Physics 231 Topic 10: Solids & Fluids Alex Brown Nov 4-9 2015 MSU Physics 231 Fall 2015 1 Key Concepts: Solids & Fluids States of Matter Density and matter states Solids and Elasticity Young’s & bulk moduli Fluid Pressure & Motion Continuity equation Bernoulli’s equation Buoyancy & Archimedes Principle Surface Tension and Viscosity Poiseuille’s law Covers chapter 10 in Rex & Wolfson MSU Physics 231 Fall 2015 2 States of matter Solid Liquid Gas Plasma difficult to deform easy to deform easy to deform easy to deform difficult to compress difficult to compress easy to compress easy to compress difficult to flow easy to flow easy to flow easy to flow not charged not charged not charged charged MSU Physics 231 Fall 2015 3 Phase Transformations MSU Physics 231 Fall 2015 4 Solids amorphous ordered MSU Physics 231 Fall 2015 crystalline 5 The Deformation of Solids Strain = ConstantStress Constant: elastic modulus stress Stress: Related to the force causing the deformation: Force per unit area causing the deformation Strain: Measure of the degree of deformation Measure of the amount of deformation For small stress, strain and stress are linearly correlated. inelastic regime breaking point elastic limit elastic regime The elastic modulus depends on: strain • Material that is deformed • Type of deformation (a different modulus is defined for different types of deformations) MSU Physics 231 Fall 2015 6 The Young’s modulus (tensile) L L0 tensile stress Y tensile strain 2 tensile stress : F/A [N/m Pascal (Pa)] tensile strain : L/L0 [ Unitless! ] F L0 F/A Y L / L0 A L (L is positive) Beyond the elastic limit an object is permanently deformed (it does not return to its original shape if the stress is removed). MSU Physics 231 Fall 2015 7 The Shear Modulus x shear stress S shear strain shear stress : F/A [N/m 2 Pascal (Pa)] shear strain : x/h [ Unitless! ] F/A Fh S x / h Ax MSU Physics 231 Fall 2015 8 Bulk Modulus V0 V0 V volume stress B volume strain volume stress : F/A [N/m 2 Pascal (Pa)] volume strain : V/V0 F / A P B V / V0 V / V0 (V is negative) P F / A pressure Compressibility: 1/(Bulk modulus) MSU Physics 231 Fall 2015 9 Some Elastic Moduli Material Young’s Modulus (N/m2) Shear Modulus (N/m2) Bulk Modulus (N/m2) Steel 20x1010 8.4x1010 16x1010 Bone 1.8x1010 8.0x1010 - Aluminum 7x1010 2.5x1010 7x1010 0.21x1010 Water MSU Physics 231 Fall 2015 10 An Example (tensile) An architect wants to design a 5m high circular pillar with a radius of 0.5 m that holds a bronze statue that weighs 1.0x104 kg. He chooses concrete for the material of the pillar (Y=1.0x1010 Pa). How much does the pillar compress? F/A Y L / L0 5m 2 M statue g /( R pillar ) L / L0 M statue gL0 L 2 YR pillar Rpillar = 0.5m Y = 1.0x1010 Pa L0 = 5m Mstatue = 1.0x104 kg L = 6.2x10-5 m MSU Physics 231 Fall 2015 11 Another Example (tensile) A builder is stacking 1 m3 cubic concrete blocks. Each block weighs 5x103 kg. The ultimate strength of concrete for compression is 2x107 Pa. How many blocks can he stack before the lowest block is crushed? The force on the low end of the lowest block is: F = n (mblockg) n = total number of blocks mblock = mass of one block = 5x103 kg g = 9.8 m/s2 Ultimate strength: 2x107= F/A = n (mblockg)/(1 m2) n = 408 blocks. MSU Physics 231 Fall 2015 12 Moving Earth’s Crust (shear) 100 m 30 m A tectonic plate in the lower crust (100 m deep) of the earth is shifted during an earthquake by 30m. What is the shear stress involved, if the upper layer of the earth does not move? (shear modulus S = 1.5x1010 Pa) shear stress F / A S shear strain x / h F/A = S(x/h) = 4.5x109 Pa MSU Physics 231 Fall 2015 13 An Example (bulk) What force per unit area needs to be applied to compress 1 m3 water by 1%? (B=0.21x1010 Pa) F / A B V / V0 V/V0 = -0.01 so, F/A = 2.1x107 Pa !!! MSU Physics 231 Fall 2015 14 Density M V density specific density 3 (kg / m ) specific material / water ( 4 o C) density (kg/m3) Specific density (kg/m3) water 1.00x103 1.00 oxygen 1.43 0.00143 lead 11.3x103 11.3 material MSU Physics 231 Fall 2015 15 Pressure Pressure = P = F/A (N/m2=Pa) Same Force, different pressure MSU Physics 231 Fall 2015 16 An Example A nail is driven into a piece of wood with a force of 700N. What is the pressure on the wood if Anail = 1 mm2? A person (weighing 700 N) is lying on a bed of such nails (his body covers 1000 nails). What is the pressure exerted by each of the nails? Pnail = F/Anail = 700N/1x10-6m2 = 7x108 Pa Pperson = F/(1000Anail) = 700/(1000x10-6 1E-6) = 7x105 Pa (about 7 times the atmospheric pressure) MSU Physics 231 Fall 2015 17 Clicker Quiz! A block (1) of mass M is lying on the floor. The contact surface between the block and the floor is A. A second block (2), of mass 2M but with a contact surface of only 0.25A is also placed on the floor. What is the ratio of the pressure exerted on the floor by block 1 to the pressure exerted on the floor by block 2 (I.e. P1/P2)? a) b) c) d) e) 1/8 ¼ ½ 1 2 MSU Physics 231 Fall 2015 18 Clicker Quiz! A block (1) of mass M is lying on the floor. The contact surface between the block and the floor is A. A second block (2), of mass 2M but with a contact surface of only 0.25A is also placed on the floor. What is the ratio of the pressure exerted on the floor by block 1 to the pressure exerted on the floor by block 2 (I.e. P1/P2)? a) b) c) d) e) 1/8 ¼ ½ 1 2 P1= F1/A1 = Mg/A P2= F2/A2 = 2Mg/(0.25A) = 8 Mg/A P1/P2 = 1/8 MSU Physics 231 Fall 2015 19 Force and Pressure Air (Pa=1.0x105 Pa) A P=0 vacuum F Fp What is the force needed to move the lid? Force due to pressure difference: F = P A If A=0.010 m2 then a force Fp = (1.0x105) x 0.010 = 1000 N (225 pounds) is needed to pull the lid. MSU Physics 231 Fall 2015 20 Magdeburg’s Hemispheres Otto von Guericke (Mayor of Magdeburg, 17th century) MSU Physics 231 Fall 2015 21 Pascal’s principle A change in pressure applied to a fluid that is enclosed is transmitted to the whole fluid and all the walls of the container that hold the fluid. MSU Physics 231 Fall 2015 22 Hydraulic Lift Pascal’s principle If we apply a small force F1, we can exert a very large Force F2. P = F1/A1 = F2/A2 If A2>>A1 then F2>>F1. MSU Physics 231 Fall 2015 23 Pressure & Depth (column of water in water) water F1 top F2 PtA Horizontal direction: P1=F1/A P2=F2/A F1=F2 (no net force) So, P1=P2 Pt=Patm atmospheric, A: surface area, M: mass Forces in the up direction Ft = - PtA (top) Fb = -Mg + PbA = - g A h + PbA (bottom) bottom W=Mg PbA Since the column of water is not moving: Ft + Fb = 0 - PtA - g Ah + PbA = 0 Pb = Pt + g h MSU Physics 231 Fall 2015 24 Pressure and Depth: Ph = P0 + g h Where: Ph = the pressure at depth h P0 = the pressure at depth 0 = density of the liquid g = 9.8 m/s2 h = depth P0 = Patm = 1.013x105 Pa = 1 atm =760 Torr From Pascal’s principle: If P0 changes then the pressures at all depths changes with the same value. MSU Physics 231 Fall 2015 25 A Submarine A submarine is built in such a way that it can stand pressures of up to 3x106 Pa (approx 30 times the atmospheric pressure). How deep can it go? Ph = P0+ g h 30x105 = (1.0x105) + (1.0x103)(9.81)h h=296 m MSU Physics 231 Fall 2015 26 Does the shape of the container matter? NO!! MSU Physics 231 Fall 2015 27 Pressures at same heights are the same P0 P0 h P=P0+gh h h P=P0+gh MSU Physics 231 Fall 2015 P=P0+gh 28 Pressure Measurement: The mercury barometer P0 = m g h m = mercury = 13.6x103 kg/m3 MSU Physics 231 Fall 2015 29 Buoyant force: B P0 ht B hb W Pt = P0 + g ht (top) Pb = P0 + g hb (bottom) Pb - Pt = g (hb-ht) = g h B = Fb – Ft = (Pb – Pt) A = g h A = g V = Mf g (weight of displaced fluid) W = Mo g (weight of object) If the object is not moving: B = W so: Mo= Mf (for a submerged object) Archimedes (287 BC) principle: the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object