1. MSU Physics 231 Fall 2015 1 Physics 231 Topic 10: Solids & Fluids Alex Brown Nov 4-9 2015

2. MSU Physics 231 Fall 2015 2 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

3. MSU Physics 231 Fall 2015 3 States of matter

4. MSU Physics 231 Fall 2015 4 Phase Transformations

5. MSU Physics 231 Fall 2015 5 Solids amorphous ordered crystalline

6. MSU Physics 231 Fall 2015 6 The Deformation of Solids 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. Strain = Constant  Stress Constant: elastic modulus The elastic modulus depends on: Material that is deformed Type of deformation (a different modulus is defined for different types of deformations) strain stress elastic regime elastic limit inelastic regime breaking point

7. MSU Physics 231 Fall 2015 7 The Young’s modulus (tensile) Beyond the elastic limit an object is permanently deformed (it does not return to its original shape if the stress is removed).

8. MSU Physics 231 Fall 2015 8 The Shear Modulus xx

9. MSU Physics 231 Fall 2015 9 Bulk Modulus Compressibility: 1/(Bulk modulus)

10. MSU Physics 231 Fall 2015 10 Some Elastic Moduli MaterialYoung’s Modulus (N/m 2 ) Shear Modulus (N/m 2 ) Bulk Modulus (N/m 2 ) Steel20x10 10 8.4x10 10 16x10 10 Bone1.8x10 10 8.0x10 10 - Aluminum7x10 10 2.5x10 10 7x10 10 Water0.21x10 10

11. MSU Physics 231 Fall 2015 11 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 1.0x10 4 kg. He chooses concrete for the material of the pillar (Y=1.0x10 10 Pa). How much does the pillar compress? 5m R pillar = 0.5m L 0 = 5m Y = 1.0x10 10 Pa M statue = 1.0x10 4 kg  L = 6.2x10 -5 m

12. MSU Physics 231 Fall 2015 12 Another Example (tensile) A builder is stacking 1 m 3 cubic concrete blocks. Each block weighs 5x10 3 kg. The ultimate strength of concrete for compression is 2x10 7 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 (m block g) n = total number of blocks m block = mass of one block = 5x10 3 kg g = 9.8 m/s 2 Ultimate strength: 2x10 7 = F/A = n (m block g)/(1 m 2 ) n = 408 blocks.

13. MSU Physics 231 Fall 2015 13 Moving Earth’s Crust (shear) 30 m 100 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 = 1.5x10 10 Pa) F/A = S(  x/h) = 4.5x10 9 Pa

14. MSU Physics 231 Fall 2015 14 An Example (bulk) What force per unit area needs to be applied to compress 1 m 3 water by 1%? (B=0.21x10 10 Pa)  V/V 0 = -0.01 so,  F/A = 2.1x10 7 Pa !!!

15. MSU Physics 231 Fall 2015 15 Density density specific density materialdensity (kg/m 3 )Specific density (kg/m 3 ) water1.00x10 3 1.00 oxygen1.430.00143 lead11.3x10 3 11.3

16. MSU Physics 231 Fall 2015 16 Pressure Pressure = P = F/A (N/m 2 =Pa) Same Force, different pressure

17. MSU Physics 231 Fall 2015 17 An Example A nail is driven into a piece of wood with a force of 700N. What is the pressure on the wood if A nail = 1 mm 2 ? 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? P nail = F/A nail = 700N/1x10 -6 m 2 = 7x10 8 Pa P person = F/(1000A nail ) = 700/(1000x10 -6 1E-6) = 7x10 5 Pa (about 7 times the atmospheric pressure)

18. 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. P 1 /P 2 )? a)1/8 b)¼ c)½ d)1 e)2

19. MSU Physics 231 Fall 2015 19 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. P 1 /P 2 )? a)1/8 b)¼ c)½ d)1 e)2 P 1 = F 1 /A 1 = Mg/A P 2 = F 2 /A 2 = 2Mg/(0.25A) = 8 Mg/A P 1 /P 2 = 1/8

20. MSU Physics 231 Fall 2015 20 Force and Pressure A P=0 vacuum F Air (P a =1.0x10 5 Pa) FpFp Force due to pressure difference: F =  P A If A=0.010 m 2 then a force F p = (1.0x10 5 ) x 0.010 = 1000 N (225 pounds) is needed to pull the lid. What is the force needed to move the lid?

