Introduction to Fluid Mechanics
Dear student, in the previous units you studied motion of solid objects. Now we turn our attention to fluids. A fluid is any substance that can flow — this includes both liquids (like water, oil, blood) and gases (like air, oxygen). Fluids are everywhere around us: the water you drink, the air you breathe, the blood flowing through your body — all are fluids.
In this unit, we will study fluids at rest (fluid statics) and fluids in motion (fluid dynamics). These concepts are very important in engineering, medicine, and everyday life. Ready? Let us start!
3.1 Fluid Statics
What is a Fluid?
A fluid is a substance that cannot sustain a shear stress (a force parallel to its surface) when at rest. Unlike solids, fluids continuously deform under shear stress. Fluids are classified into:
- Liquids: Have a definite volume but no definite shape (take the shape of their container). They are nearly incompressible.
- Gases: Have neither definite volume nor definite shape. They are highly compressible.
Density
Density is one of the most basic properties of a fluid. It tells us how much mass is packed into a given volume.
where ρ (rho) = density, m = mass, V = volume.
Unit: kg/m³. Water has a density of 1000 kg/m³. Air has a density of about 1.29 kg/m³.
Relative Density (Specific Gravity)
Relative density has no unit (it is a ratio). For example, if a liquid has a relative density of 0.8, its density is 0.8 × 1000 = 800 kg/m³.
Pressure
Pressure is defined as the force acting perpendicularly per unit area:
Unit: Pascal (Pa) = N/m². Other common units: atmosphere (atm), bar, mmHg.
1 atm = 1.013 × 10⁵ Pa = 760 mmHg = 1.013 bar
1 bar = 10⁵ Pa
Have you noticed why a sharp knife cuts better than a blunt one? The sharp knife has a very small edge area, so the same force produces a much larger pressure. Think about it!
Worked Example 1
A block of metal of mass 500 kg rests on a surface of area 2 m². What pressure does it exert on the surface?
F = mg = 500 × 10 = 5000 N
$$P = \frac{F}{A} = \frac{5000}{2} = 2500 \text{ Pa}$$
- Density ρ = m/V; unit: kg/m³.
- Pressure P = F/A; unit: Pa (N/m²).
- 1 atm = 1.013 × 10⁵ Pa = 760 mmHg.
- Pressure is a scalar quantity (not a vector!).
- Same force over smaller area = larger pressure.
Practice Question 1: The density of mercury is 13,600 kg/m³. What is the mass of 0.002 m³ of mercury?
Practice Question 2: A person weighing 600 N stands on one foot of area 0.02 m². What pressure is exerted on the ground?
Practice Question 3: The relative density of engine oil is 0.9. What is its density in kg/m³?
3.2 Pressure in Fluids at Rest
Pressure Due to a Fluid Column
Dear student, when you swim to the bottom of a deep pool, you feel pressure in your ears. Why? Because the weight of the water above you pushes down. The deeper you go, the more water is above you, and the greater the pressure.
Consider a column of fluid of height h, density ρ, and cross-sectional area A:
This is the gauge pressure — the pressure due to the fluid alone. If we include atmospheric pressure on top, we get the absolute pressure:
Important Properties of Fluid Pressure
- Fluid pressure acts equally in all directions at a given depth (Pascal’s law of fluid pressure).
- Pressure depends only on depth, not on the shape or area of the container.
- Pressure increases linearly with depth.
Worked Example 2
What is the pressure at the bottom of a swimming pool that is 3 m deep? (ρ_water = 1000 kg/m³, g = 10 m/s², P_atm = 1.013 × 10⁵ Pa)
Gauge pressure: P = ρgh = 1000 × 10 × 3 = 30000 Pa = 30 kPa
Absolute pressure: P_abs = P_atm + ρgh = 101300 + 30000 = 131300 Pa
Worked Example 3
A water tank is 5 m tall and filled with water. A small hole is made on the side 2 m below the water surface. What is the gauge pressure at the hole?
The depth at the hole is h = 2 m (measured from the water surface, NOT from the bottom).
$$P = \rho g h = 1000 \times 10 \times 2 = 20000 \text{ Pa} = 20 \text{ kPa}$$
Important: Always measure depth from the fluid surface, not from the bottom!
Pascal’s Principle
This principle is the basis for the hydraulic press (also called hydraulic lift), which is used in car repair shops, heavy machinery, and braking systems.
A small force on a small area can produce a large force on a large area! This is called mechanical advantage.
Worked Example 4
In a hydraulic press, the small piston has an area of 10 cm² and the large piston has an area of 500 cm². A force of 200 N is applied to the small piston. Find: (a) the force on the large piston, (b) the mechanical advantage.
(a) Force on large piston:
$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$ $$F_2 = F_1 \times \frac{A_2}{A_1} = 200 \times \frac{500}{10} = 200 \times 50 = 10000 \text{ N}$$
(b) Mechanical advantage:
$$MA = \frac{F_2}{F_1} = \frac{10000}{200} = 50$$
But wait — do we get something for nothing? No! The large piston moves a shorter distance. If the small piston moves down by d₁, the large piston moves up by d₂, and by conservation of energy: F₁d₁ = F₂d₂.
Worked Example 5
In the hydraulic press above, if the small piston moves down by 25 cm, how far does the large piston move up?
By conservation of energy (work input = work output):
F₁d₁ = F₂d₂
200 × 0.25 = 10000 × d₂
50 = 10000 × d₂
d₂ = 50/10000 = 0.005 m = 0.5 cm
The large piston moves only 0.5 cm — 50 times less distance, but with 50 times more force!
