/m³ floats in water. Sketch the equilibrium and find (a) the submerged depth $h$ , (b) the height above water.

Solution. Weight $W = ρ_{cube} L^{3} g$ , buoyant force $F_{b} = ρ_{water} (L^{2} h) g$ . At equilibrium, set $F_{b} = W$ :

$ρ_{water} L^{2} h g = ρ_{cube} L^{3} g ⟹ h = L \frac{ρ _{cube}}{ρ _{water}} = 0.10 \frac{600}{1000} = 0.060 m .$

Height above water $L - h = 0.10 - 0.06 = 0.04$ m.

Example problem 5

Solution.

Volume of sphere $V = \frac{4}{3} π R^{3} = \frac{4}{3} π (0.2)^{3} = 0.0335$ m³.
Weight of sphere $W = ρ_{sphere} V g = 2000 \cdot 0.0335 \cdot 9.8 = 656$ N.
Buoyant force $F_{b} = ρ_{oil} V g = 800 \cdot 0.0335 \cdot 9.8 = 263$ N.
Net downward force $= W - F_{b} = 656 - 263 = 393$ N. Thus the scale reads $393 N$ .

Reasoning practice

(a) Why is the hydrostatic pressure at a given depth the same in a very wide lake as in a narrow canal?

Explanation: Archimedes’ principle gives the buoyant force:

$F_{b} = ρ_{water} g V_{disp} .$

For a floating object in equilibrium, the net force is zero, so the buoyant force must equal the object’s weight $W = m g$ :

$ρ_{water} g V_{disp} = m g ⟹ V_{disp} = \frac{m}{ρ _{water}} .$

(c) A sealed container full of fluid is shaken vigorously. Will the pressure at the bottom change compared to when it is at rest? Explain.

Explanation:

Example problem 6

A wooden block (density $ρ = 600$ kg/m³) of volume $0.005$ m³ floats in water. Determine the submerged volume and the height floating above the surface.

Weight of block:

$W = ρ_{block} V g = 600 \times 0.005 \times 9.8 = 29.4 N .$
Buoyant force at equilibrium equals weight:

$F_{b} = ρ_{water} V_{disp} g = 29.4.$
Solve for submerged volume:

$V_{disp} = \frac{29.4}{1000 \times 9.8} = 0.0030 m^{3} .$
Fraction submerged $= V_{disp} / V = 0.0030/0.005 = 0.60$ . Thus 60% submerged, 40% above water.

Example problem 7

When the iron sphere is fully submerged in mercury, its buoyant force exceeds its weight:

$F_{b} = ρ_{Hg} V g \approx 1370 N, W = mg = 784 N,$

$T = 0 (string slack).$

Achievable AP Physics 1

9. Fluids

Our AP Physics 1 course is currently in development and is a work-in-progress.

Static fluids

12 min read

Font

Discuss

Feedback

Fluids at rest (liquids and gases) exert pressure on any surface they touch and produce buoyant forces on objects immersed in them.

How the intrinsic property of density leads to hydrostatic pressure
How that pressure transmits via Pascal’s principle
How displaced fluid produces buoyancy via Archimedes’ principle

The goal is to build a clear conceptual picture and then support it with algebra and worked examples.

Density & specific weight

Definition & physical meaning.

The mass density $ρ$ of a substance tells you how much mass is packed into a given volume:

$ρ = \frac{m}{V}, [ρ] = kg/ m^{3} .$

Higher density means more mass per unit volume (for example, lead).
Lower density means less mass per unit volume (for example, air).

Multiplying density by the acceleration of gravity $g$ gives the specific weight:

$γ = ρ g, [γ] = N/ m^{3},$

Specific weight tells you the weight of one unit volume of the material under gravity.

Example Problem 1

A block of oak has mass $m = 0.80$ kg and volume $V = 1.0 \times 1 0^{- 3}$ m³. Calculate its density and specific weight.

Compute density:

$ρ = \frac{m}{V} = \frac{0.80}{1.0 \times 1 0 ^{- 3}} = 800 kg/ m^{3} .$
Compute specific weight:

$γ = ρ g = 800 \times 9.8 = 7.84 \times 1 0^{3} N/ m^{3} .$

Discussion.

Hydrostatic pressure

In a fluid at rest, pressure increases with depth because deeper layers must support the weight of the fluid above them.

Pressure definition:

$p = \frac{F}{A}, [p] = Pa = N/ m^{2} .$

You can feel this effect when you dive deeper in a pool: the pressure on your ears increases because there’s more water above you.

Derivation of the hydrostatic law. Consider a vertical column of fluid with cross-sectional area $A$ and height $h$ .

