# General Physics III PHY 112

```General Physics III PHY 112
Course Contents
• Molecular treatment of properties of matter, Elasticity; Hooke’s
Law, Young’s, shear and bulk moduli.
• Hydrostatics; Pressure, buoyancy. Archimedes Principles.
• Hydrodynamics; Streamlines Bernoulli and continuity equations.
Turbulence, Reynold’s number. Viscosity; laminar flow, Poiseuille’s
equation.
• Surface tension; adhesion, cohesion, capillarity, drops and bubbles.
Texts: (1) Fundamental of Physics by Halliday, Resnick and Walter
(2) College Physics by Serway.
Molecular treatment of the properties of matter
All matter consists of some distribution of atoms or molecules.
The three states of matter are solids, liquid and gas.
Solid :
definite shape and volume
well ordered arrangement of molecules.
vibration of molecules about a fixed point
crystalline or amorphous.
Liquid:
definite volume but no definite shape
molecules are at distance apart and are free to move about
Weaker attractive force compare to solid
Occupy the shape of the container.
Gas:
no definite shape and volume
molecules are farther apart
force attraction is very weak
they are free to move about in random motion.
Atomic arrangement in matter
.
(a) Well ordered
arrangement of
atoms
in solid
(b) Random arrangement
of atom in amorphous
solid
(c) Erratic motion
of atoms
Stress and Material Deformation
• Forces put stresses on solids, and stresses can strain, deform, and
break them. Fluids under pressure can perform work. Flowing gases
cause pressure differences that can be used to do work.
• Solid objects are deformable, i.e it is possible to change their shapes
and sizes (or both) by applying external forces.
• However, strong intermolecular forces resist deformation.
•
The deformation of solid occurs due to stress.
Stress is a quantity that is proportional to the force causing a deformation in
the material (elongation or compression).
Strain
•
Strain is a measure of the degree of fractional deformation due to applied
stress on a material.
X
Strain 
X
•
strain has no unit since it the ratio of two similar units.
•
For sufficiently small stresses, stress α strain. Stress = Y strain.
•
A sufficiently large force will permanently deform or damage the object.
Object tend to regain its shape and size once the force is removed.
•
Y is called the Elastic Modulus and it depends material being deformed and
nature of the deformation. A material with high Y can be taken to stiff and
difficult to deform.
Elastic modulus and types of deformation
• 1. Young’s modulus: resistance of a solid to a change in its length.
• 2. Shear modulus: resistance to motion of the planes of a solid sliding past
each other.
• 3. Bulk modulus: resistance of solids or liquids to changes in their volume
Young’s Modulus: Elasticity in Length
When an external force F is applied perpendicular to the cross section A of bar
having a length L, internal forces in the bar resist distortion (“stretching”), but the
bar attains an equilibrium in which its length increased by ∆L.
The tensile stress which is the ratio of the magnitude of the external force F to the
cross-sectional area A occurs.
The tensile strain in this case is defined as the ratio of the change in length ∆L to the
original length L .
Elasticity (Cont.)
• Therefore, Young’s modulus by a combination of these
two ratios:
F
Stress
Y
 A (N/m²)
Strain L
L
• Note that Young’s modulus is typically used to characterize a rod or wire
stressed under either tension or compression.
• The elastic limit of a substance is defined as the maximum stress that can
be applied to the substance before it becomes permanently deformed.
• It is possible to exceed the elastic limit of a substance by applying a
sufficiently large stress
Typical values for the elastic modulus for
various materials
Stress versus strain curve
Plastic deformation
A material is said to be ductile if it can be stressed well beyond its elastic limit without
breaking.
A brittle material is one that breaks soon after the elastic limit is reached
Hooke’s Law
A elastic object regains it original shape and length after an external force is removed.
The ability of an object to do this is called elasticity.
Hooke’s states that for an elastic material, the extension, e produced is directly proportional
to the force applied F. F α e, F= ke k is constant of proportionality (N/m).
For a variable force, the work done to produce the extension e is
Fe
W 
2
By subtitution F= ke,
ke2
W 
2
Shear Modulus: Elasticity of Shape
.
Shear stress is the deformation occurs
when an object is subjected to a force
tangential to one of its faces while the
opposite face is held fixed by another
force.
The shear stress = F/A
The shear strain = x/h,
Hence, shear modulus S (N/m²) is given by
Bulk Modulus: Volume Elasticity
Bulk modulus characterizes the response of a substance to uniform
squeezing or to a reduction in pressure when the object is placed in a
partial vacuum.
• When a solid is under uniform pressure
(forces), it undergoes a change in volume
but no change in shape.
The volume stress (Pressure) P = F/A (Nm²)
If the pressure on an object changes by an
∆P = ∆F/A, then the object will experience
a volume change ∆V.
The volume strain = ∆V/V .
The bulk modulus B
Bulk Modulus cont.