Measurement of Pressure
Barometer
A barometer measures atmospheric pressure. It was invented by Evangelista Torricelli. A simple mercury barometer consists of a glass tube filled with mercury, inverted into a mercury dish.
At standard atmospheric pressure, h = 760 mm of mercury (760 mmHg).
Manometer
A manometer measures the pressure of a gas in a container. It consists of a U-tube containing a liquid (usually mercury or water).
Worked Example 6
A mercury barometer reads 750 mmHg on a day when the atmospheric pressure is 1.013 × 10⁵ Pa. Calculate the density of mercury. (g = 10 m/s²)
h = 750 mm = 0.75 m
$$P_{\text{atm}} = \rho g h$$ $$1.013 \times 10^5 = \rho \times 10 \times 0.75$$ $$\rho = \frac{1.013 \times 10^5}{7.5} = 13507 \text{ kg/m}^3$$
(Note: The actual density of mercury is 13,600 kg/m³. The slight difference is because 760 mmHg gives exactly 1 atm, but here the reading is 750 mm.)
Worked Example 7
A U-tube manometer containing water is connected to a gas container. The water level in the open arm is 20 cm higher than in the arm connected to the container. If atmospheric pressure is 1.01 × 10⁵ Pa, find the gas pressure. (ρ_water = 1000 kg/m³, g = 10 m/s²)
Δh = 20 cm = 0.2 m
Since the open arm is higher, the gas pressure pushes the liquid down more than the atmosphere does. Therefore:
$$P_{\text{gas}} = P_{\text{atm}} + \rho g \Delta h$$ $$P_{\text{gas}} = 1.01 \times 10^5 + 1000 \times 10 \times 0.2$$ $$P_{\text{gas}} = 101000 + 2000 = 103000 \text{ Pa} = 1.03 \times 10^5 \text{ Pa}$$
- Pressure at depth h: P = ρgh (gauge) or P = P_atm + ρgh (absolute).
- Pressure depends on depth, NOT on the shape of the container.
- Pascal’s principle: pressure applied to a confined fluid is transmitted equally throughout.
- Hydraulic press: F₂ = F₁(A₂/A₁), but d₂ = d₁(A₁/A₂).
- Barometer: P_atm = ρ_Hg × g × h; standard h = 760 mmHg.
- Manometer: P_gas = P_atm ± ρgΔh (sign depends on which side is higher).
- Always convert mm or cm to meters before calculating!
Practice Question 4: The pressure at a point 10 m below the surface of a lake is 2 × 10⁵ Pa. What is the atmospheric pressure above the lake? (ρ = 1000 kg/m³, g = 10 m/s²)
2 × 10⁵ = P_atm + 1000 × 10 × 10
2 × 10⁵ = P_atm + 1 × 10⁵
P_atm = 2 × 10⁵ − 1 × 10⁵ = 1 × 10⁵ Pa
Practice Question 5: In a hydraulic brake, the force on the brake pad is 2400 N when the driver applies 60 N on the pedal. If the pedal piston has area 3 cm², what is the area of the brake piston?
60/3 = 2400/A₂
A₂ = 2400 × 3/60 = 2400 × 0.05 = 120 cm²
Practice Question 6: A barometer reads 720 mmHg. Express this pressure in Pa. (ρ_Hg = 13600 kg/m³, g = 10 m/s²)
P = ρgh = 13600 × 10 × 0.72 = 97920 Pa
3.3 Archimedes’ Principle
The Discovery
Dear student, there is a famous story about the ancient Greek scientist Archimedes. He was asked by the king to check whether a crown was made of pure gold or was a mixture. While taking a bath, he noticed that the water rose when he got in, and he suddenly realized how to solve the problem. He reportedly jumped out and shouted “Eureka!” (I found it!).
What he discovered is one of the most important principles in fluid mechanics.
where F_b = buoyant force, ρ_fluid = density of the fluid, V_displaced = volume of fluid displaced.
Why Does Buoyancy Occur?
Pressure in a fluid increases with depth. So the bottom of an immersed object experiences a greater upward pressure than the downward pressure on the top. This difference in pressure creates a net upward force — the buoyant force.
Floating and Sinking
Whether an object floats or sinks depends on the comparison between the buoyant force and the weight of the object:
- Floating (equilibrium): Weight of object = Buoyant force → ρ_object × V_object × g = ρ_fluid × V_displaced × g
- Sinking: Weight of object > Buoyant force → ρ_object > ρ_fluid
- Rising (when released underwater): Weight of object < Buoyant force → ρ_object < ρ_fluid
• If ρ_object < ρ_fluid → it FLOATS (partially submerged)
• If ρ_object = ρ_fluid → it FLOATS (fully submerged, neutrally buoyant)
• If ρ_object > ρ_fluid → it SINKS
Fraction Submerged for a Floating Object
For a floating object, weight = buoyant force:
The fraction submerged equals the ratio of densities!
Worked Example 8
A block of wood of volume 0.005 m³ and density 600 kg/m³ floats in water. Find: (a) the buoyant force on the block, (b) the volume of the block submerged, (c) the fraction of the block above water.