The volume is $V = A h$ .
The mass is $m = ρ V = ρ A h$ .
The weight is

$W = m g = (ρ V) g = ρ A h g .$

That weight is supported by the pressure force at the bottom, $p A$ :

$p A = W ⟹ p = ρ g h .$

This is the gauge pressure due only to the fluid above the point.

Absolute pressure. If the fluid surface is open to the atmosphere, the total (absolute) pressure at depth $h$ is

$p_{abs} = p_{atm} + ρ g h .$

Example problem 2

What is the pressure 20 m below the surface of freshwater? Use $ρ = 1000$ kg/m³, $g = 9.8$ m/s² and $p_{atm} = 1.01 \times 1 0^{5}$ Pa.

Compute gauge pressure:

$p_{gauge} = ρ g h = 1000 \times 9.8 \times 20 = 1.96 \times 1 0^{5} Pa .$
Add atmospheric pressure:

$p_{abs} = 1.01 \times 1 0^{5} + 1.96 \times 1 0^{5} = 2.97 \times 1 0^{5} Pa .$

Interpretation.

At 20 m depth, the water contributes nearly twice atmospheric pressure on its own. Any diver or submarine must withstand that extra pressure.

Pascal’s principle & hydraulic machines Pascal’s principle statement

Any change in pressure applied to an enclosed incompressible fluid is transmitted equally in all directions throughout the fluid.

Since $p_{1} = p_{2}$ , we have

$\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}} ⟹ F_{2} = F_{1} \frac{A _{2}}{A _{1}} .$

Example Problem 3

A hydraulic jack has small piston area $A_{1} = 0.02$ m² and large piston area $A_{2} = 0.80$ m². Find the force $F_{1}$ needed to lift a car of weight $F_{2} = 12000$ N.

Use $F_{2} = F_{1} (A_{2} / A_{1})$ .
Solve:

$F_{1} = F_{2} \frac{A _{1}}{A _{2}} = 12000 \times \frac{0.02}{0.80} = 300 N .$

Sidenote

Practical tip

Hydraulic systems trade force for distance: the small piston must move farther than the large piston rises.

Buoyancy & archimedes’ principle Physical origin of buoyancy

Here

$p_{top} = p_{atm} + ρ g h_{top}, p_{bot} = p_{atm} + ρ g h_{bot} .$

The net upward force on the block is

$F_{b} = p_{bot} A - p_{top} A = ρ g (h_{bot} - h_{top}) A = ρ g V_{disp},$

where $V_{disp} = A Δ h$ is the volume of fluid displaced.

Archimedes’ principle.

$F_{b} = ρ_{fluid} V_{disp} g .$

This holds whether the object is fully or partially submerged.

Example Problem 4

A cube of side $L = 0.10$ m and density $ρ_{cube} = 600$ kg/m³ floats in water. Sketch the equilibrium and find (a) the submerged depth $h$ , (b) the height above water.

Solution. Weight $W = ρ_{cube} L^{3} g$ , buoyant force $F_{b} = ρ_{water} (L^{2} h) g$ . At equilibrium, set $F_{b} = W$ :

$ρ_{water} L^{2} h g = ρ_{cube} L^{3} g ⟹ h = L \frac{ρ _{cube}}{ρ _{water}} = 0.10 \frac{600}{1000} = 0.060 m .$

Height above water $L - h = 0.10 - 0.06 = 0.04$ m.

Example problem 5

Solution.

Volume of sphere $V = \frac{4}{3} π R^{3} = \frac{4}{3} π (0.2)^{3} = 0.0335$ m³.
Weight of sphere $W = ρ_{sphere} V g = 2000 \cdot 0.0335 \cdot 9.8 = 656$ N.
Buoyant force $F_{b} = ρ_{oil} V g = 800 \cdot 0.0335 \cdot 9.8 = 263$ N.
Net downward force $= W - F_{b} = 656 - 263 = 393$ N. Thus the scale reads $393 N$ .

Reasoning practice

(a) Why is the hydrostatic pressure at a given depth the same in a very wide lake as in a narrow canal?

Explanation: Archimedes’ principle gives the buoyant force:

$F_{b} = ρ_{water} g V_{disp} .$

For a floating object in equilibrium, the net force is zero, so the buoyant force must equal the object’s weight $W = m g$ :

$ρ_{water} g V_{disp} = m g ⟹ V_{disp} = \frac{m}{ρ _{water}} .$

(c) A sealed container full of fluid is shaken vigorously. Will the pressure at the bottom change compared to when it is at rest? Explain.

Explanation:

Example problem 6

A wooden block (density $ρ = 600$ kg/m³) of volume $0.005$ m³ floats in water. Determine the submerged volume and the height floating above the surface.