We can characterize a volume (“bulk”) compression in terms of the bulk modulus, which is
defined as
VP
B
V
A negative sign is inserted in this defining equation so that B is a positive number. An
increase in P results to decrease in volume.
The reciprocal of the bulk modulus is called the compressibility, K, of the material.
K = 1/B.
•
Shear modulus and Young’s modulus are not given for liquids because a liquid does not
sustain a shearing stress nor a tensile stress (it flows instead).
Fluid Mechanics
. Among the three states of matter, both liquids and gases are fluids.
A fluid is a collection of molecules that are randomly arranged and held
together by weak cohesive forces and by forces exerted by the walls of a
container.
• Fluids do not sustain shearing stresses or tensile stresses. the force exerted by
a fluid on an object is always perpendicular to the surfaces of the object
Fluid mechanics covers:
hydrostatics (fluid at rest)
hydrodynamics (fluid on motion).
Pressure
• If F is the magnitude of the force exerted on the piston and A is the surface
area of the piston, pressure P of the
fluid at the level to which the device has been
submerged is defined as the ratio F/A:
P = F/A (Nm²) or Pascal (P)
Note that pressure is a scalar quantity because it is
proportional to the magnitude of the force on the piston.
The pressure exerted by a fluid varies linearly with depth.
Also, atmospheric pressure decreases with increasing altitude; it is for this
reason that aircraft flying at high altitudes must have pressurized cabins.
Pressure (cont.)
. The density of a substance is defined as its mass per unit volume ρ= M/V.
(kgm³).
A fluid of density at rest and open to the atmosphere, assume that is constant;
this means that the fluid is incompressible.
PA is the upward force exerted by the outside fluid on
the bottom of the cylinder.
PoA is the downward force exerted by the atmosphere
on the top of the cylinder.
The mass of liquid in the cylinder is M = ρV= ρAh
At equilibrium, the net force acting on the clyinder
must be zero.
Pressure cont.
Considering the upward to be positive direction,
∑Fy = PA – PoA –Mg = O
PA - PoA –ρAhg = 0
PA - PoA = ρAhg
P = Po – ρhg
This implies that the pressure is the same at all points having the same depth,
independent of the shape of the container.
The pressure P at a depth h below the surface of a liquid open to the atmosphere is
greater than atmospheric pressure by an amount ρgh.
Po = 1.00atm = 1.013 x 10⁵ Pa.
Pressure Measuring Devices
.
(a) Open tube manometer
(b) Barometer
One end of a U-shaped tube containing a liquid is open to the atmosphere, and the
other end is connected to a system of unknown pressure P (absolute pressure) and
The difference in pressure P - Po is equal to ρgh. Its value is normally appears on a
pressure gauge.
For barometer, Po = ρgh. Since the closed end of the tube is nearly a vacuum, and so
its pressure can be taken as zero (P = 0).
Pascal’s Principle
• The pressure in a fluid depends on depth and on the value of
Po, any increase in pressure at the surface must be transmitted
to every other point in the fluid.
• Pascal’s law: A change in the pressure applied to a fluid is transmitted
undiminished to every point of the fluid and to the walls of the container.
An important application of Pascal’s law is
the hydraulic press.
Pressure must be the same on both sides ,
therefore, P= F₁/A ₁= F ₂ /A ₂.
Since liquid is neither added nor removed, its
volume is the same, hence A₁d ₁ = A ₂ d₂.
Hydraulic brakes, car lifts, hydraulic jacks,
and forklifts all make use of this principle
BUOYANT FORCES AND ARCHIMEDES’S PRINCIPLE
The upward force exerted by water on any immersed object is called a buoyant force.
.
The buoyant force acts vertically upward through the point that was the center of gravity
of the displaced fluid.
The manner in which buoyant forces act is summarized by Archimedes’s
principle, which states that the magnitude of the buoyant force always equals
the weight of the fluid displaced by the object.
This cube is acted upon by gravitational force Fg (downward
force) and Bouyant force B (upward force).
At equilibrum, Fg = B.
The pressure at the bottom of the cube is greater than the
pressure at the top by an amount gh, where h is the length
of any side of the cube.
Buoyant Force (cont.)
. The pressure at the bottom of the cube is greater than the pressure at the top
by an amount ρgh, where h is the length of any side of the cube.
• The pressure difference P between the bottom and top faces of the cube is
equal to the buoyant force per unit area of those faces i.e.
∆P = B/A
B = (∆P)A = ρghA = ρgV where V is the volume of the cube
Note that mass of the fluid in the cube M = ρV
B = Fg = ρVg = Mg
Mg = weight of the fluid in the cube.
• Thus, the buoyant force is a result of the pressure differential on a
submerged or partly submerged object.
Buoyant Force (Cont.)
Case 1: Totally Submerged Object.
When an object is totally submerged in a fluid of
density ρf, the magnitude of the upward buoyant
force is B = ρfV₀g . V₀ is the volume of the object.