Since the block floats: Weight = Buoyant force
(a) Buoyant force:
$$F_b = W = \rho_{\text{wood}} \, V \, g = 600 \times 0.005 \times 10 = 30 \text{ N}$$
(b) Volume submerged:
$$F_b = \rho_{\text{water}} \, V_{\text{disp}} \, g$$ $$30 = 1000 \times V_{\text{disp}} \times 10$$ $$V_{\text{disp}} = \frac{30}{10000} = 0.003 \text{ m}^3$$
(c) Fraction above water:
Fraction submerged = 0.003/0.005 = 0.6 (i.e., 60%)
Fraction above water = 1 − 0.6 = 0.4 (40%)
Shortcut: fraction submerged = ρ_wood/ρ_water = 600/1000 = 0.6 ✓
Worked Example 9
An iron cube of side 10 cm and density 7800 kg/m³ is completely submerged in water. Find: (a) the buoyant force, (b) the net force on the cube (does it sink or float?). (ρ_water = 1000 kg/m³, g = 10 m/s²)
Side = 10 cm = 0.1 m, V = (0.1)³ = 0.001 m³
(a) Buoyant force:
$$F_b = \rho_{\text{water}} \, V \, g = 1000 \times 0.001 \times 10 = 10 \text{ N}$$
(b) Weight of cube:
$$W = \rho_{\text{iron}} \, V \, g = 7800 \times 0.001 \times 10 = 78 \text{ N}$$
Net downward force = W − F_b = 78 − 10 = 68 N (it sinks!)
Since ρ_iron (7800) > ρ_water (1000), we expected it to sink.
Worked Example 10
A piece of metal weighs 50 N in air and 45 N when fully submerged in water. Find: (a) the buoyant force, (b) the volume of the metal, (c) the density of the metal.
(a) Buoyant force:
$$F_b = W_{\text{air}} – W_{\text{water}} = 50 – 45 = 5 \text{ N}$$
(b) Volume of metal:
$$F_b = \rho_{\text{water}} \, V \, g$$ $$5 = 1000 \times V \times 10$$ $$V = \frac{5}{10000} = 0.0005 \text{ m}^3 = 500 \text{ cm}^3$$
(c) Density of metal:
$$\rho = \frac{m}{V} = \frac{W}{Vg} = \frac{50}{0.0005 \times 10} = \frac{50}{0.005} = 10000 \text{ kg/m}^3$$
This density (10000 kg/m³) is close to the density of some steels.
Apparent Weight
When an object is immersed in a fluid, it appears to weigh less. This apparent weight is:
Worked Example 11
A stone weighs 80 N in air and 60 N when submerged in a liquid of density 1200 kg/m³. Find the volume of the stone.
F_b = W_air − W_liquid = 80 − 60 = 20 N
$$F_b = \rho_{\text{liquid}} \, V \, g$$ $$20 = 1200 \times V \times 10$$ $$V = \frac{20}{12000} = 0.00167 \text{ m}^3 = 1670 \text{ cm}^3$$
- Buoyant force: F_b = ρ_fluid × V_displaced × g.
- F_b = weight of fluid displaced — ALWAYS use the FLUID density, not the object density.
- Floating: ρ_object < ρ_fluid; fraction submerged = ρ_object/ρ_fluid.
- Sinking: ρ_object > ρ_fluid.
- Apparent weight: W_app = W_true − F_b.
- F_b depends on V_displaced, NOT on the depth of immersion (as long as fully submerged).
- A common exam trick: use weight difference to find volume, then find density.
Practice Question 7: An iceberg floats in seawater with 90% of its volume submerged. What is the density of the ice? (ρ_seawater = 1030 kg/m³)
0.90 = ρ_ice/1030
ρ_ice = 0.90 × 1030 = 927 kg/m³
Practice Question 8: A metal block of mass 4 kg and volume 0.0015 m³ is placed in water. Does it float or sink? If it sinks, what is the apparent weight?
Since 2667 > 1000 (ρ_water), it sinks.
F_b = ρ_water × V × g = 1000 × 0.0015 × 10 = 15 N
W = mg = 4 × 10 = 40 N
W_apparent = 40 − 15 = 25 N
Practice Question 9: A wooden beam of density 500 kg/m³ has a volume of 0.02 m³. What weight can be placed on top of it before it is completely submerged in water?
F_b (max) = ρ_water × V × g = 1000 × 0.02 × 10 = 200 N
W_beam = ρ_wood × V × g = 500 × 0.02 × 10 = 100 N
For equilibrium: W_beam + W_added = F_b
100 + W_added = 200
W_added = 100 N
So a maximum weight of 100 N can be placed before it is fully submerged.
3.4 Fluid Flow
Types of Fluid Flow
- Streamline (laminar) flow: The fluid flows in smooth, parallel layers. Each particle follows the same path as the one before it. This is orderly flow.
- Turbulent flow: The fluid flows in irregular, chaotic patterns with eddies and swirls. This happens at high velocities or with obstacles.
Can you think of examples? Water flowing slowly from a tap is laminar; when you open the tap fully, it becomes turbulent!
Equation of Continuity
When an incompressible fluid flows through a pipe, the mass flow rate must be the same at every cross-section (what goes in must come out — conservation of mass):
where A = cross-sectional area, v = flow velocity. This is the equation of continuity.
This explains why water shoots out faster from a narrow nozzle — the same volume must pass through a smaller opening in the same time!
Worked Example 12
Water flows through a pipe of cross-sectional area 0.04 m² with a velocity of 2 m/s. The pipe narrows to an area of 0.01 m². Find the velocity in the narrow section.
$$A_1 v_1 = A_2 v_2$$ $$0.04 \times 2 = 0.01 \times v_2$$ $$v_2 = \frac{0.08}{0.01} = 8 \text{ m/s}$$
The velocity increases by a factor of 4 when the area decreases by a factor of 4.
