MECHANICAL PROPERTIES OF FLUIDS-Notes

In the study of fluids, pressure is defined as the normal force exerted by a fluid per unit area of the surface with which it is in contact. Unlike solids, a fluid cannot sustain shear stress in equilibrium; hence, the force it exerts on any surface is always perpendicular to that surface.

Mathematically, pressure \(P\) is expressed as

where \(F\) is the normal force acting on the surface and \(A\) is the area over which this force acts.
The SI unit of pressure is pascal (Pa), where \[1\,Pa=1\,N\ m^{−2}\]

Nature of Pressure in Fluids

Pressure in a fluid has certain distinctive characteristics that differentiate it from force:

• Pressure is a scalar quantity, even though it arises due to force.
• At a given point inside a fluid at rest, pressure acts equally in all directions.
• Pressure does not depend on the shape of the container but depends on the depth, density of the fluid, and gravitational acceleration.

Consider a liquid of uniform density \(rho\) contained in a vessel. Let us imagine a small horizontal area \(A\) at a depth \(h\) below the free surface of the liquid.

The liquid column directly above this area has:

Height \(h\)
Volume \(V=Ah\)
Mass \(m=\rho Ah\)

The weight of this liquid column is

This weight acts vertically downward and is responsible for the pressure at depth \(h\). Therefore, the pressure \(P\) at that depth is:

If the liquid surface is exposed to atmospheric pressure \(P_0\), then the total pressure at depth \(h\) becomes:

Important Aspects of Pressure

• Pressure increases linearly with depth in a liquid.
• Greater the density of the fluid, greater is the pressure at the same depth.
• Atmospheric pressure plays a crucial role in phenomena such as drinking with a straw, suction cups, and fluid flow.
• Pressure explains several practical observations like leakage of water jets from holes at different heights in a tank.
• In fluids at rest, pressure always acts normal to any surface.

Pascal’s Law states that when an external pressure is applied to an enclosed fluid at rest, that pressure is transmitted undiminished and equally in all directions throughout the fluid and to the walls of the container.

Physical Meaning of Pascal’s Law

A fluid at rest cannot sustain shear stress. Therefore, any additional pressure applied at one point does not remain localized. Instead, it spreads uniformly throughout the fluid because the fluid particles adjust their positions until equilibrium is restored. As a result, every part of the fluid experiences the same increase in pressure.

Proof of Pascal’s Law (Conceptual Proof)

Consider a liquid enclosed in a rigid, sealed container. Imagine a small element of the fluid in the shape of a cube. Since the fluid is at rest, this element must be in mechanical equilibrium.

If an external pressure is applied at any point on the fluid, the fluid particles transmit the force to neighboring particles through molecular interactions. If the pressure transmitted in one direction were different from that in another direction, the fluid element would experience a net force and begin to accelerate.

Since the fluid remains at rest, no such acceleration occurs. Hence, the increase in pressure must be equal in all directions. This establishes that pressure applied to a confined fluid is transmitted undiminished throughout the fluid, which is precisely Pascal’s Law.

Mathematical Expression of Pascal’s Law

Let a pressure \(P\) be applied on a confined fluid.

If a force \(F_1\) acts on a small area \(A_1\), the pressure produced is

According to Pascal’s Law, the same pressure acts throughout the fluid. Therefore, on another area \(A_2\), the force \(F_2\) is given by

Equating the two expressions:

This relation shows that a small force applied on a small area can produce a much larger force on a larger area.

Important Aspects of Pascal’s Law

• Pascal’s Law applies only to fluids at rest, not to flowing fluids.
• The fluid must be completely enclosed for uniform pressure transmission.
• Pressure, not force, is transmitted equally throughout the fluid.
• The law does not violate conservation of energy; a gain in force is accompanied by a loss in displacement.
• Pascal’s Law explains the working of hydraulic systems used in everyday life.

Limitations of Pascal’s Law

• It does not apply if the fluid is compressible to a significant extent.
• It assumes the fluid is ideal and in equilibrium.
• Any leakage or presence of air bubbles reduces the effectiveness of pressure transmission.

