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Solids and fluids. State of matter. Solid ~Definite shape ~Definite volume Liquid No definite shape ~Definite volume Gas No definite shape No definite volume Plasma Dark matter ? Dark energy?. Deformation of Solids.
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Solids and fluids PHY231
State of matter • Solid • ~Definite shape • ~Definite volume • Liquid • No definite shape • ~Definite volume • Gas • No definite shape • No definite volume • Plasma • Dark matter ? • Dark energy? PHY231
Deformation of Solids • Large forces can permanently deform or break a solid object (e.g. car after a crash) • In most cases, the object returns to its original shape after the force is removed • Elastic behavior • Elastic properties of solids • Stress: Force per unit area causing deformation • Strain: Measure of the amount of deformation • Stress is proportional to strain PHY231
Elastic deformation of solids • Many types of deformations are possible. In any case, there is an elastic regime where the stress (deforming force/unit area) is proportional to the strain (~deformation) • For each type of deformation there is a specific elastic modulus. Elastic modulus are also different for different solids • Three primary types of deformation and modulus • Young’s modulus (Y): Elasticity in length • Shear modulus (S): Elasticity of shape • Bulk modulus (B): Elasticity of Volume
Large Y means object is hard to elongate/shorten Young’s modulus (Y) • Elasticity in length • Solid deforms because of the external perpendicular force F (tensile force) applied on its cross section A • Internal force increases until it balances F. At equilibrium one quantifies • Tensile stress = F/A • Tensile strain = DL/L0 • During Elastic regime [Y]=N/m2=Pa PHY231
Mountaineer • For safety in climbing, a mountaineer uses a nylon rope 50 m long and 1.0 cm in diameter. • When supporting a 90-kg climber, the rope elongates by 1.6 m. • What is its Young’s modulus ? PHY231
1 Gpa = 109 Pa = 109 N/m2 Swing • The swing in a child play yard uses steel wires that have ultimate strength (=maximum stress) of 0.50 GPa. For safety, a maximum load of 200 kg is foreseen. • What should be the minimum diameter of the steel wires? • Assume the swing has two wires under tensile stress. PHY231
Shear modulus (S) • Elasticity of shape • Shear stress • One face held fixed • Parallel force applied to opposite face • No change in volume • Shear stress = F/A • Shear strain = Dx/h [S]=N/m2=Pa Large S means object is hard to bend PHY231
example • A 80 kg man stands at the very end of a planck 2.0cm thick and 15.0 cm wide. The planck is 2.0 m long and held firmly at the other end. If the plank bows down by 5.0 cm, what is the shear modulus of the wood? PHY231
Bulk modulus (B) • Volume elasticity • Solid subjects to uniform force per unit area • Example: Object immersed in a fluid • Volume stress = DP= DF/A • = Change in Pressure • Volume strain = DV/V [B] =N/m2=Pa • Large B means object is hard to compress/expand PHY231
Canon ball • A lead canon ball was fired from a pirate ship but missed its target. It sinks to the bottom of the ocean. By what fraction will the volume of the canon ball shrink if the pressure increases by 20 MPa and the bulk modulus of lead is 40 Gpa ? PHY231
Density • For object with uniform composition and Mass M and Volume V SI unit [r] = kg/m3 • Solid and liquid are typically 1000 times denser than gas • Atoms are ~ 10 times closer in solids and liquids than in gas PHY231
Metal spheres • Two metal spheres (A and B) are of same volume but B is 3 times denser than A. Which one is heavier? • A • B • Same mass A B PHY231
Metal spheres • Two metal spheres (A and B) are of same volume but B is 3 times denser than A (rB=3rA). Which one is heavier? • A • B • Same mass A B PHY231
A cube of liquid water of M = 1.10 kg is held in a cylindrical plastic box having 6.00 cm for radius and 10.0 cm height. • The water is cooled until it is completely transformed into ice. Is the box large enough to contain the ice cube? • Yes • No H=10 cm R= 6.0 cm PHY231
Yes • No H=10 cm R= 6.0 cm PHY231
Pressure • F is the magnitude of a force exerted perpendicular to a given surface of area A • The pressure P is the ratio force over surface SI unit [P] = N/m2 = Pa (Pascal) US 1 psi = 6894.757… Pa ~ 6.89 kPa Deformation of solids depends on F/A , so the larger the area where the force is applied the smaller the risk to exceed the ultimate strength of an object On the other hand, if you want a large pressure you must apply a force over a small area. That’s why nails for example can go through materials.
Man standing • A man is standing on a hard floor, his mass is 90 kg and his feet are 30 cm long and 10 cm wide. • Assuming is weight is distributed evenly over the contact surface between his feet and the floor, what is the pressure he exerts on the floor? PHY231
Vacuum and pressure • The ambient air-pressure is surprisingly strong. Imagine you remove all the air of a cylindrical container with a lid of Alid=100 cm2 surface. • The surrounding air pushes and seals the container. If the ambient air pressure is 105 Pa, how much force is needed (on each side) to open the container? Pout=0.10 MPa Pin=0.0 Pa PHY231
Pressure in a static fluid (=hydrostatic pressure) • Assume a fluid in static equilibrium in a container Horizontal equilibrium Vertical equilibrium All points at same depth have same pressure Pressure increase with depth by rgh
Bottom of the ocean • Depth of the oceans reaches more than 10,000 m at some locations. Assuming we carry the previous container with a 100 cm2 lid with vacuum inside at the bottom of the ocean. • What would be the force due to the water pressure on the lid? (rwater=1000 kg/m3) PHY231
We found a force of 106 N on the lid. It’s equivalent to have 100,000kg = 100 tons on the lid. • A H3 hummer weights 6000 lbs or ~ 2700 kg. • 100 tons is equivalent to ~36 H3 piled up on the lid !!