Worked Example 13
Water flows through a pipe at 3 m/s. The pipe diameter is reduced from 6 cm to 3 cm. Find the flow velocity in the narrower section.
A = πd²/4, so A₁/A₂ = d₁²/d₂² = 6²/3² = 36/9 = 4
$$A_1 v_1 = A_2 v_2 \Rightarrow v_2 = v_1 \times \frac{A_1}{A_2} = 3 \times 4 = 12 \text{ m/s}$$
Shortcut: When diameter is halved, area becomes 1/4, so velocity becomes 4 times.
Bernoulli’s Equation
Daniel Bernoulli developed an equation that relates pressure, velocity, and height in a flowing fluid. It is based on the conservation of energy for a flowing fluid.
Each term represents a type of energy per unit volume:
- P — pressure energy per unit volume
- ½ρv² — kinetic energy per unit volume
- ρgh — gravitational potential energy per unit volume
For a horizontal pipe (h = constant):
Applications of Bernoulli’s Principle
1. Lift on an airplane wing (airfoil): The top surface of the wing is curved, so air flows faster over the top than the bottom. Faster air = lower pressure on top. The higher pressure underneath pushes the wing upward → lift.
2. Spray bottle / perfume atomizer: When you squeeze the bulb, air rushes across the top of the tube, creating low pressure. The liquid rises up the tube and is sprayed out.
3. Venturi meter: A device used to measure fluid flow rate. A constriction in the pipe creates a pressure difference that can be measured.
Worked Example 14
Water flows horizontally through a pipe. At a point where the cross-sectional area is 0.05 m², the pressure is 200 kPa and the velocity is 2 m/s. At another point where the area is 0.02 m², find the pressure. (ρ_water = 1000 kg/m³)
First find v₂ using continuity:
$$A_1 v_1 = A_2 v_2 \Rightarrow v_2 = \frac{0.05 \times 2}{0.02} = 5 \text{ m/s}$$
Now apply Bernoulli’s equation (horizontal, so h is constant):
$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$ $$200000 + \frac{1}{2}(1000)(4) = P_2 + \frac{1}{2}(1000)(25)$$ $$200000 + 2000 = P_2 + 12500$$ $$P_2 = 202000 – 12500 = 189500 \text{ Pa} = 189.5 \text{ kPa}$$
The pressure decreased where the velocity increased — as Bernoulli’s principle predicts!
Worked Example 15
Water flows through a horizontal pipe that narrows from diameter 10 cm to 5 cm. If the pressure in the wide section is 150 kPa and the flow speed is 1.5 m/s, find the pressure in the narrow section. (ρ = 1000 kg/m³)
Area ratio: A₁/A₂ = (d₁/d₂)² = (10/5)² = 4
v₂ = 4 × v₁ = 4 × 1.5 = 6 m/s
$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$ $$150000 + \frac{1}{2}(1000)(2.25) = P_2 + \frac{1}{2}(1000)(36)$$ $$150000 + 1125 = P_2 + 18000$$ $$P_2 = 151125 – 18000 = 133125 \text{ Pa} \approx 133.1 \text{ kPa}$$
Viscosity
Viscosity is the internal friction in a fluid that opposes the relative motion between adjacent layers. Think of it as the “thickness” or “stickiness” of a fluid.
- High viscosity: honey, engine oil, glycerin — flow slowly
- Low viscosity: water, petrol, air — flow easily
Viscosity causes energy loss in fluid flow (the fluid needs extra force to keep flowing). In many of our calculations, we have assumed “ideal fluids” — non-viscous, incompressible, and streamline. Real fluids are more complex.
- Equation of continuity: A₁v₁ = A₂v₂ (conservation of mass for incompressible fluid).
- Where area is smaller, velocity is larger (and vice versa).
- Bernoulli’s equation: P + ½ρv² + ρgh = constant.
- Bernoulli’s principle: higher velocity → lower pressure (and vice versa).
- For horizontal flow: P₁ + ½ρv₁² = P₂ + ½ρv₂².
- Applications: airplane lift, spray bottles, venturi meter.
- Viscosity = internal friction in fluids; causes energy loss.
- Bernoulli’s equation assumes: ideal fluid, streamline flow, steady state.
Practice Question 10: A pipe carries water. At one point, the pipe diameter is 8 cm and the water velocity is 3 m/s. At another point, the diameter is 4 cm. What is the velocity at the second point?
v₂ = v₁ × A₁/A₂ = 3 × 4 = 12 m/s
Practice Question 11: In a horizontal pipe, the pressure drops from 250 kPa to 200 kPa where the velocity increases from 4 m/s to 10 m/s. Is this consistent with Bernoulli’s equation? (ρ = 800 kg/m³)
Left side: 250000 + ½(800)(16) = 250000 + 6400 = 256400
Right side: 200000 + ½(800)(100) = 200000 + 40000 = 240000
256400 ≠ 240000 → Not consistent.
This means the fluid has significant viscosity, or there is turbulent flow, or the pipe is not horizontal. Bernoulli’s equation does not apply perfectly here.
Practice Question 12: Explain how Bernoulli’s principle explains the lift force on an airplane wing.
3.5 Safety and High Pressure
Dangers of High Pressure
Dear student, high-pressure systems are very useful but also very dangerous if not handled properly. Think about the following:
- Pressure cookers: Cook food faster by increasing the boiling point of water through high pressure. But if the safety valve fails, the cooker can explode!