Weight of block:

$W = ρ_{block} V g = 600 \times 0.005 \times 9.8 = 29.4 N .$
Buoyant force at equilibrium equals weight:

$F_{b} = ρ_{water} V_{disp} g = 29.4.$
Solve for submerged volume:

$V_{disp} = \frac{29.4}{1000 \times 9.8} = 0.0030 m^{3} .$
Fraction submerged $= V_{disp} / V = 0.0030/0.005 = 0.60$ . Thus 60% submerged, 40% above water.

Example problem 7

When the iron sphere is fully submerged in mercury, its buoyant force exceeds its weight:

$F_{b} = ρ_{Hg} V g \approx 1370 N, W = mg = 784 N,$

$T = 0 (string slack).$

Density & Specific Weight\

Density ( $ρ = m / V$ ): mass per unit volume, units kg/m³
Specific weight ( $γ = ρ g$ ): weight per unit volume, units N/m³
Higher density = more mass/weight per volume; lower density = less

Hydrostatic Pressure\

Pressure in fluid at rest increases with depth: $p = ρ g h$
Absolute pressure at depth: $p_{abs} = p_{atm} + ρ g h$
Pressure depends only on depth and fluid density, not container shape/width

Pascal’s Principle & Hydraulic Machines\

Pressure change applied to enclosed fluid transmitted equally in all directions
Hydraulic systems: $F_{2} = F_{1} (A_{2} / A_{1})$ , force multiplied by area ratio
Small input force over small area produces large output force over large area

Buoyancy & Archimedes’ Principle\

Buoyant force arises from pressure difference between bottom and top surfaces
Archimedes’ principle: $F_{b} = ρ_{fluid} V_{disp} g$
Floating equilibrium: object displaces fluid equal to its own weight

Reasoning Practice\

Hydrostatic pressure at a given depth same regardless of container width/shape
Floating objects (regardless of shape) displace fluid equal to their own weight
Shaking a sealed fluid container: average pressure at depth unchanged after settling
Equal-area pistons: force applied to one piston transmitted equally to the other via fluid

Key Example Takeaways\

Fraction submerged = $ρ_{object} / ρ_{fluid}$ for floating objects
Buoyant force reduces apparent weight of submerged objects
If buoyant force exceeds object weight, supporting string goes slack (object floats up)

Next | 9.2 Dynamic fluids

Static fluids

Fluids at rest (liquids and gases) exert pressure on any surface they touch and produce buoyant forces on objects immersed in them.

How the intrinsic property of density leads to hydrostatic pressure
How that pressure transmits via Pascal’s principle
How displaced fluid produces buoyancy via Archimedes’ principle

The goal is to build a clear conceptual picture and then support it with algebra and worked examples.

Density & specific weight

Definition & physical meaning.

The mass density $ρ$ of a substance tells you how much mass is packed into a given volume:

$ρ = \frac{m}{V}, [ρ] = kg/ m^{3} .$

Higher density means more mass per unit volume (for example, lead).
Lower density means less mass per unit volume (for example, air).

Multiplying density by the acceleration of gravity $g$ gives the specific weight:

$γ = ρ g, [γ] = N/ m^{3},$

Specific weight tells you the weight of one unit volume of the material under gravity.

Example Problem 1

A block of oak has mass $m = 0.80$ kg and volume $V = 1.0 \times 1 0^{- 3}$ m³. Calculate its density and specific weight.

Compute density:

$ρ = \frac{m}{V} = \frac{0.80}{1.0 \times 1 0 ^{- 3}} = 800 kg/ m^{3} .$
Compute specific weight:

$γ = ρ g = 800 \times 9.8 = 7.84 \times 1 0^{3} N/ m^{3} .$

Discussion.

Hydrostatic pressure

In a fluid at rest, pressure increases with depth because deeper layers must support the weight of the fluid above them.

Pressure definition:

$p = \frac{F}{A}, [p] = Pa = N/ m^{2} .$

You can feel this effect when you dive deeper in a pool: the pressure on your ears increases because there’s more water above you.

Derivation of the hydrostatic law. Consider a vertical column of fluid with cross-sectional area $A$ and height $h$ .

The volume is $V = A h$ .
The mass is $m = ρ V = ρ A h$ .
The weight is

$W = m g = (ρ V) g = ρ A h g .$

That weight is supported by the pressure force at the bottom, $p A$ :

$p A = W ⟹ p = ρ g h .$

This is the gauge pressure due only to the fluid above the point.

Absolute pressure. If the fluid surface is open to the atmosphere, the total (absolute) pressure at depth $h$ is

$p_{abs} = p_{atm} + ρ g h .$

Example problem 2

What is the pressure 20 m below the surface of freshwater? Use $ρ = 1000$ kg/m³, $g = 9.8$ m/s² and $p_{atm} = 1.01 \times 1 0^{5}$ Pa.

