Unit Seven: Solids and Fluids

GE253 Physics Unit Seven:Solids and Fluids John Elberfeld JElberfeld@itt-tech.edu 518 872 2082

Schedule • Unit 1 – Measurements and Problem Solving • Unit 2 – Kinematics • Unit 3 – Motion in Two Dimensions • Unit 4 – Force and Motion • Unit 5 – Work and Energy • Unit 6 – Linear Momentum and Collisions • Unit 7 – Solids and Fluids • Unit 8 – Temperature and Kinetic Theory • Unit 9 – Sound • Unit 10 – Reflection and Refraction of Light • Unit 11 – Final

Chapter 7 Objectives • Distinguish between stress and strain, and use elastic moduli to compute dimensional changes. • Explain the pressure-depth relationship and state Pascal's principle and describe how it is used in practical applications. • Relate the buoyant force and Archimedes' principle and tell whether an object will float in a fluid, on the basis of relative densities. • Identify the simplifications used in describing ideal fluid flow and use the continuity equation and Bernoulli's equation to explain common effects of ideal fluid flow. • Describe the source of surface tension and its effect and discuss fluid viscosity.

Reading Assignment • Read and study College Physics, by Wilson and Buffa, Chapter 7, pages 219 to 250 • Be prepared for a quiz on this material

Written Assignments • Do the homework on the handout. • You must show all your work, and carry through the units in all calculations • Use the proper number of significant figures and, when reasonable, scientific notation

Introduction • Until now, we have beendealing with “ideal” objects • Now we will look at objects that can stretch, bend, break, flow, and compress when forces are applied

Introduction • In this week, we will study real, macroscopic objects, such as wires, blocks, and containers of fluids. • We will keep the concept of force and see what happens to macroscopic objects when forces act on them and see what forces macroscopic objects exert.

Stress and Strain • The stress of taking physics may cause a strain on your brain • But in physics, these terms have different and precise meanings. • We will learn to use these terms as we study the effects of forces on solids and fluids – in this unit, not to accelerate objects, but to deform them.

Stress and Strain • Stress (force/area) causes strain (% deformation of some type)

Stress and Strain • In physics, stress is related to the force applied to an object • Strain is related to how the object is deformed as a result. • Both also depend on the size of the object.

Stress • The diagram shows a wire or rod of length Lo and area A. • When you apply a force F on the wire, it stretches by a distance . • When you apply a force F in the opposite direction on a rod, the rod is compressed by a distance . • The greater the force F, the greater is the amount of stretching or compressing. A F L0

Stress • It is clear that a thick wire is harder to stretch than a thin one. • For this reason, you can define the concept of stress as: • Stress = F / A • The units of stress are N/m2.

Strain • The longer the wire or rod, the greater is the amount of stretching. For this reason, you can define strain as: • Strain = | ΔL | / L • The vertical lines stand for absolute value, which means the same strain results from positive ΔL due to stretching or negative Δ L due to compressing.

Strain • A strain of 0.05 on a wire means that the wire has been stretched to a length 5% greater than its original length.

Young’s Modulus • NOTE: This graph shows Young’s Modulus: how much a wire will stretch if you apply different forces on it. Young’sModulusRegion

Graph • Engineers need to know the force need to stretch material a certain distance • What force cause the bridge to sag into the waves? • Graphs show strain on the X axis and stress on the Y axis

Young’s Modulus • Young’s Modulus works in the straight line part of the graph only • The ratio of stress to strain in a material is known as Young’s modulus: • Y = (F/A) / ( ΔL / L) • Where, F is force; A is cross sectional area perpendicular to the applied force; and L is length of the object being stressed. • NOTE: Young’s Modulus is an ELASTIC Modulus because materials bounce back to their original shape in that part of the graph.

Young’s Modulus • The bigger the modulus, the stronger the material, and the more force it takes to cause a specific deformation. • Because material bounce back to their original shapes, moduli (plural) are elastic • The units of Young’s modulus are the same as those of stress, N/m2. • Young’s modulus for steel is 20 x 1010 N/m2and for bone is 1.5 x 1010 N/m2 for example.

