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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 is a measure of the degree of fractional deformation due to applied stress on a material.


 


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:


Stress Strain




• 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




By subtitution F= ke,



2 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


 




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).


• 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


The upward force exerted by water on any immersed object is called a buoyant force.


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 = F g = 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 ρ f , the magnitude of the upward buoyant force is B = ρ f V₀g . V₀ is the volume of the object.

If M and ρ₀ are the mass and density of the object respectively, Its weight F g = Mg = ρ₀V₀g Then, the net force = Buoyant Force – Weight Net Force = B – F g = ( ρ f ρ₀) V₀g If ρ₀ < ρ f , F g < B → up acceleration of the object.

If ρ₀ > ρ f , F g > 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 V f 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 = ρ f V f g. The weight of the object F g = Mg = ρ ₀ V ₀ g. Note that B = F g → ρ f V f g = ρ ₀ V ₀ g. 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


. In a non viscous fluid, internal friction is neglected. An object moving through the fluid experiences no viscous force.

* The flow is


. 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


. 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 A 1 in the time t must equal the mass that flows out through A 2 in the same interval i.e

∆M 1 = ∆M 2 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


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,


, 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)