Compute gauge pressure:

$p_{gauge} = ρ g h = 1000 \times 9.8 \times 20 = 1.96 \times 1 0^{5} Pa .$
Add atmospheric pressure:

$p_{abs} = 1.01 \times 1 0^{5} + 1.96 \times 1 0^{5} = 2.97 \times 1 0^{5} Pa .$

Interpretation.

At 20 m depth, the water contributes nearly twice atmospheric pressure on its own. Any diver or submarine must withstand that extra pressure.

Pascal’s principle & hydraulic machines Pascal’s principle statement

Any change in pressure applied to an enclosed incompressible fluid is transmitted equally in all directions throughout the fluid.

Since $p_{1} = p_{2}$ , we have

$\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}} ⟹ F_{2} = F_{1} \frac{A _{2}}{A _{1}} .$

Example Problem 3

A hydraulic jack has small piston area $A_{1} = 0.02$ m² and large piston area $A_{2} = 0.80$ m². Find the force $F_{1}$ needed to lift a car of weight $F_{2} = 12000$ N.

Use $F_{2} = F_{1} (A_{2} / A_{1})$ .
Solve:

$F_{1} = F_{2} \frac{A _{1}}{A _{2}} = 12000 \times \frac{0.02}{0.80} = 300 N .$

Sidenote

Practical tip

Hydraulic systems trade force for distance: the small piston must move farther than the large piston rises.

Buoyancy & archimedes’ principle Physical origin of buoyancy

Here

$p_{top} = p_{atm} + ρ g h_{top}, p_{bot} = p_{atm} + ρ g h_{bot} .$

The net upward force on the block is

$F_{b} = p_{bot} A - p_{top} A = ρ g (h_{bot} - h_{top}) A = ρ g V_{disp},$

where $V_{disp} = A Δ h$ is the volume of fluid displaced.

Archimedes’ principle.

$F_{b} = ρ_{fluid} V_{disp} g .$

This holds whether the object is fully or partially submerged.

Example Problem 4

A cube of side $L = 0.10$ m and density $ρ_{cube} = 600$ kg/m³ floats in water. Sketch the equilibrium and find (a) the submerged depth $h$ , (b) the height above water.

Solution. Weight $W = ρ_{cube} L^{3} g$ , buoyant force $F_{b} = ρ_{water} (L^{2} h) g$ . At equilibrium, set $F_{b} = W$ :

$ρ_{water} L^{2} h g = ρ_{cube} L^{3} g ⟹ h = L \frac{ρ _{cube}}{ρ _{water}} = 0.10 \frac{600}{1000} = 0.060 m .$

Height above water $L - h = 0.10 - 0.06 = 0.04$ m.

Example problem 5

Solution.

Volume of sphere $V = \frac{4}{3} π R^{3} = \frac{4}{3} π (0.2)^{3} = 0.0335$ m³.
Weight of sphere $W = ρ_{sphere} V g = 2000 \cdot 0.0335 \cdot 9.8 = 656$ N.
Buoyant force $F_{b} = ρ_{oil} V g = 800 \cdot 0.0335 \cdot 9.8 = 263$ N.
Net downward force $= W - F_{b} = 656 - 263 = 393$ N. Thus the scale reads $393 N$ .

Reasoning practice

(a) Why is the hydrostatic pressure at a given depth the same in a very wide lake as in a narrow canal?

Explanation: Archimedes’ principle gives the buoyant force:

$F_{b} = ρ_{water} g V_{disp} .$

For a floating object in equilibrium, the net force is zero, so the buoyant force must equal the object’s weight $W = m g$ :

$ρ_{water} g V_{disp} = m g ⟹ V_{disp} = \frac{m}{ρ _{water}} .$

(c) A sealed container full of fluid is shaken vigorously. Will the pressure at the bottom change compared to when it is at rest? Explain.

Explanation:

Example problem 6

A wooden block (density $ρ = 600$ kg/m³) of volume $0.005$ m³ floats in water. Determine the submerged volume and the height floating above the surface.

Weight of block:

$W = ρ_{block} V g = 600 \times 0.005 \times 9.8 = 29.4 N .$
Buoyant force at equilibrium equals weight:

$F_{b} = ρ_{water} V_{disp} g = 29.4.$
Solve for submerged volume:

$V_{disp} = \frac{29.4}{1000 \times 9.8} = 0.0030 m^{3} .$
Fraction submerged $= V_{disp} / V = 0.0030/0.005 = 0.60$ . Thus 60% submerged, 40% above water.

Example problem 7

When the iron sphere is fully submerged in mercury, its buoyant force exceeds its weight:

$F_{b} = ρ_{Hg} V g \approx 1370 N, W = mg = 784 N,$

$T = 0 (string slack).$