Practice • The femur (upper leg bone) is the longest and strongest bone in the body. • Let us assume that a typical femur is circular and has a radius of 2.0 cm. • How much force is required to extend the bone by 0.010%?

Calculations • Y = (F/A) / ( ΔL / L) • From the chart: Ybone = 1.5 x 1010N/m2 • R = 2cm(1m/100cm) = .02m • A = R2 = (.02m)2 =1.26 x 10-3m2 • ΔL / L = .01% = .01/100 = 1 x 10-4 • Y ( ΔL / L) = (F/A) • F = A Y ( ΔL / L) • F = 1.26 x 10-3m2 1.5 x 1010N/m2 1 x 10-4 • F = 1.89 x 103 N (about 425 pounds)

Practice • A mass of 16 kg is suspended from a steel wire of 0.10-cm-diameter. • By what percentage does the length of the wire increase? • Young’s modulus for steel is 20 x 1010 N/m2.

Calculation • Y = (F/A) / ( ΔL / L) • F = W = mg =16kg 9.8m/s2 = 157 N • R = .1cm(1m/100cm)/2 = .0005m • A = R2 = (.0005m)2 =7.85 x 10-7m2 • ( ΔL / L) = (F/A) / Y = (F/ R2) / Y • (ΔL/L) = (157N/ 7.85 x 10-7m2) / 20x1010N/m2 • (ΔL/L) = .001 = 0.1 % • NOTE: The change in length is inversely proportional to the SQUARE of the radius!

Shear Modulus • To distort a rectangular solid, apply a force on one of its faces in a direction parallel to the face. • Simultaneously, you must also apply a force on the opposite face in the opposite direction. • The diagram on the screen shows how you can distort an object by applying force.

Shear • The diagram on the screen shows that the amount of distortion can be measured by the new angle Ø , which the faces make. • Shearing stress is defined as F/A, where F is the tangential force and A is the area of the surface that the force acts on. • Shearing strain is the angle Ø . • Similar to linear distortions, shearing strain is directly proportional to shearing stress. • The constant of proportionality is called the shear modulus (S): • S = (F/A) / Φ

Shear • Shear modulus has the same units as Young’s modulus, N/m2. • For many substances, shear modulus is approximately one-third of the Young’s modulus.

Bulk Modulus • You can also distort a rectangular solid by applying forces perpendicular to its surfaces. • This stress causes the solid to become smaller and is known as volume stress or pressure. • Volume stress is defined as F/A, where F is the force perpendicular to the surfaces, and A is the area of the surface.

Bulk Modulus • Volume strain is the change in volume divided by the original volume. • Similar to shearing stress and stress, strain is directly proportional to volume stress. • The constant of proportionality is called bulk modulus (B) and is defined as: • B = (F/A) / ( ΔV / V)

Bulk Modulus • Solids are usually surrounded by air that exerts a compressing force on their surface. • Pressure = Force/Area • Therefore, when you apply a force F, you increase this force, or more precisely, the volume stress. • Bulk modulus is usually written as: • B = (F/A) / ( ΔV / V) • B = Δp / ( ΔV / V) • Δp is the increase in pressure above normal air pressure

Table 1.0 x 10 9 4.5 x 109 26 x 109 2.2 x 109

Compare Bulk Modulus • NOTE: • B = Δp / ( ΔV / V) • For a gas, it takes the smallest pressure to create a change in volume, so gasses have the smallest bulk modulus • Solids require a big pressure to have a change in volume, so they have the biggest bulk modulus • Bigger modulus implies a stronger material

Practice • By how much must you change the pressure on a liter of water to compress it by 0.10%?

Calculation • B = Δp / ( ΔV / V) • B = 2.2 x 109N/m2 • ( ΔV / V) = .1% (1/100%) = .001 • Δp = B ( ΔV / V) • Δp = 2.2 x 109 N/m2 .001 = 2.2 x 106 N/m2 • Because there is always air pressure, this is the INCREASE in pressure to cause the change in volume

Pressure in a Fluid • The molecules in a solid are tightly bound. • In a liquid or a gas, however, the molecules are in motion and free to move. • The diagram shows a cubical container filled with gas. • The molecules of the gas are represented by the red dots. • The blue arrows show the direction they are moving in.