- Boilers and steam engines: Generate steam at very high pressure. Boiler explosions have caused many accidents in history.
- Gas cylinders: Store gases under high pressure. A damaged cylinder can become a dangerous projectile.
- Dams: Hold back huge volumes of water. The pressure at the base of a tall dam is enormous.
- Scuba diving: Divers experience increased pressure underwater. Rising too quickly causes decompression sickness (“the bends”).
Safety Measures
- Always use safety valves in pressure vessels to release excess pressure.
- Regularly inspect pressure equipment for cracks, corrosion, and wear.
- Never exceed the rated pressure of a container or system.
- Use pressure gauges to monitor pressure at all times.
- Follow proper warm-up and cool-down procedures for boilers.
- For deep-sea diving, follow decompression protocols to avoid the bends.
- Dams must be designed to withstand the maximum possible water pressure at their base.
Why Are Dams Thicker at the Bottom?
The pressure in a fluid increases with depth: P = ρgh. At the base of a dam, the water pressure is maximum. So the dam must be strongest (thickest) at the bottom to resist this enormous force. This is why dams have a triangular cross-section — wider at the bottom, narrower at the top.
Worked Example 16
A dam is 50 m tall and holds back water. What is the pressure at the base of the dam? (ρ = 1000 kg/m³, g = 10 m/s²)
$$P = \rho g h = 1000 \times 10 \times 50 = 500000 \text{ Pa} = 500 \text{ kPa}$$
This is about 5 times atmospheric pressure! This is why dams must be very strong at the base.
- Pressure increases linearly with depth — bottom of dams/containers experience the most pressure.
- Safety valves are essential in all pressure vessels.
- Decompression sickness in divers is caused by dissolved gases coming out of solution when pressure decreases too quickly.
- Pressure cookers work by raising the boiling point of water through increased pressure.
- Dams are thicker at the bottom because P = ρgh is largest there.
Practice Question 13: A dam holds back water to a depth of 80 m. What is the gauge pressure at: (a) 20 m below the surface, (b) at the base of the dam?
(b) P = ρgh = 1000 × 10 × 80 = 800000 Pa = 800 kPa
The pressure at the base is 4 times the pressure at 20 m depth.
Unit 3 Summary
In this unit, you learned five major topics:
- Fluid Statics: Density (ρ = m/V), pressure (P = F/A), and basic properties of fluids.
- Pressure in Fluids at Rest: Pressure at depth (P = ρgh), Pascal’s principle, hydraulic press, barometers, and manometers.
- Archimedes’ Principle: Buoyant force equals weight of fluid displaced. Floating and sinking conditions. Apparent weight.
- Fluid Flow: Equation of continuity (A₁v₁ = A₂v₂), Bernoulli’s equation (P + ½ρv² + ρgh = constant), and applications including lift.
- Safety and High Pressure: Dangers of high pressure, safety measures, dam design.
Quick Revision Notes — Unit 3: Fluid Mechanics
3.1 & 3.2 — Fluid Statics and Pressure — Key Formulas
| Quantity | Formula |
|---|---|
| Density | ρ = m/V |
| Relative density | RD = ρ_substance / ρ_water |
| Pressure | P = F/A |
| Pressure at depth h | P = ρgh (gauge) |
| Absolute pressure at depth | P_abs = P_atm + ρgh |
| Pascal’s principle | F₁/A₁ = F₂/A₂ |
| Hydraulic press force | F₂ = F₁(A₂/A₁) |
| Barometer | P_atm = ρ_Hg × g × h |
| Manometer (open side higher) | P_gas = P_atm + ρgΔh |
| Manometer (open side lower) | P_gas = P_atm − ρgΔh |
- Always use depth from the FLUID SURFACE, not from the bottom.
- Convert all units to SI before calculating (cm → m, mm → m).
- In hydraulic press questions, also check for displacement: d₁A₁ = d₂A₂.
- Know that 1 atm = 760 mmHg = 1.013 × 10⁵ Pa.
3.3 — Archimedes’ Principle — Key Formulas
| Quantity | Formula |
|---|---|
| Buoyant force | F_b = ρ_fluid × V_displaced × g |
| Floating condition | ρ_object < ρ_fluid (then W = F_b) |
| Sinking condition | ρ_object > ρ_fluid |
| Fraction submerged (floating) | V_disp/V_obj = ρ_obj/ρ_fluid |
| Apparent weight | W_app = W_true − F_b |
| Finding volume from weight loss | V = (W_air − W_fluid) / (ρ_fluid × g) |
| Finding object density | ρ_obj = W_air / (V × g) |
- Use the FLUID density in F_b = ρ_fluid × V × g — NOT the object’s density!
- V_displaced = V_object ONLY when fully submerged.
- When partially submerged (floating), V_displaced < V_object.
- Weight loss method: F_b = W_air − W_submerged is the most common exam approach.
- For floating objects: fraction submerged = ρ_obj / ρ_fluid. This shortcut saves time!
3.4 — Fluid Flow — Key Formulas
| Quantity | Formula |
|---|---|
| Equation of continuity | A₁v₁ = A₂v₂ |
| Bernoulli’s equation (general) | P + ½ρv² + ρgh = constant |
| Bernoulli’s (horizontal) | P₁ + ½ρv₁² = P₂ + ½ρv₂² |
| Volume flow rate | Q = Av |
| Area from diameter | A = πd²/4 |
- Continuity equation: smaller area → higher velocity (inversely proportional).