21. MSU Physics 231 Fall 2015 21 Magdeburg’s Hemispheres Otto von Guericke (Mayor of Magdeburg, 17th century)

22. MSU Physics 231 Fall 2015 22 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.

23. MSU Physics 231 Fall 2015 23 Pascal’s principle P = F 1 /A 1 = F 2 /A 2 If A 2 >>A 1 then F 2 >>F 1. Hydraulic Lift If we apply a small force F 1, we can exert a very large Force F 2.

24. MSU Physics 231 Fall 2015 24 Pressure & Depth (column of water in water) Horizontal direction: P 1 =F 1 /A P 2 =F 2 /A F 1 =F 2 (no net force) So, P 1 =P 2 Forces in the up direction F t = - P t A (top) F b = -Mg + P b A = -  g A h + P b A (bottom) Since the column of water is not moving: F t + F b = 0 - P t A -  g Ah + P b A = 0 P b = P t +  g h water F1F1 F2F2 top PtAPtA PbAPbA W=Mg P t =P atm atmospheric, A: surface area, M: mass bottom

25. MSU Physics 231 Fall 2015 25 Pressure and Depth: P h = P 0 +  g h Where: P h = the pressure at depth h P 0 = the pressure at depth 0  = density of the liquid g = 9.8 m/s 2 h = depth P 0 = P atm = 1.013x10 5 Pa = 1 atm =760 Torr From Pascal’s principle: If P 0 changes then the pressures at all depths changes with the same value.

26. MSU Physics 231 Fall 2015 26 A Submarine A submarine is built in such a way that it can stand pressures of up to 3x10 6 Pa (approx 30 times the atmospheric pressure). How deep can it go? P h = P 0 +  g h 30x10 5 = (1.0x10 5 ) + (1.0x10 3 )(9.81)h h=296 m

27. MSU Physics 231 Fall 2015 27 Does the shape of the container matter? NO!!

28. MSU Physics 231 Fall 2015 28 Pressures at same heights are the same P0P0 P=P 0 +  gh h P0P0 h h

29. MSU Physics 231 Fall 2015 29 Pressure Measurement: The mercury barometer P 0 =  m g h  m =  mercury = 13.6x10 3 kg/m 3

30. MSU Physics 231 Fall 2015 30 Buoyant force: B htht hbhb P t = P 0 +  g h t (top) P b = P 0 +  g h b (bottom) P b - P t =  g (h b -h t ) =  g  h B = F b – F t = (P b – P t ) A =  g  h A =  g V = M f g (weight of displaced fluid) W = M o g (weight of object) If the object is not moving: B = W so: M o = M f (for a submerged object) P0P0 Archimedes (287 BC) principle: the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object B W

31. MSU Physics 231 Fall 2015 31 Comparing densities B =  g V buoyant force W= M o g =  o g V weight of object  o =  W = B stationary  o >  W > B object goes down  o <  W < B object goes up B W

32. MSU Physics 231 Fall 2015 32 A floating object (B=W) h A B W W = M o g (weight of object) B = weight of the fluid with density  displaced by the object = M d g =  V d g Thus V d = M o /  volume displaced For a block with top area A and h = height of the object under water, then V d = A h h = M o / (  A) = (  o V o ) / (  A)

33. MSU Physics 231 Fall 2015 33 P=P 0 +  g h h = distance between liquid surface and the point where you measure P P0P0 P h B =  V o g = M f 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 =  Pressure at depth h Buoyant force for submerged object Buoyant force for floating object h B 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. V d = M o /  (any object) h =  o V o /(  A) = M o /(  A) (for a block area A) W=mg