Pressure • When a gas molecule with momentum p hits the wall and bounces back, it exerts an impulse equal to 2p on the wall. • To calculate the force exerted by a single molecule, you need to know how long the collision lasted. • The effect of a single molecule is very small, but the effect of a room full, like in a tornado, can be huge.

Measuring Pressure • The diagram shows a device that measures the pressure of a gas. • Because gas molecules can move freely, the pressure of a gas will be the same no matter where you place the gauge. • Furthermore, any changes in pressure applied to the container will be transmitted throughout the gas volume by the moving molecules.

Pascal’s Principle • Because liquid molecules are also free to move, the same principles also apply to liquids and thus all fluids. • Pascal’s principle states this effect: • Pressure applied to an enclosed fluid is transmitted undiminished to every point in the fluid and to the walls of the container.

Atmospheric Pressure • When a gas is in a container, it exerts uniform pressure throughout the container. • In the case of atmosphere, however, the air stays at the Earth’s surface because of the force of gravity. • Air pressure is greatest at sea-level and decreases as the altitude increases. • The pressure of our atmosphere at sea-level is: • This amount of pressure is defined as 1 atmosphere (atm).

Density • A liquid is much more dense than a gas. • As a result, the effect of gravity on a liquid is much greater than that on a gas in the same sized container on the earth. • In ordinary sized containers, a gas exerts uniform pressure throughout the container it is in, but a liquid does not. • To refresh your memory, density is the mass of an object or system of particles divided by the volume it occupies. • ρ = m / V

Depth and Pressure in a Liquid • The diagram explains the relationship between pressure and depth in a liquid. • It shows a container of water with an imaginary rectangular column of water. • The column of water has a surface area A, and a weight mg. • Hence, a person holding the column of water experiences a force mg, and a pressure mg/A,which is the pressure of the water at the bottom of the container.

Demonstration

Pressure • You can derive the equation for pressure using the relationship between mass and density: • ρ = m/V => m = ρ V • V = A h => m = ρ Ah • p = F/A = mg/A = ρ Ah g / A • p = ρ g h • The total pressure of the liquid at the bottom of the container is: • p = p0 + ρ g h where p0 = normal air pressure

Density

Practice • What is the total pressure exerted on the back of a scuba diver in a lake at a depth of 8.00 m?

Total Pressure • Pressing down on the divers back is all the water above him, plus all the AIR above the water. • pwater = ρ g h = 1000kg/m3 x 9.8m/s2 x 8m • pwater = 7.84 x 104 N/m2 = 7.84 x 104 Pa • pAir = 1 Atmosphere = 101kPa • Ptotal = 7.84 x 104 Pa + 101kPa • Ptotal = 1.79 x 105 Pa

Barometers • A barometer is a device that measures atmospheric pressure. • To make a barometer at home, turn a glass upside down under water. • When you lift it straight up, it will remain filled with water. • Air pressure holdsthe water up

Barametric Pressure • At the surface of the Earth, the mercury in a barometer rises to a height of 760 mm. • This means atmospheric pressure is: • The density of mercury is taken from the density table shown previously. • Here we have introduced another unit of pressure called the torr, named after a famous scientist, and representing the pressure corresponding to 1 mm of mercury.

Pascal’s Principle • The pressure of a gas in a container is the same at every point in the gas and on the walls of the container. • Pressure in a liquid varies with depth • The diagram on the screen shows a piston that exerts a force F over an area A on a body of water. • This creates a pressure p that can be experienced throughout the fluid.

Pressure • Total pressure is the sum of the pressure from the weight of the fluid AND the added pressure from the piston

Hydraulic Lift • Two pistons are connected to a hydraulic lift. • When you press the input piston, the output piston rises. • The pressure on the input piston is the same as that on the output piston. • Keep in mind that the force is given by the pressure times the area. F = P A • The output piston has a much larger area than the input piston. • Therefore, the output piston exerts a much larger force than the input piston, and you can easily lift an automobile. F = P A

Lift a Car • p1 = p2 • F1/A1 = F0/A0 • F0 = F1 (A0 /A1 )

Unit Seven: Solids and Fluids