- Bernoulli: higher velocity → lower pressure. Memorize this relationship!
- When solving Bernoulli problems: FIRST use continuity to find the unknown velocity, THEN apply Bernoulli.
- Bernoulli assumes ideal fluid (no viscosity, incompressible, streamline).
- For circular pipes: A ∝ d², so v ∝ 1/d².
Important Definitions
- Fluid: A substance that flows and deforms continuously under shear stress (liquids and gases).
- Density: Mass per unit volume of a substance (ρ = m/V).
- Pressure: Force per unit area, acting perpendicular to the surface (P = F/A).
- Gauge pressure: Pressure measured relative to atmospheric pressure (P = ρgh).
- Absolute pressure: Total pressure including atmospheric pressure (P = P_atm + ρgh).
- Pascal’s principle: Pressure applied to a confined fluid is transmitted equally throughout.
- Buoyant force (upthrust): The upward force exerted by a fluid on an immersed object.
- Archimedes’ principle: The buoyant force equals the weight of fluid displaced.
- Streamline flow: Smooth, orderly flow in parallel layers.
- Turbulent flow: Chaotic, irregular flow with eddies.
- Equation of continuity: A₁v₁ = A₂v₂, expressing conservation of mass.
- Bernoulli’s principle: In a flowing fluid, higher velocity corresponds to lower pressure.
- Viscosity: Internal friction in a fluid that resists flow between layers.
Common Mistakes to Avoid
- Pressure depth: Measuring depth from the wrong point. Always measure from the fluid surface!
- Unit conversion: Forgetting to convert cm or mm to meters. (10 cm = 0.1 m, NOT 10 m!)
- Archimedes’ F_b: Using the object’s density instead of the fluid’s density in F_b = ρVg.
- Archimedes’ volume: Assuming V_displaced = V_object when the object is floating (it is NOT — only the submerged part counts).
- Continuity equation: Writing v₁/v₂ = A₁/A₂ instead of v₁/v₂ = A₂/A₁ (inverse relationship!).
- Bernoulli’s equation: Forgetting that higher v means lower P, not higher P.
- Manometer sign: Getting the sign wrong: if the open side is higher, P_gas > P_atm (add ρgΔh).
- Gauge vs absolute: Confusing gauge pressure with absolute pressure. Gauge does NOT include atmospheric pressure.
- Hydraulic press: Forgetting that the small piston moves a greater distance than the large piston.
- Bernoulli assumptions: Applying Bernoulli to viscous fluids or turbulent flow (it only works for ideal, streamline flow).
Important Constants to Remember
- ρ_water = 1000 kg/m³
- ρ_mercury = 13,600 kg/m³
- ρ_air = 1.29 kg/m³ (approximately)
- ρ_seawater ≈ 1030 kg/m³
- g = 10 m/s² (in most exam problems)
- 1 atm = 1.013 × 10⁵ Pa = 760 mmHg
- 1 bar = 10⁵ Pa
Challenge Exam Questions — Unit 3
Section A: Multiple Choice Questions
Question 1: A hydraulic press has pistons of areas 0.02 m² and 0.4 m². If a force of 100 N is applied to the small piston, the force on the large piston is:
F₂ = F₁ × A₂/A₁ = 100 × 0.4/0.02 = 100 × 20 = 2000 N
Question 2: A block of wood floats in water with 60% of its volume submerged. The density of the wood is:
Fraction submerged = ρ_wood/ρ_water
0.6 = ρ_wood/1000
ρ_wood = 600 kg/m³
Question 3: Water flows through a pipe. If the diameter is doubled while the flow rate stays the same, the velocity becomes:
A = πd²/4. If d doubles, A becomes 4 times larger.
From continuity: A₁v₁ = A₂v₂, so v₂ = v₁ × A₁/A₂ = v₁ × 1/4.
The velocity becomes one-fourth.
Question 4: According to Bernoulli’s principle, where the velocity of a fluid is high, the pressure is:
Bernoulli’s principle: P + ½ρv² = constant. If v increases, ½ρv² increases, so P must decrease to keep the sum constant. Higher velocity → lower pressure.
Question 5: A piece of metal weighs 30 N in air and 25 N in water. Its density is:
F_b = 30 − 25 = 5 N
V = F_b/(ρ_water × g) = 5/(1000 × 10) = 0.0005 m³
ρ = W/(Vg) = 30/(0.0005 × 10) = 30/0.005 = 6000 kg/m³
Question 6: The height of the mercury column in a barometer at a place is 700 mmHg. The atmospheric pressure at that place is approximately:
P = ρ_Hg × g × h = 13600 × 10 × 0.7 = 95200 Pa ≈ 0.95 × 10⁵ Pa
Closest to 0.93 × 10⁵ Pa.
Section B: Fill in the Blanks
Question 7: The SI unit of pressure is ________.
Question 8: The buoyant force on an object immersed in a fluid depends on the ________ of the fluid and the ________ of fluid displaced.
F_b = ρ_fluid × V_displaced × g
Question 9: In a hydraulic press, if the area ratio A₂/A₁ is 20, the mechanical advantage is ________.
Mechanical advantage = F₂/F₁ = A₂/A₁ = 20
Question 10: Standard atmospheric pressure is ________ mmHg, which equals ________ Pa.
Question 11: According to the equation of continuity, if the cross-sectional area of a pipe is halved, the fluid velocity becomes ________ times.
A₁v₁ = A₂v₂ → if A₂ = A₁/2, then v₂ = 2v₁
Question 12: The pressure at the bottom of a dam is ________ (greater/less) than at the top because pressure ________ with depth.