Pressure with depth Shape of the container doesn’t matter, only depth Type of fluid matters PHY231
Pascal’s principle • A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container • Neglecting the height difference between left and right pistons we have: PHY231
Hydraulic press • What force F is needed to keep the H3 in equilibrium? Neglect height difference and use A1=20cm2, A2=200cm2, MH3=2700 kg 2.6 kN 0.44 kN 26 kN x1 x
Pressure measurements • Barometer • Measures P0 • Open tube manometer • Measures P of a fluid Absolute pressure = P Gauge pressure = P-P0
Buoyant force: • An object in a fluid is buoyed by a force B with magnitude equal to the weight of the fluid displaced by the object =mfg robject=rfluid robject<rfluid robject>rfluid Same mass of fluid displaced = Same B in all cases
Floating object • A floating object of mass mo=roVo is in equilibrium and has displaced a volume of fluid Vf mo mf Vf equilibrium The floating object has displaced a mass of fluid equal to its own total mass
A sphere of 5 cm radius is submerged into water. What will be the buoyant force once the object is totally submerged into water? B mw and Vw are mass and volume of water displaced
An object is weighted and found to have T1=mog=10 N. It is then totally submerged into water and reweighted, the scale indicates T2=5.0 N • What is the density of the object compared to water? Mw and Vw are mass and volume of water displaced. Volume of water displaced is equal to volume of the object Vo T1 B T2 -m0 g -m0 g PHY231
Weight in water • An object is weighted and found to have T1=mog=10 N. It is then totally submerged into water and reweighted, the scale indicates T2=8.0 N • What is the density of the object? • A) 2rw • B) 3rw • C) 4rw • D) 5rw
Weight in water • An object is weighted and found to have T1=mog=10 N. It is then totally submerged into water and reweighted, the scale indicates T2=8.0 N • What is the density of the object? • A) 2rw • B) 3rw • C) 4rw • D) 5rw
Boat (homework) Pond without anything, volume V0 In each case you need to understand what is the water supporting (boat, man etc…) through the buoyant force before and after an action. PHY231
B B Vboatimmersed Vboatimmersed -manchor g -mboat g -mboat g Vanchorimmersed -manchor g Water supports boat and anchor • 1) Lower anchor but not touching the bottom of the pond • Water still supports boat and anchor. • Total volume of water displaced will still be the same. • Height of the water won’t change
B1 B2 -manchor g -mboat g -mboat g N = manchor g -manchor g Water supports boat and anchor 2) Lower anchor and touch bottom of the pond - Water only supports boat, anchor supported by reaction force at the bottom - Total volume of water displaced will be smaller, height of water goes down.
Vglass of water VD1 VD2 Vglass of water 3) Take a glass of water and drink it while you are on the boat - Water must support boat and water drunk - The boat sinks so that the additional volume of water displaced is equal to the volume of water drunk - Level of pond stays the same Water supports boat
Fluids in motion • We will first consider an ideal fluid • Nonviscous (no internal friction) • Incompressible • Steady motion (time independent) • No turbulence PHY231
!! V=Volume v=velocity Equation of continuity • Consider the steady flow of a fluid in a pipe • During time Dt, the amount of fluid Dm1 entering is equal to the amount leaving Dm2 Continuity equation • Tells you that: • If pipe gets smaller, velocity increases • If pipe gets larger, velocity decreases
Bernouilli’s equation • Consequence of energy conservation for an ideal fluid PHY231
Constricted pipe • The inside diameters D1 of the larger portions of a horizontal pipe shown below are 2.5 cm. Water flows to the right at a rate of FR=1.8*10-4 m3/s. What is the diameter D2 of the constriction at the center of the pipe ? H2=5.0cm H2=10cm D1 D2 FR PHY231
H2=5.0cm H2=10cm D1 D2 FR
Water tank P0 • If h=0.500 m and Y=3.00 m. At what distance X does the water hits the floor? • (assume water falls down very slowly at the top of the tank) v1~0 v2 y1 P0 Y y2 X PHY231
Viscous flow • So far we have considered fluids without viscosity (no internal friction in the fluid) • Non-viscous fluid: • Velocity uniform across the pipe • Viscous fluid: • Velocity maximum at the center ofthe pipe PHY231
Poiseuille’s law • Considering a pipe of radius R and length L with pressures P1 and P2 at its ends (and P1>P2). • For a viscous fluid, the rate of flow is inversely proportional to the viscosity h PHY231
IV bag • You are asked to set-up an IV bag at the hospital to inject 500 cm3 of a glucose solution into the arm of a patient in 30 minutes. • You are given a needle of length L=2.5 cm and diameter d=0.41 mm • Assuming the patient blood pressure is 105 Pa, at what height h should you hang the IV bag above the patient’s arm? (neglect the transition between the tube and the needle) • rsolution = 1.0 kg/m3 • hsolution = 1.0 mPa.s • Patm=105 Pa
P0 P1 P2 PHY231