Atmospheric Pressure — Definition

Atmospheric pressure is the pressure exerted by the Earth’s atmosphere due to the weight of the air column above a given surface. Although air appears light and invisible, it has mass and is acted upon by gravity. As a result, every surface on Earth experiences a continuous force due to the air above it.

Pressure, being force per unit area, allows this force to act uniformly on all exposed surfaces.

Origin of Atmospheric Pressure (Physical Explanation)

Air molecules are constantly moving and colliding with surfaces. These collisions transfer momentum to the surface, producing pressure. At any point near the Earth’s surface, the atmosphere above acts like a very tall column of fluid. The weight of this air column is responsible for atmospheric pressure.

As altitude increases, the height of the air column decreases, leading to a reduction in atmospheric pressure.

Standard Atmospheric Pressure

At sea level, atmospheric pressure has a well-defined standard value used for measurement and comparison:

This value plays a reference role in fluid mechanics, particularly when measuring pressures inside liquids and gases.

Gauge Pressure - Definition

Gauge pressure is the pressure measured relative to atmospheric pressure. It represents the excess pressure of a fluid above the surrounding atmospheric pressure.

In many practical situations, we are not concerned with total pressure but only with how much pressure exceeds the atmospheric value.

Derivation of Gauge Pressure in a Liquid

Consider a point at depth \(h\) inside a liquid open to the atmosphere.

Pressure due to the liquid column: \[P_{liquid}=\rho g h\]

Total pressure at that point: \[P_{abs}=P_{atm}+\rho g h\]

Thus, gauge pressure is: \[\boxed{\;\mathbf{P_g=\rho g h}\;}\]

This shows that gauge pressure depends only on depth, density, and gravity, not on atmospheric pressure itself.

Important Aspects

• Atmospheric pressure is the reference level for most pressure measurements.
• Gauge pressure ignores atmospheric pressure and focuses on usable pressure.
• Pressure measuring devices like manometers and pressure gauges are calibrated in gauge pressure.
• Human activities such as breathing, drinking with a straw, and fluid flow depend directly on atmospheric pressure.
• Absolute pressure is essential in thermodynamic calculations, while gauge pressure is more practical in engineering applications.

Streamline flow, also known as laminar flow, is a type of fluid motion in which every particle of the fluid follows a smooth, well-defined and fixed path, called a streamline. In this flow, the velocity of the fluid at any given point remains constant with time, and the motion occurs in orderly layers without intermixing or random disturbances.

Concept of a Streamline

A streamline is an imaginary curve drawn in a fluid such that the tangent at any point gives the direction of velocity of the fluid particle at that point.

Two important properties follow naturally

• No two streamlines can intersect, because a fluid particle cannot have two velocities at the same point.
• The closer the streamlines, the greater is the speed of the fluid.

Definition

Viscosity is the intrinsic property of a fluid by virtue of which it opposes the relative motion between its adjacent layers while flowing. When one layer of a fluid moves faster than another, an internal resisting force arises, tending to reduce the velocity difference. This resistance is called viscous resistance, and the property responsible for it is viscosity.

In simple terms, viscosity measures the internal friction of a fluid.

• Fluids with high viscosity (e.g., honey, glycerine) resist flow strongly.
• Fluids with low viscosity (e.g., water, air) flow easily.

Physical Origin of Viscosity

The origin of viscosity differs in liquids and gases:

• In liquids:
  viscosity mainly arises due to intermolecular attraction. When layers slide past each other, attractive forces resist their relative motion.
• In gases:
  viscosity originates from the random molecular motion and the exchange of momentum between layers moving at different velocities.

Thus, viscosity is a macroscopic manifestation of microscopic molecular interactions.

Newton’s Law of Viscous Flow (Statement)

For a fluid flowing steadily in parallel layers (laminar flow), the viscous force acting between adjacent layers is directly proportional to:

• The area of contact between the layers
• The velocity gradient perpendicular to the direction of flow

Mathematically,

Introducing the constant of proportionality, we write:

where
\(\eta\) is the coefficient of viscosity of the fluid.