If M and ρ₀ are the mass and density of the object respectively,
Its weight Fg = Mg = ρ₀V₀g
Then, the net force = Buoyant Force – Weight
Net Force = B – Fg = (ρf - ρ₀) V₀g
If ρ₀ < ρf, Fg < B → up acceleration of the object.
If ρ₀ > ρf, Fg > B → downward acceleration of the object (object sinks in the fluid).
Buoyant Force (cont.)
Case 2: Floating Object
Consider an object of volume V₀ in static equilibrium floating on a fluid—that is,
an object that is only partially submerged.
If Vf is the volume of the fluid displaced by the object (this volume is the same as
the volume of that part of the object that is beneath the fluid level).
The buoyant force B = ρfVfg.
The weight of the object Fg = Mg = ρ₀V₀g.
Note that B = Fg → ρfVfg = ρ₀V₀g.
Then,
Hydrodynamics (Fluid in Motion)
• When a fluid is in motion, its flow can be characterized by streamline flow
(steady or laminar ) or turbulent flow.
Streamline flow : particles pass a particular
point move along exactly the same smooth
path followed by previous particles passing
that point. The velocity of the fluid at any
point remains constant in time.
Turbulent flow: Above a certain critical
speed, fluid flow becomes turbulent. It is
irregular flow characterized by small
whirlpool-like regions.
Fluid viscosity
Viscosity is commonly used in the description of fluid flow to characterize
the degree of internal friction in the fluid.
It is associated with the resistance that two adjacent layers of fluid have
when moving relative to each other.
Viscosity causes part of the kinetic energy of a fluid to be converted to
internal energy.
This mechanism is similar to the one by which an object sliding on a rough
horizontal surface loses kinetic energy.
A fluid such as kerosene has a lower viscosity than does crude oil or
molasses.
Characteristics of an ideal fluid
* The fluid is nonviscous. In a non viscous fluid, internal friction is neglected.
An object moving through the fluid experiences no viscous force.
* The flow is steady. In steady (laminar) flow, the velocity, density and
pressure of the fluid at each point remain constant.
* The fluid is incompressible. The density of an incompressible fluid is
constant.
* The flow is irrotational. The fluid has no angular momentum about any
point.
Equation of Continuity
Consider a fluid flowing through a pipe of non
uniform size.
.
For a steady flow, the mass that flows into the
bottom of the pipe through A1 in the time t must
equal the mass that flows out through A2 in the same
interval i.e
∆M1 = ∆M2
ρ₁A₁v₁ = ρ₂A₂v₂
For an incompressible fluid ρ₁ = ρ₂, → A₁v₁ = A₂v₂
This expression is called the equation of continuity.
The product of the cross-sectional area of the pipe
and the fluid speed at that cross section is a constant.
Av has dimensions of volume per unit time, is called
the flow rate.
Bernoulli’s Equation
It relates the pressure of a fluid to its speed and elevation. Bernoulli’s equation is not a
freestanding law of physics; rather, it’s a consequence of energy conservation as
applied to an ideal fluid.
.
Assume that the fluid is incompressible, nonviscous, and flows in a nonturbulent,
steady-state manner and the flow through a nonuniform pipe, at the lower end of the
pipe, the work done by the fluid is
W₁ = F₁∆x = P ₁A ₁ ∆x₁ = P ₁V
At the upper portion,
W₂ = -F₂∆x = P₂A₂ ∆x₂ = -P₂V
(negative implies work is opposite its displacement)
Volume V is the same because of continuity equation.
The net work done = P ₁V - P₂V.
Note: part of the goes into changing the kinetic
Energy while the rest goes into the gravitational
enegy of the fluid.
Bernoulli’s Equation (cont.)
The change in K.E and P.E of the fluid of mass M
∆K. E. = ½ Mv₂² - ½ Mv₁²
∆P.E = mgy₂ - mgy₁
By subtitution,
P ₁V - P₂V = ½ Mv₂² - ½ Mv₁² + mgy₂ - mgy₁
If we divide each term by V and recall that ρ = m/V, this expression becomes.
P₁– P₂ = 1/2ρv₂² + 1/2ρv₁² + ρgy₂ - ρgy₁
Rearranging
P₁ + 1/2ρv₁² + ρgy₁ = P₂ + 1/2ρv₂² + ρgy₂
This is Bernoulli’s equation, often expressed as
P + 1/2ρv² + ρgy = constant
• Bernoulli’s equation states that the sum of the pressure P, the kinetic
energy per unit volume, 1/2ρv², and the potential energy per unit
volume, ρgy, has the same value at all points along a streamline.
Bernoulli’s equation (cont.)
For a Venturi tube (a horizontal
constricted pipe y₁=y₂),
Bernoulli’s equation reduces to
P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂²
Then the pressure change ∆P = P₁ - P₂
∆P = 1/2ρ(v₂² - v₁²) + ρg(y₂ - y₁)
SURFACE TENSION, CAPILLARY ACTION,
AND VISCOUS FLUID FLOW
.
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