MSU Physics 231 Fall 2015 30 Comparing densities B = g V B W W = M o g = o g V buoyant force weight of object o = W=B stationary o > W>B object goes down o < W<B object goes up MSU Physics 231 Fall 2015 31 A floating object (B=W) A B W h W = Mo g (weight of object) B = weight of the fluid with density displaced by the object = Md g = Vd g Thus Vd = Mo / volume displaced For a block with top area A and h = height of the object under water, then Vd = A h h = Mo / ( A) = (oVo ) / ( A) MSU Physics 231 Fall 2015 32 P0 Pressure at depth h P = P0+ g h h = distance between liquid surface and the point where you measure P h P Buoyant force for submerged object B = Vo g = Mf g The buoyant force equals the weight of the amount of fluid that can be put in the volume taken by the object. If object is not moving: B = W o = Buoyant force for floating object The buoyant force equals the weight of the amount of fluid that can be put in the part of the volume of the object that is under water. Vd = Mo/ (any object) h = oVo/( A) = Mo/( A) (for a block area A) MSU Physics 231 Fall 2015 W=mg B h W=mg 33 quiz W=Mg A block of weight Mg is placed in water and found to stay submerged as shown in the picture. The water is then replaced by another liquid of lower density. What will happen if the block is placed in the liquid of lower density? a) the block will float on the surface of the liquid b) the block will be partially submerged and partially above the liquid c) the block will again be submerged as shown in the picture d) the block will sink to the bottom initially B = Vo g W= Mg B=W lower density liquid: W remains the same, B becomes smaller the block will sink to the bottom B < W MSU Physics 231 Fall 2015 34 Problem An air mattress 2m long 0.5m wide and 0.08m thick and has a mass of 2.0 kg. A) How deep will it sink in water? B) How much weight can you put on top of the mattress before it sinks? A) h = Mo/( A) = 2.0/[1.0x103 (2)(0.5)] = 0.002 m = 2 mm B) if the objects sinks the mattress is just completely submerged: h = thickness of mattress. h = Mo/( A) 0.08 = (Mweight+2.0)/[1.0x103(2)(0.5)] so Mweight=78 kg MSU Physics 231 Fall 2015 35 T1 T2 o = density of object = density of fluid V = volume of object T1 = Mog = oVg T2 = Mog – Mfg = oVg - Vg (T2 /T1 ) = R = 1 - ( /o ) = o (1-R) (o / ) = 1/(1-R) = specific density when the fluid is water = w MSU Physics 231 Fall 2015 36 Density of a liquid An object with a density of 2395 kg/m3 and mass of 0.0194 kg is hung from a scale and submerged in a liquid. The weight read from the scale is 0.128 N. What is the density of the liquid? T1 = Mog = (0.0194)(9.81) = 0.190 T2 = 0.128 R = T2 /T1 = 0.674 = o (1-R) = (2395) (1-0.674) = 781 kg/m3 MSU Physics 231 Fall 2015 37 Tricky… h l MSU Physics 231 Fall 2015 38 a) Fb=Fballoon Without the balloon: B = W = Vd g Vd = volume of water displaced B W With the balloon B + Fb= W The balloon is trying to pull the boat out of the water. Vd g = W - Fb Vd = (W - Fb)/( g) If Fb = 0 then Vd increases (the boat sinks deeper) As a result, the water level rises. MSU Physics 231 Fall 2015 39 b) water thrown out (B = W) Vd = Mo / volume displaced Initially there is water in the boat: Vd = (mw + mboat) / = (mw/ ) + (mboat/) = Vwater thrown out + (mboat/ ) When the water is thrown out of the boat: Vd = mboat/ (same Eq. As above but with mw = 0) So by throwing the water, the displaced volume reduces by the volume of the water thrown out of the boat. The boat should rise and the water level go down BUT the volume of water is thrown back into the lake so it is filled up again! Answer: water level is unchanged MSU Physics 231 Fall 2015 40 c) Anchor thrown out (W = B) Vd = Mo / volume displaced So Vd = (mboat + manchor) / If the anchor is thrown overboard the gravitational force is still acting on it and nothing chances. If the anchor hits the ground, what happens? Vd decreases – the water level falls. MSU Physics 231 Fall 2015 41 Fluid flow - equation of continuity x2 x1 v1 1 A1, 2 v2 A2, the mass flowing into area 1 (M1) must be the same as the mass flowing into area 2 (M2), else mass would accumulate in the pipe. (liquid is incompressible: 1 = 2 = ) M1= M2 (M = V = Ax) A1 x1 = A2 x2 (x = v t) A1v1 t = A2v2 t A1v1 = A2v2 Av = constant MSU Physics 231 Fall 2015 42 Bernoulli’s equation P2,A2,y2 v2 P1 + ½ v12 + g y1= P2 + ½ v22 + g y2 P1,A1,y1 P + ½ v2 + g y = constant v1 P: pressure : density v: velocity y: height g: gravitational acceleration P: pressure ½v2: kinetic Energy per unit volume gy: potential energy per unit volume MSU Physics 231 Fall 2015 43 (1) P + ½ v2 + g y = constant (2) A v = constant (1) Since y = constant P + ½ v2 = constant A. Incompressible fluid, so density is constant equal B. (2) AA>AB so vA<vB greater (1) vA < vB so PA > PB equal less than C. Must be the same, else the liquid would get ‘stuck’ between A and B, o See B. MSU Physics 231 Fall 2015 44 Moving cans P0 Top view P case: 1: no blowing 2: blowing P0 Before air is blown in between the cans, the cans remain at rest and the air in between the cans is at rest (0 velocity) P = Po When air is blown in between the cans, the velocity is not equal to 0. (ignore the y part) Bernoulli’s law: P + ½v2 = P0 so P = P0 - ½v2 So P < P0 Because of the pressure difference left and right of each can, they move inward MSU Physics 231 Fall 2015 45 A Hole in a Tank P0 y Ph = P0 + gh h Ph + g y = P0+ g (y+h) If h=1m and y=3m what is x? Assume that the holes are small and the water level doesn’t drop noticeably. x MSU Physics 231 Fall 2015 46 If h=1 m and y=3 m what is x? P0 y B h Use Bernoulli’s law A PA + ½vA2 + gyA= PB + ½vB2 + gyB at A: PA = P0 vA=? yA = y at B: PB = P0 vB=0 yB = y + h vA2 = 2gh vA = 4.43 m/s x1 MSU Physics 231 Fall 2015 47 vA In the horizontal direction: x(t) = x0 + v0xt + ½at2 = 0 + 4.43t + 0 3m 0 x1 In the vertical direction: y(t) = y0 + v0yt + ½at2 = 3 - 0.5gt2 = 0 when the water hits the ground, so t = 0.78 s so x(t=0.78) = (4.43)(0.78) = 3.45 m MSU Physics 231 Fall 2015 48 Viscosity Viscosity: stickiness of a fluid One layer of fluid feels a large resistive force when sliding along another one or along a surface of for example a tube. PHY 231 MSU Physics 231 Fall 2015 49 49 Viscosity Contact surface A F v moving F=Av/d d =coefficient of viscosity unit: Ns/m2 or poise=0.1 Ns/m2 fixed T (oC) Fluid water blood oil 20 37 30 MSU Physics 231 Fall 2015 Viscosity (Ns/m2) 1.0x10-3 2.7x10-3 250x10-3 50 Poiseuille’s Law (not covered in homework or exams) P1 R v P2 How fast does a fluid flow through a tube? L Rate of flow Q= v/t= R4(P1-P2) 8L MSU Physics 231 Fall 2015 (unit: m3/s) 51 Example Flow rate Q Tube length: L=3 m =1.50 Ns/m2 P=105 Pa P=106 Pa What should the radius of the tube be to allow for Q = 0.5 m3/s ? Rate of flow Q= R4(P1-P2) 8L R=[ (8Q L) / (P1-P2) ]1/4 = 0.05 m MSU Physics 231 Fall 2015 52 Quiz A object with weight of W (in N) is resting on a table with K legs each having a contact surface S (m2) with the floor. The weight of the table is V (in N). The pressure P exerted by each of the legs on the floor is: a) b) c) d) e) (W+V)/S W/S (W+V)/(KS) W/(KS) (W+V)g/S with g=9.81 m/s2 MSU Physics 231 Fall 2015 53 Clicker Question A object with weight of W (in N) is resting on a table with K legs each having a contact surface S (m2) with the floor. The weight of the table is V (in N). The pressure P exerted by each of the legs on the floor is: a) b) c) d) e) (W+V)/S W/S (W+V)/(KS) W/(KS) (W+V)g/S with g=9.81 m/s2 P=F/A F=V+W A=KS MSU Physics 231 Fall 2015 54 Measurement of Pressure The open-tube manometer. The pressure at A and B is the same: PA = PB P = P0 + gh so h = (P-P0)/(g) If the pressure P = 1.1 atm, what is h? (the liquid is water) h = (1.10-1.00)x105/(1.0x103x9.81) = 1.0 m MSU Physics 231 Fall 2015 55