Summary of Key Points

• Viscosity is internal resistance to flow in fluids.
• Newton’s law of viscosity relates force with velocity gradient.
• Coefficient of viscosity quantifies fluid resistance.
• Viscous force always opposes relative motion.
• Temperature affects viscosity differently in liquids and gases.
• Viscosity is central to understanding real fluid motion.

Stokes’ Law states that when a small spherical body moves slowly through a viscous fluid, the resistive force (viscous drag) acting on it is directly proportional to the radius of the sphere, the velocity of the sphere, and the coefficient of viscosity of the fluid.

Mathematically,

where
\(\eta\) = coefficient of viscosity of the fluid,
\(r\) = radius of the spherical body,
\(v\) = velocity of the body relative to the fluid.

Proof through Force Balance: Terminal Velocity

Consider a sphere of radius \(r\) and density \(\rho_s\) falling through a liquid of density \(\rho_l\).

Forces Acting on the Sphere

Weight (downward):

Viscous drag (upward, opposing motion):

Condition for Terminal Velocity

At terminal velocity v_t, net force is zero:

Substituting,

Rearranging,

Important Aspects of Stokes’ Law

• Validity Conditions:
  
  • The body must be spherical.
  • The motion must be slow (low Reynolds number).
  • The fluid must be viscous and incompressible.
  • Flow should be laminar, not turbulent.
• Linear Dependence on Velocity:
  Drag force increases directly with speed, unlike air drag at high speeds.
• Role of Viscosity:
  Higher viscosity → greater resistance → smaller terminal velocity.
• Size Dependence:
  Larger spheres experience greater viscous force.
• Foundation for Terminal Velocity:
  Stokes’ Law explains why small droplets (fog, mist) fall slowly.
• Experimental Use:
  Used to determine viscosity of liquids by observing terminal velocity.

Surface tension is the property of a liquid by virtue of which its free surface behaves like a stretched elastic membrane, tending to acquire the minimum possible area. Quantitatively, it is defined as the tangential force per unit length acting along the surface of a liquid at right angles to an imaginary line drawn on the surface.

Mathematically,

where
\(T\) = surface tension,
\(F\) = tangential force acting along the surface,
\(L\) = length over which the force acts.

Surface energy is the energy possessed by the free surface of a liquid due to the unbalanced molecular forces acting on surface molecules. Quantitatively, it is defined as the work required to increase the surface area of a liquid by unit area, while keeping temperature constant.

Mathematically,

Surface energy represents the stored energy of a liquid surface, which makes the surface tend to contract and acquire minimum area.

Molecular Basis of Surface Energy

Molecules inside a liquid are attracted equally in all directions by neighboring molecules, resulting in zero net force and minimum potential energy. However, molecules at the surface experience unbalanced inward cohesive forces, raising their potential energy.

Thus:
Surface molecules are in a higher energy state

• Extra energy is required to bring molecules from the interior to the surface
• This excess energy stored at the surface is called surface energy.

Proof: Surface Energy Equals Surface Tension per Unit Area

Consider a rectangular wire frame with a thin liquid film and a movable wire of length \({L}\).
Surface tension acting on the wire = \(\mathbf{2}{TL}\) (two surfaces)
If the wire moves by a small distance \({dx}\),

This establishes that surface energy per unit area is numerically equal to surface tension

Important Aspects of Surface Energy

• Unit and Dimensions
  • SI unit: joule (J)
  • Surface energy per unit area: J m⁻²
  • Dimensions: [MT⁻²]
• Relation with Surface Tension
  • Surface energy per unit area equals surface tension
  • Both arise from molecular cohesion
• Dependence on Temperature
  • Surface energy decreases as temperature increases
  • Approaches zero at critical temperature
• Effect of Impurities
  • Detergents and soaps reduce surface energy
  • This helps spreading and cleaning action
• Double Surface in Films
  • Soap films have two surfaces
  • Surface energy is twice that of a single liquid surface
• Difference from Internal Energy
  • Surface energy is associated only with surface molecules
  • Internal energy involves all molecules

Applications and Physical Significance

• Formation and stability of soap bubbles
• Floating of small objects like needles
• Capillary rise and fall
• Wetting and non-wetting of surfaces
• Biological processes involving membranes

The angle of contact is defined as the angle between the tangent to the liquid surface at the point of contact and the solid surface, measured inside the liquid.