P = ρgh: as h increases, P increases.
Section C: Short Answer Questions
Question 13: State Pascal’s principle and explain how it is applied in a hydraulic brake system.
Application in hydraulic brakes: When the driver presses the brake pedal, a small force is applied to a small piston in the master cylinder. This creates pressure that is transmitted through the brake fluid to larger pistons at each wheel. The larger pistons exert a much greater force on the brake pads, which press against the brake discs/drums, slowing the car. The mechanical advantage (F₂/F₁ = A₂/A₁) allows a small foot force to produce a large braking force.
Question 14: Explain why an iron nail sinks in water but a ship made of iron floats.
A ship, although made of iron, has a hollow structure filled with air. The overall density of the ship (total mass/total volume including the air inside) is LESS than the density of water. The large volume of the ship displaces enough water to create a buoyant force equal to the ship’s weight. The shape of the ship is designed to maximize the displaced volume while minimizing the mass, ensuring it floats.
Question 15: Explain how a barometer measures atmospheric pressure. Why is mercury used instead of water?
Why mercury instead of water?
(1) Mercury is much denser (13,600 vs 1000 kg/m³), so the column height is manageable (760 mm vs about 10.3 m for water).
(2) Mercury does not evaporate easily at room temperature, so the vacuum at the top is maintained.
(3) Mercury does not wet glass, so the column has a well-defined meniscus.
Question 16: State Bernoulli’s principle and give two practical applications.
Applications:
1. Airplane lift: The curved upper surface of the wing causes air to flow faster over the top, creating lower pressure above the wing. The pressure difference between top and bottom creates an upward lift force.
2. Venturi meter: Used to measure the flow rate of a fluid. A constriction in the pipe increases velocity and decreases pressure. The pressure difference between the wide and narrow sections is measured and used to calculate the flow rate.
3. Spray bottle: Squeezing the bulb sends fast-moving air across the top of a tube dipped in liquid. The low pressure above the tube draws liquid up, which is then sprayed out as a fine mist.
Question 17: What is viscosity? How does it affect fluid flow?
Effects on fluid flow:
(1) Viscosity causes energy loss in fluid flow — the fluid needs external force (like a pump) to maintain flow.
(2) It creates a velocity gradient: the fluid layer next to a solid surface moves slowest (zero at the wall — no-slip condition), while the center moves fastest.
(3) High viscosity promotes laminar (smooth) flow; low viscosity at high speed promotes turbulent flow.
(4) Bernoulli’s equation does NOT account for viscosity — it assumes ideal (non-viscous) fluids. For real viscous fluids, energy is lost and Bernoulli’s equation overestimates the pressure recovery.
Section D: Step-by-Step Calculation Questions
Question 18: A hydraulic lift is used to lift a car of mass 1500 kg. The area of the large piston is 0.15 m² and the area of the small piston is 0.005 m². Find: (a) the force that must be applied to the small piston, (b) how far the small piston must move to lift the car by 0.1 m. (g = 10 m/s²)
(a) Force on small piston:
$$F_1 = F_2 \times \frac{A_1}{A_2} = 15000 \times \frac{0.005}{0.15} = 15000 \times \frac{1}{30} = 500 \text{ N}$$
(b) Distance moved by small piston:
$$F_1 d_1 = F_2 d_2$$ $$500 \times d_1 = 15000 \times 0.1$$ $$d_1 = \frac{1500}{500} = 3 \text{ m}$$
The operator must push the small piston down 3 m to lift the car by only 0.1 m!
Question 19: A metal cube of side 5 cm weighs 8.5 N in air and 7.5 N when fully submerged in a liquid of unknown density. Find: (a) the density of the liquid, (b) the density of the metal cube. (g = 10 m/s²)
(a) Density of liquid:
F_b = 8.5 − 7.5 = 1 N
$$F_b = \rho_{\text{liquid}} \, V \, g$$ $$1 = \rho_{\text{liquid}} \times 0.000125 \times 10$$ $$\rho_{\text{liquid}} = \frac{1}{0.00125} = 800 \text{ kg/m}^3$$
(b) Density of metal:
$$\rho_{\text{metal}} = \frac{W}{Vg} = \frac{8.5}{0.000125 \times 10} = \frac{8.5}{0.00125} = 6800 \text{ kg/m}^3$$
Question 20: Water flows through a horizontal pipe that narrows from diameter 12 cm to 6 cm. At the wide section, the pressure is 180 kPa and the velocity is 1.5 m/s. Find: (a) the velocity in the narrow section, (b) the pressure in the narrow section. (ρ = 1000 kg/m³)
(a) Velocity in narrow section:
$$v_2 = v_1 \times \frac{A_1}{A_2} = 1.5 \times 4 = 6 \text{ m/s}$$
(b) Pressure in narrow section (Bernoulli, horizontal):
$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$ $$180000 + \frac{1}{2}(1000)(2.25) = P_2 + \frac{1}{2}(1000)(36)$$ $$180000 + 1125 = P_2 + 18000$$ $$P_2 = 181125 – 18000 = 163125 \text{ Pa} \approx 163.1 \text{ kPa}$$
The pressure dropped from 180 kPa to 163.1 kPa because velocity increased from 1.5 to 6 m/s.
