Unit 8: Fluids

Internal Structure and Density

• The three common phases of matter are solid, liquid, and gas. A solid has a fixed shape and volume, a liquid has a fixed volume but no fixed shape, and a gas has neither. Fluids include both liquids and gases, since they flow to fit the shape of their containers.
• Density measures how much mass is contained in a given volume. It is calculated as \[ \rho = \frac{m}{V} \] and is typically expressed in \( \text{kg/m}^3 \). Higher density means more mass packed into the same volume, which has important implications for buoyancy and pressure.
• Density helps explain why some objects float while others sink. An object less dense than the fluid will float, while one more dense will sink. This ties directly to buoyant force, which is studied later in this unit.

Pressure

• Pressure is defined as the force applied per unit area: \[ P = \frac{F_\perp}{A} \] where \(F_\perp\) is the force perpendicular to the surface. Pressure is a scalar quantity, meaning it has magnitude but no direction.
• Pressure in fluids acts equally in all directions at a given depth, which is why divers feel pressure from all sides underwater. The SI unit of pressure is the pascal (\(Pa\)), where \(1 \, Pa = 1 \, N/m^2\).
• The absolute pressure at depth is given by \[ P_{\text{absolute}} = P_0 + \rho g h \] where \(P_0\) is atmospheric pressure, \(\rho\) is fluid density, \(g\) is gravity, and \(h\) is depth. Gauge pressure, measured by most instruments, is \[ P_{\text{gauge}} = \rho g h \] which excludes atmospheric pressure.
• Pressure differences in fluids explain phenomena like ears popping during a dive, blood pressure readings, and how submarines withstand high pressures deep underwater. This connects fluid statics with human physiology and engineering.

Fluids and Newton’s Laws

• The buoyant force arises from pressure differences at different depths in a fluid. It always acts upward and is equal in magnitude to the weight of the fluid displaced: \[ F_B = \rho_f V_f g \] where \(\rho_f\) is fluid density, \(V_f\) is displaced volume, and \(g\) is gravity.
• If the buoyant force equals the weight of the object, the object floats at equilibrium. If the buoyant force is less than the weight, the object sinks. This concept connects Newton’s Second Law with fluid statics, since the net force determines acceleration.
• An object floats when the volume of displaced fluid is less than the object’s total volume, and it is fully submerged when the displaced fluid equals the object’s entire volume. This explains why ships float despite being made of dense steel: their overall density, including hollow space, is less than that of water.
• The buoyant force results from the pressure difference between the top and bottom of an object submerged in a fluid. Because pressure increases with depth, the upward force on the bottom surface is greater than the downward force on the top surface. This imbalance produces the net upward buoyant force described by Archimedes’ Principle.
• Newton’s Second Law applies to floating and sinking objects by analyzing the net force. If the buoyant force is greater than the weight of the object, the net force is upward and the object accelerates upward. If the buoyant force is less than the weight, the object accelerates downward, sinking until it reaches equilibrium or the bottom.
• At equilibrium, the buoyant force equals the object’s weight. This explains why an object floats partially submerged: it displaces just enough fluid so that the buoyant force balances its weight. For example, an ice cube floats with about 90% of its volume underwater, since the density of ice is slightly less than that of water.
• Newton’s Third Law is also at play. As the fluid exerts an upward buoyant force on the object, the object exerts an equal and opposite downward force on the fluid. This reciprocal relationship ensures forces are balanced within the fluid-object system, maintaining physical consistency.
• Buoyancy and Newton’s Laws extend beyond water. Hot air balloons rise because the buoyant force of less dense heated air exceeds the balloon’s weight. Submarines control depth by adjusting the density of water in their ballast tanks, showing how Newton’s laws explain controlled floating and sinking in engineering.

Fluids and Conservation Laws

• Ideal fluid flow assumes the fluid is nonviscous, incompressible, steady, and irrotational. These conditions simplify analysis and allow us to use powerful conservation equations. While real fluids may not meet all these conditions, the ideal fluid model is accurate for many everyday cases.
• The continuity equation states that \[ A_1 v_1 = A_2 v_2 \] meaning that the product of cross-sectional area and velocity is constant for incompressible flow. This explains why water speeds up when it passes through a narrow section of a hose.
• Bernoulli’s equation describes the conservation of mechanical energy in a fluid: \[ P_1 + \frac{1}{2}\rho v_1^2 + \rho g y_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g y_2 \] It links pressure, velocity, and height, showing that increases in fluid speed correspond to decreases in pressure.
• Bernoulli’s Principle explains why airplanes generate lift and why roofs can blow off in high winds: faster airflow reduces pressure. This principle connects fluid mechanics to aerodynamics and meteorology.
• Torricelli’s Theorem, derived from Bernoulli’s principle, gives the speed of fluid exiting an opening: \[ v = \sqrt{2 g \Delta y} \] It shows that the exit speed depends only on the vertical height of the fluid above the hole, not the shape of the container.

Example Problem 1: Floating and Submerged Object

Context: A wooden block of mass 3.0 kg and volume 0.004 m³ is placed in water (density = 1000 kg/m³). The block floats at equilibrium.