It is usually denoted by \(\theta\).

This angle decides whether a liquid wets a solid surface or does not wet it

The angle of contact arises due to the competition between two intermolecular forces:

• Adhesive force between liquid and solid molecules
• Cohesive force between liquid molecules themselves

The balance between these forces fixes the shape of the liquid surface near the solid and hence determines the value of \(\theta\).

Relation of Angle of Contact with Surface Tension

The angle of contact is governed by the equilibrium of surface tensions at the line of contact between:

• Solid–air surface
• Solid–liquid surface
• Liquid–air surface

This balance determines the curvature of the liquid surface and hence the angle \(\theta\). Therefore, angle of contact is an outcome of surface tension and molecular forces, not an independent property.

Angle of Contact and Capillary Action

• if \(\theta \lt 90^\circ\):
  liquid rises in capillary (water in glass tube)
• if \(\theta \gt 90^\circ\):
  liquid falls in capillary (mercury in glass tube)

Thus, angle of contact plays a central role in capillarity

Liquids and gases, when free from external constraints, arrange themselves to minimize surface energy. This natural tendency gives rise to drops (liquid enclosed by air) and bubbles (gas enclosed by a liquid film). Both are governed by surface tension, which creates excess pressure across curved surfaces. NCERT treats drops and bubbles as key applications of surface tension and surface energy.

At a curved surface, surface molecules experience unbalanced cohesive forces that try to reduce surface area. Curvature demands an internal pressure difference to balance this contracting tendency. The sharper the curvature (smaller radius), the larger the required pressure difference.

Definition of a Liquid Drop

A liquid drop is a small quantity of liquid surrounded by air, having one free surface. Due to surface tension, it tends to become spherical so as to minimize surface area and hence surface energy.

Excess Pressure Inside a Liquid Drop (Derivation)

Consider a spherical liquid drop of radius rand surface tension T.
Surface area: \(A=4\pi r^2\)
If the radius increases by dr, the increase in area:
\[dA=8\pi r\mathrm{d}r\]

Increase in surface energy:

Work done by excess pressure \Delta Pin expanding the drop:

Equating work done to increase in surface energy:

Result: The excess pressure inside a liquid drop is inversely proportional to its radius.

Definition of a Bubble

A bubble consists of a gas enclosed by a thin liquid film. Unlike a drop, a bubble has two free surfaces:

• Inner surface (liquid–gas)
• Outer surface (liquid–air)

Excess Pressure Inside a Soap Bubble (Derivation)

Let a soap bubble have radius rand surface tension T.
Total surface area (two surfaces):

If radius increases by dr,

Increase in surface energy:

Work done by excess pressure \(\Delta P\):

Result: The excess pressure inside a soap bubble is twice that inside a liquid drop of the same radius.

The two thigh bones (femurs), each of cross-sectional area10 cm2 support the upper part of a human body of mass 40 kg. Estimate the average pressure sustained by the femurs.

Solution

Given:
Cross-sectional area of each femur bone, \(A_1 = 10~\text{cm}^2\)
Mass of upper part of the body, \(m = 40~\text{kg}\)
Acceleration due to gravity, \(g = 9.8~\text{m s}^{-2}\)

Total cross-sectional area of the two femurs is
\[ \begin{aligned} A &= 2A_1 \\&= 2 \times 10~\text{cm}^2 \\&= 2 \times 10 \times 10^{-4}~\text{m}^2\\ &= 20 \times 10^{-4}~\text{m}^2 \end{aligned} \]

The weight of the upper part of the body is
\[ \begin{aligned} W &= mg \\&= 40 \times 9.8 \\&= 392~\text{N} \end{aligned} \]

The average pressure sustained by the femurs is given by
\[ \begin{aligned} P_{\text{av}} &= \dfrac{F}{A}\\\\ &= \dfrac{392}{2 \times 10 \times 10^{-4}}\\\\ &= \dfrac{392}{20 \times 10^{-4}}\\\\ &\approx 2 \times 10^{5}~\text{N m}^{-2} \end{aligned} \]

Therefore, the average pressure sustained by the femurs is \(\displaystyle 2 \times 10^{5}~\text{N m}^{-2}\).