Question 21: A balloon of volume 0.01 m³ is filled with helium gas of density 0.18 kg/m³. It is surrounded by air of density 1.29 kg/m³. Find: (a) the buoyant force on the balloon, (b) the total weight of the balloon (including helium) if the balloon material weighs 0.5 N, (c) the net upward force. (g = 10 m/s²)
$$F_b = \rho_{\text{air}} \, V \, g = 1.29 \times 0.01 \times 10 = 0.129 \text{ N}$$
(b) Total weight:
Weight of helium = ρ_He × V × g = 0.18 × 0.01 × 10 = 0.018 N
Weight of balloon material = 0.5 N
Total weight = 0.018 + 0.5 = 0.518 N
(c) Net upward force:
F_net = F_b − W = 0.129 − 0.518 = −0.389 N
The net force is downward (−0.389 N)! The balloon is too heavy to rise because the balloon material alone weighs 0.5 N, which is more than the buoyant force of 0.129 N. In practice, helium balloons work because they have very thin, lightweight material and much larger volumes.
Question 22: A U-tube manometer containing mercury (ρ = 13,600 kg/m³) is connected to a gas container. The mercury level in the arm connected to the gas is 50 mm lower than in the open arm. If atmospheric pressure is 1.013 × 10⁵ Pa, find the absolute pressure of the gas. (g = 10 m/s²)
Δh = 50 mm = 0.05 m
$$P_{\text{gas}} = P_{\text{atm}} + \rho_{\text{Hg}} \, g \, \Delta h$$ $$P_{\text{gas}} = 1.013 \times 10^5 + 13600 \times 10 \times 0.05$$ $$P_{\text{gas}} = 101300 + 6800 = 108100 \text{ Pa} = 1.081 \times 10^5 \text{ Pa}$$
Question 23: A solid sphere of density 4000 kg/m³ and volume 0.0003 m³ is dropped into a tank containing two layers of liquid: the upper layer is oil (ρ = 800 kg/m³, depth 0.5 m) and the lower layer is water (ρ = 1000 kg/m³, depth 1 m). Find: (a) the buoyant force when the sphere is fully in the oil layer, (b) the buoyant force when the sphere is fully in the water layer, (c) will the sphere float at the interface or sink to the bottom? (g = 10 m/s²)
$$F_{b,\text{oil}} = \rho_{\text{oil}} \, V \, g = 800 \times 0.0003 \times 10 = 2.4 \text{ N}$$
(b) In water (fully submerged):
$$F_{b,\text{water}} = \rho_{\text{water}} \, V \, g = 1000 \times 0.0003 \times 10 = 3 \text{ N}$$
(c) Does it sink or float?
Weight of sphere = ρ_sphere × V × g = 4000 × 0.0003 × 10 = 12 N
Even in water, the maximum buoyant force (3 N) is much less than the weight (12 N). Since ρ_sphere (4000) > ρ_water (1000), the sphere sinks to the bottom of the tank.
Question 24: A water tank has a side wall that is 4 m wide and 3 m deep. Calculate: (a) the total force on the side wall due to water pressure, (b) the depth at which the center of pressure acts. (ρ = 1000 kg/m², g = 10 m/s²)
The average pressure on the wall: P_avg = ρg(h/2) = 1000 × 10 × 1.5 = 15000 Pa
(a) Total force on the wall:
$$F = P_{\text{avg}} \times A = 15000 \times (4 \times 3) = 15000 \times 12 = 180000 \text{ N} = 180 \text{ kN}$$
(b) Center of pressure depth:
For a rectangular vertical wall, the center of pressure is at 2h/3 from the surface:
$$h_{cp} = \frac{2}{3} \times 3 = 2 \text{ m from the surface}$$
(This is below the midpoint because pressure increases with depth — more force acts on the lower part.)
Question 25: A venturi meter has a main pipe diameter of 20 cm and a throat diameter of 10 cm. A manometer connected between the main pipe and throat shows a mercury height difference of 5 cm. Find the volume flow rate of water through the pipe. (ρ_water = 1000 kg/m³, ρ_Hg = 13600 kg/m³, g = 10 m/s²)
A₂ = π(0.1)²/4 = 0.00785 m²
A₁/A₂ = 4
Pressure difference from manometer:
ΔP = (ρ_Hg − ρ_water)gΔh = (13600 − 1000) × 10 × 0.05 = 12600 × 0.5 = 6300 Pa
From Bernoulli + continuity:
$$\Delta P = P_1 – P_2 = \frac{1}{2}\rho_{\text{water}}(v_2^2 – v_1^2)$$ $$6300 = \frac{1}{2}(1000)(v_2^2 – v_1^2)$$ $$12.6 = v_2^2 – v_1^2$$
From continuity: v₂ = 4v₁
$$12.6 = 16v_1^2 – v_1^2 = 15v_1^2$$ $$v_1^2 = 0.84 \Rightarrow v_1 = 0.917 \text{ m/s}$$
Volume flow rate:
$$Q = A_1 v_1 = 0.0314 \times 0.917 = 0.0288 \text{ m}^3\text{/s} = 28.8 \text{ L/s}$$
End of Challenge Questions
Dear student, you have completed all the challenge questions for Unit 3! Remember these tips for your exam:
- Always draw a diagram — especially for manometer, hydraulic press, and buoyancy problems.
- Identify whether you need gauge pressure or absolute pressure.
- In Archimedes problems, clearly identify what is V_displaced.
- For Bernoulli problems, always find the unknown velocity FIRST using continuity, then apply Bernoulli.
- Check your units at every step — this prevents most calculation errors.
- Practice makes perfect! Keep solving problems until these formulas feel natural.