• (a) Determine the weight of the block.
  \[ W = mg = (3.0)(9.8) = 29.4\, \text{N} \]
• (b) Calculate the buoyant force on the block when floating.
  At equilibrium, buoyant force = weight: \[ F_B = W = 29.4\, \text{N} \]
• (c) Find the volume of water displaced.
  Using \(F_B = \rho V_f g\): \[ V_f = \frac{F_B}{\rho g} = \frac{29.4}{(1000)(9.8)} = 0.0030\, \text{m}^3 \]
• (d) What fraction of the block’s volume is submerged?
  \[ \frac{V_f}{V} = \frac{0.0030}{0.0040} = 0.75 \]
  So, 75% of the block is submerged.
• (e) If the block were pushed completely underwater and released, explain whether it would rise or sink.
  If fully submerged, buoyant force would be: \[ F_B = \rho V g = (1000)(0.004)(9.8) = 39.2\, \text{N} \]
  Since 39.2 N > 29.4 N, the block would accelerate upward until 75% of its volume is submerged again.

Example Problem 2: Continuity and Bernoulli’s Principle

Context: Water flows through a horizontal pipe that narrows from a diameter of 10 cm to 4 cm. At the wider section, the pressure is 180,000 Pa and the flow speed is 2.0 m/s. Assume the density of water is 1000 kg/m³.

• (a) Calculate the flow speed in the narrow section using the continuity equation.
  Continuity: \[ A_1v_1 = A_2v_2 \] \[ v_2 = \frac{A_1}{A_2}v_1 = \left(\frac{0.10^2}{0.04^2}\right)(2.0) = 12.5\, \text{m/s} \]
• (b) Use Bernoulli’s equation to calculate the pressure in the narrow section.
  \[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \] Substituting: \[ 180000 + \frac{1}{2}(1000)(2.0^2) = P_2 + \frac{1}{2}(1000)(12.5^2) \] \[ 180000 + 2000 = P_2 + 78125 \Rightarrow P_2 ≈ 103000\, \text{Pa} \]
• (c) Explain why the pressure decreased.
  According to Bernoulli’s Principle, an increase in fluid speed leads to a decrease in pressure. The fluid speeds up in the narrow pipe, converting pressure energy into kinetic energy.

Example Problem 3: Torricelli’s Theorem

Context: A large open tank is filled with water to a depth of 3.0 m. A small hole is drilled at the bottom. Assume atmospheric pressure acts at the water’s surface.

• (a) Use Torricelli’s Theorem to calculate the speed of water exiting the hole.
  \[ v = \sqrt{2g\Delta y} = \sqrt{2(9.8)(3.0)} ≈ 7.67\, \text{m/s} \]
• (b) If the hole has an area of 2.0 cm², calculate the volume flow rate.
  \[ Q = Av = (2.0 \times 10^{-4})(7.67) ≈ 1.53 \times 10^{-3}\, \text{m}^3/\text{s} \]
• (c) How long would it take for the tank to lose 0.50 m³ of water at this rate (assuming constant depth)?
  \[ t = \frac{V}{Q} = \frac{0.50}{1.53 \times 10^{-3}} ≈ 327\, \text{s} \, (\text{about 5.5 minutes}) \]
• (d) Explain why the assumption of constant depth is unrealistic.
  As water drains, the depth decreases, reducing exit velocity. In reality, flow rate decreases with time, so the actual draining time would be longer than the calculated result.

Common Misconceptions in Fluids

• Many students believe that the buoyant force depends on the depth of the object in the fluid. In reality, the buoyant force depends only on the volume of fluid displaced, not the depth of submersion. This is why a submarine floating at 10 m depth experiences the same buoyant force as it does at 100 m depth, provided it displaces the same volume of water.
• It is a common mistake to assume that heavier objects always sink and lighter objects always float. Whether an object floats or sinks depends on its density relative to the fluid, not simply its weight. A large steel ship floats because its overall density (including air spaces) is less than the density of water.
• Students often think that pressure acts only downward in a fluid, but pressure actually acts equally in all directions at a given depth. This is why divers feel squeezed from all sides and why submarines must withstand pressure from every angle. Forgetting this leads to errors in pressure calculations and force diagrams.
• When applying Bernoulli’s equation, many students forget that it only applies to ideal fluids with steady, incompressible, and nonviscous flow. In real fluids with turbulence or viscosity, Bernoulli’s principle is only an approximation. Misapplying it to real-life situations without checking assumptions can lead to incorrect conclusions.
• A frequent error is to confuse gauge pressure with absolute pressure. Gauge pressure excludes atmospheric pressure, while absolute pressure includes it. For example, a car tire reading of 220 kPa is gauge pressure — the actual pressure inside the tire is atmospheric pressure plus the gauge reading.
• Some students believe that increasing the diameter of a pipe always increases flow rate. According to the continuity equation, increasing cross-sectional area decreases velocity if flow rate remains constant. This counterintuitive result is essential for solving fluid dynamics problems correctly.