What is the pressure on a swimmer 10 m below the surface of a lake?

Solution

Given:
Depth of swimmer, \(h = 10~\text{m}\)
Atmospheric pressure, \(P_{a} = 1.01 \times 10^{5}~\text{N m}^{-2}\)
Density of water, \(\rho = 1 \times 10^{3}~\text{kg m}^{-3}\)
Acceleration due to gravity, \(g = 10~\text{m s}^{-2}\)

The total pressure at a depth \(h\) in a liquid is given by
\[ P = P_{a} + h \rho g \]

Substituting the values, we get
\[ P = 1.01 \times 10^{5} + 10 \times 1 \times 10^{3} \times 10 \]

Simplifying,
\[ \begin{aligned} P &= 1.01 \times 10^{5} + 1.0 \times 10^{5}\\ &= 2.01 \times 10^{5}~\text{N m}^{-2} \end{aligned} \]

Hence, the pressure on the swimmer \(10~\text{m}\) below the surface of the lake is \(\displaystyle 2.01 \times 10^{5}~\text{N m}^{-2}\).

The density of the atmosphere at sea level is 1.29 \(\mathrm{kg/m^3}\). Assume that it does not change with altitude. Then how high would the atmosphere extend?

Solution

Given:
Density of atmosphere at sea level, \(\rho = 1.29~\text{kg m}^{-3}\)
Atmospheric pressure at sea level, \(P = 1.01 \times 10^{5}~\text{Pa}\)
Acceleration due to gravity, \(g = 9.8~\text{m s}^{-2}\)

Using the relation between pressure, density, gravity and height,
\[ P = h \rho g \]

Therefore, the height \(h\) of the atmosphere is
\[ h = \dfrac{P}{\rho g} \]

Substituting the values, we get
\[ \begin{aligned} h &= \dfrac{1.01 \times 10^{5}}{1.29 \times 9.8}\\\\ &= 7989~\text{m}\\\\ &\approx 8~\text{km} \end{aligned} \]

Hence, if the density of air remained constant, the atmosphere would extend up to about \(8~\text{km}\) above sea level.

At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) What is the gauge pressure? (c) Find the force acting on the window of area 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sea level atmospheric pressure. (The density of sea water is \(\mathrm{1.03 × 10^3\ kg\ m^{-3},\ g = 10\ m\ s^{–2}}\)).

Solution

Given:
Depth, \(h = 1000~\text{m}\)
Density of sea water, \(\rho = 1.03 \times 10^{3}~\text{kg m}^{-3}\)
Acceleration due to gravity, \(g = 10~\text{m s}^{-2}\)
Atmospheric pressure, \(P_{0} = 1.01 \times 10^{5}~\text{N m}^{-2}\)
Area of submarine window, \(A = 20~\text{cm} \times 20~\text{cm} = 400~\text{cm}^{2} = 0.04~\text{m}^{2}\)

(a) Absolute pressure:
The absolute pressure at depth \(h\) is given by
\[ P = P_{0} + h \rho g \]

Substituting the values,
\[ \begin{aligned} P &= 1.01 \times 10^{5} + 1000 \times 1.03 \times 10^{3} \times 10\\ &= 1.01 \times 10^{5} + 103 \times 10^{5}\\ &= 104.01 \times 10^{5}~\text{N m}^{-2}\\ &\approx 104~\text{atm} \end{aligned} \]

(b) Gauge pressure:
The gauge pressure is the pressure due to water column alone,
\[ \begin{aligned} P_{g} &= h \rho g\\ &= 1000 \times 1.03 \times 10^{3} \times 10\\ &= 1.03 \times 10^{7}~\text{N m}^{-2}\\ &= 103 \times 10^{5}~\text{N m}^{-2} \end{aligned} \]

(c) Force on submarine window:
The interior pressure is sea level atmospheric pressure \(P_{0}\), so the net pressure difference is
\[ \begin{aligned} \Delta P &= P - P_{0} = h \rho g \\ &= 103 \times 10^{5}~\text{N m}^{-2} \end{aligned} \]

Force acting on the window is
\[ \begin{aligned} F &= \Delta P \times A\\ &= 103 \times 10^{5} \times 0.04\\ &= 4.12 \times 10^{5}~\text{N} \end{aligned} \]

Two syringes of different cross-sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively. (a) Find the force exerted on the larger piston when a force of 10 N is applied to the smaller piston. (b) If the smaller piston is pushed in through 6.0 cm, how much does the larger piston move out?

Solution

Given:
Diameter of smaller piston, \(d_{1} = 1.0~\text{cm}\), so radius \(r_{1} = 0.5~\text{cm} = 0.5 \times 10^{-2}~\text{m}\)
Diameter of larger piston, \(d_{2} = 3.0~\text{cm}\), so radius \(r_{2} = 1.5~\text{cm} = 1.5 \times 10^{-2}~\text{m}\)
Force on smaller piston, \(F_{1} = 10~\text{N}\)
Displacement of smaller piston, \(L_{1} = 6.0~\text{cm} = 0.06~\text{m}\)

(a) Force on larger piston:
By Pascal's law, pressure is the same on both pistons, so
\[ \begin{aligned} \dfrac{F_{2}}{A_{2}} &= \dfrac{F_{1}}{A_{1}} \\\\ \text{or} \\\\ F_{2} &= F_{1} \dfrac{A_{2}}{A_{1}} \end{aligned} \]

Area of smaller piston, \(A_{1} = \pi r_{1}^{2} = \pi \left( \dfrac{1}{2} \times 10^{-2} \right)^{2}\)
Area of larger piston, \(A_{2} = \pi r_{2}^{2} = \pi \left( \dfrac{3}{2} \times 10^{-2} \right)^{2}\)
Therefore,
\[ \begin{aligned} \dfrac{A_{2}}{A_{1}} &= \dfrac{\left( \dfrac{3}{2} \times 10^{-2} \right)^{2}}{\left( \dfrac{1}{2} \times 10^{-2} \right)^{2}} \\\\&= \left( \dfrac{3}{1} \right)^{2} \\\\&= 9 \end{aligned} \]

Hence,
\[ F_{2} = 10 \times 9 = 90~\text{N} \]

(b) Displacement of larger piston:
Since volume of water displaced is the same,
\[ \begin{aligned} A_{1} L_{1} &= A_{2} L_{2} \\\\ \text{or} \\\\\ L_{2} &= L_{1} \dfrac{A_{1}}{A_{2}} \end{aligned} \]

As \(\dfrac{A_{1}}{A_{2}} = \dfrac{1}{9}\),
\[ \begin{aligned} L_{2} &= 0.06 \times \dfrac{1}{9} \\\\ &= 0.00667~\text{m} \\\\ &\approx 0.67~\text{cm} \end{aligned} \]

In a car lift compressed air exerts a force F1 on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm (Fig 9.7). If the mass of the car to be lifted is 1350 kg, calculate F1 . What is the pressure necessary to accomplish this task? \(\mathrm{(g = 9.8\ m\ s^{-2})}\).

Solution

Given:
Radius of small piston, \(r_{1} = 5.0~\text{cm} = 0.05~\text{m}\)
Radius of large piston, \(r_{2} = 15~\text{cm} = 0.15~\text{m}\)
Mass of car, \(m = 1350~\text{kg}\)
Acceleration due to gravity, \(g = 9.8~\text{m s}^{-2}\)

Weight of car (force on large piston),
\[ \begin{aligned} F_{2} &= mg \\ &= 1350 \times 9.8 \\ &= 13230~\text{N} \end{aligned} \]

By Pascal's principle, pressure is the same on both pistons,
\[ \begin{aligned} \dfrac{F_{1}}{A_{1}} = \dfrac{F_{2}}{A_{2}} \quad \text{or} \\\\ F_{1} = F_{2} \dfrac{A_{1}}{A_{2}} \end{aligned} \]

Area of small piston, \(A_{1} = \pi r_{1}^{2} = \pi (0.05)^{2}\)
Area of large piston, \(A_{2} = \pi r_{2}^{2} = \pi (0.15)^{2}\)
Therefore,
\[ \begin{aligned} \dfrac{A_{1}}{A_{2}} &= \dfrac{(0.05)^{2}}{(0.15)^{2}} \\\\ &= \dfrac{0.0025}{0.0225} \\\\ &= \dfrac{25}{225} \\\\ &=\dfrac{1}{9} \end{aligned} \]

Hence, force required on small piston,
\[ \begin{aligned} F_{1} &= 13230 \times \dfrac{1}{9} \\ &= 1470~\text{N} \\ &\approx 1.5 \times 10^{3}~\text{N} \end{aligned} \]

Pressure necessary,
\[ \begin{aligned} P &= \dfrac{F_{1}}{A_{1}} \\\\ &= \dfrac{1470}{\pi (0.05)^{2}} \\\\ &= \dfrac{1470}{0.007854} \\\\ &\approx 1.87 \times 10^{5}~\text{Pa} \end{aligned} \]

A fully loaded Boeing aircraft has a mass of \(\mathrm{3.3 × 10^5\ kg}\). Its total wing area is \(\mathrm{500\ m^2}\). It is in level flight with a speed of 960 km/h. (a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is \(\mathrm{\rho = 1.2\ kg\ m^{-3}}\)]

Solution

Given:
Mass of aircraft, \(m = 3.3 \times 10^{5}~\text{kg}\)
Total wing area, \(A = 500~\text{m}^{2}\)
Speed of aircraft, \(v = 960~\text{km h}^{-1} = 960 \times \dfrac{5}{18} = 267~\text{m s}^{-1}\)
Density of air, \(\rho = 1.2~\text{kg m}^{-3}\)
Acceleration due to gravity, \(g = 9.8~\text{m s}^{-2}\)

The weight of the aircraft is balanced by the upward force due to pressure difference between the lower and upper surfaces of the wings,
\[ \Delta P \times A = mg \]

Therefore,
\[ \begin{aligned} \Delta P &= \dfrac{mg}{A}\\\\ &= \dfrac{3.3 \times 10^{5} \times 9.8}{500}\\\\ &= 6.47 \times 10^{3}~\text{N m}^{-2}\\\\ &\approx 6.5 \times 10^{3}~\text{N m}^{-2} \end{aligned} \]

Using Bernoulli's equation between air just below and just above the wing and neglecting any height difference,
\[ P_{1} + \dfrac{1}{2}\rho v_{1}^{2} = P_{2} + \dfrac{1}{2}\rho v_{2}^{2} \] \[ \Delta P = P_{1} - P_{2} = \dfrac{1}{2}\rho \left( v_{2}^{2} - v_{1}^{2} \right) \]

This can be written as
\[ \Delta P = \dfrac{1}{2}\rho (v_{2} + v_{1})(v_{2} - v_{1}) \] \[ v_{2} - v_{1} = \dfrac{2 \Delta P}{\rho (v_{2} + v_{1})} \]

Let the average speed of air over the wing be
\[ \begin{aligned} v_{\text{av}} &= \dfrac{v_{1} + v_{2}}{2} \\ &\approx 267~\text{m s}^{-1} \end{aligned} \] so that \(v_{1} + v_{2} \approx 2 v_{\text{av}} = 534~\text{m s}^{-1}\).
Then,
\[ \begin{aligned} v_{2} - v_{1} =& \dfrac{2 \times 6.5 \times 10^{3}}{1.2 \times 534}\\\\ &\approx 20.3~\text{m s}^{-1} \end{aligned} \]

The fractional increase in speed of air above the wing relative to below is approximately
\[ \begin{aligned} \dfrac{v_{2} - v_{1}}{v_{\text{av}}} &\approx \dfrac{20.3}{267}\\\\ &\approx 0.08 = 8\% \end{aligned} \]

Thus, the required pressure difference is about \(6.5 \times 10^{3}~\text{N m}^{-2}\) and the speed of air over the upper surface needs to be roughly \(8\%\) higher than over the lower surface.

A metal block of area \(\mathrm{0.10\ m^2}\) is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in Fig. 9.13. A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released the block moves to the right with a constant speed of \(\mathrm{0.085\ m\ s^{-1}}\). Find the coefficient of viscosity of the liquid.

Solution

The motion of the metal block is observed to be uniform, which implies that the net force acting on the block–mass system is zero. Hence, the pulling force due to the hanging mass is exactly balanced by the viscous resistive force offered by the liquid film between the block and the table.

The tension in the string is equal to the weight of the hanging mass. Therefore,

The viscous force acting on the block due to the liquid film is given by the relation for viscous drag between parallel layers,

Here, \(\eta\) is the coefficient of viscosity of the liquid, \(A = 0.10\ \text{m}^2\) is the area of contact, \(v = 0.085\ \text{m s}^{-1}\) is the constant speed of the block, and \(d = 0.30\ \text{mm} = 3.0 \times 10^{-4}\ \text{m}\) is the thickness of the liquid film.

Since the block moves with constant speed, the viscous force balances the tension in the string. Hence,

Substituting the known values,

Therefore, the coefficient of viscosity of the liquid is approximately \(3.5 \times 10^{-3}\ \text{Pa s}\).

The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20oC is \(\mathrm{6.5\ cm\ s^{-1}}\). Compute the viscosity of the oil at \(\mathrm{20^\circ\,C}\). Density of oil is \(\mathrm{1.5 ×10^3\ kg\ m^{-3}}\), density of copper is \(\mathrm{8.9 × 10^3\ kg\ m^{-3}}\).

Solution

When the copper ball falls through the oil and attains terminal velocity, the net force acting on it becomes zero. At this stage, the downward gravitational force reduced by buoyancy is exactly balanced by the viscous drag acting upward on the ball.

For a spherical body moving with terminal velocity in a viscous fluid, Stokes’ law applies. According to this law, the viscous force is proportional to the velocity and is given by \(6\pi \eta r v_t\), where \(\eta\) is the coefficient of viscosity, \(r\) is the radius of the sphere, and \(v_t\) is the terminal velocity.

At terminal velocity, the force balance condition can be written as:

Simplifying the above expression to obtain viscosity,

The given quantities are now substituted. The radius of the copper ball is \(r = 2.0\ \text{mm} = 2.0 \times 10^{-3}\ \text{m}\). The terminal velocity is \(v_t = 6.5\ \text{cm s}^{-1} = 6.5 \times 10^{-2}\ \text{m\ s}^{-1}\). The density of copper is \(\rho_c = 8.9 \times 10^{3}\ \text{kg m}^{-3}\), and the density of oil is \(\rho_o = 1.5 \times 10^{3}\ \text{kg m}^{-3}\).

Hence, the coefficient of viscosity of the oil at \(20^\circ\text{C}\) is approximately \(1.0\ \text{Pa s}\).

The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at temperature of the experiments is \(\mathrm{7.30 × 10^{-2}\ N\,m^{-1}}\). 1 atmospheric pressure = \(\mathrm{1.01 × 10^5\ Pa}\), density of water = \(\mathrm{1000\ kg/m^3,\ g = 9.80\ m\ s^{-2}}\). Also calculate the excess pressure.

Solution

When air is blown through the capillary tube immersed in water, a hemispherical bubble is formed at the lower end. To create this bubble, the pressure inside the tube must overcome both the hydrostatic pressure due to the water column and the additional pressure arising from surface tension at the curved surface of the bubble.

The radius of the capillary tube is half of its diameter. Hence,

The hydrostatic pressure at a depth \(h = 8.00\ \text{cm} = 0.080\ \text{m}\) below the water surface is given by \(\rho g h\).

For a hemispherical bubble formed in a liquid, the excess pressure due to surface tension is given by \(\displaystyle \frac{2T}{r}\).

Thus, the total pressure inside the bubble above atmospheric pressure is the sum of the hydrostatic pressure and the surface tension pressure.

Therefore, the absolute pressure required inside the tube is obtained by adding this excess pressure to atmospheric pressure.

Hence, the pressure required in the tube to blow the hemispherical bubble is \(1.02 \times 10^{5}\ \text{Pa}\), and the corresponding excess pressure is \(9.30 \times 10^{2}\ \text{Pa}\).