34. MSU Physics 231 Fall 2015 34 quiz 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 =  V o g W = M g B = W lower density liquid: W remains the same, B becomes smaller the block will sink to the bottom B < W W=Mg

35. MSU Physics 231 Fall 2015 35 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 = M o /(  A) = 2.0/[1.0x10 3 (2)(0.5)] = 0.002 m = 2 mm B) if the objects sinks the mattress is just completely submerged: h = thickness of mattress. h = M o /(  A) 0.08 = (M weight +2.0)/[1.0x10 3 (2)(0.5)] so M weight =78 kg

36. MSU Physics 231 Fall 2015 36 T1T1 T2T2 T 1 = M o g =  o Vg T 2 = M o g – M f g =  o Vg -  Vg (T 2 /T 1 ) = R = 1 - (  /  o )  =  o (1-R) (  o /  ) = 1/(1-R) = specific density when the fluid is water  =  w  o = density of object  = density of fluid V = volume of object

37. MSU Physics 231 Fall 2015 37 Density of a liquid An object with a density of 2395 kg/m 3 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? T 1 = M o g = (0.0194)(9.81) = 0.190 T 2 = 0.128 R = T 2 /T 1 = 0.674  =  o (1-R) = (2395) (1-0.674) = 781 kg/m 3

38. MSU Physics 231 Fall 2015 38 Tricky… h l

39. MSU Physics 231 Fall 2015 39 Without the balloon: B = W =  V d g V d = volume of water displaced With the balloon B + F b = W The balloon is trying to pull the boat out of the water.  V d g = W - F b V d = (W - F b )/(  g) If F b = 0 then V d increases (the boat sinks deeper) As a result, the water level rises. W a) BF b =F balloon

40. MSU Physics 231 Fall 2015 40 b) water thrown out (B = W) V d = M o /  volume displaced Initially there is water in the boat: V d = (m w + m boat ) /  = (m w /  ) + (m boat /  ) = V water thrown out + (m boat /  ) When the water is thrown out of the boat: V d = m boat /  (same Eq. As above but with m w = 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

41. MSU Physics 231 Fall 2015 41 c)Anchor thrown out (W = B) V d = M o /  volume displaced So V d = (m boat + m anchor ) /  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? V d decreases – the water level falls.

42. MSU Physics 231 Fall 2015 42 Fluid flow - equation of continuity A1,A1, A2,A2, v1v1 v2v2 1 2 the mass flowing into area 1 (  M 1 ) must be the same as the mass flowing into area 2 (  M 2 ), else mass would accumulate in the pipe. (liquid is incompressible:  1 =  2 =  )  M 1 =  M 2 (M =  V =  Ax)  A 1  x 1 =  A 2  x 2 (  x = v  t) A 1 v 1  t = A 2 v 2  t Av = constant A 1 v 1 = A 2 v 2 x1x1 x2x2

43. MSU Physics 231 Fall 2015 43 Bernoulli’s equation P: pressure ½  v 2 : kinetic Energy per unit volume  gy: potential energy per unit volume P 1,A 1,y 1 P 2,A 2,y 2 v1v1 v2v2 P 1 + ½  v 1 2 +  g y 1 = P 2 + ½  v 2 2 +  g y 2 P + ½  v 2 +  g y = constant P: pressure y: height  : density g: gravitational v: velocity acceleration

44. MSU Physics 231 Fall 2015 44 (1) P + ½  v 2 +  g y = constant (2) A v = constant equal A.Incompressible fluid, so density is constant B. (2) A A >A B so v A <v B (1) v A P B greater C. Must be the same, else the liquid would get ‘stuck’ between A and B, o equal less than See B. (1) Since y = constant P + ½  v 2 = constant

45. MSU Physics 231 Fall 2015 45 Moving cans P0P0 P0P0 Top view 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 = P o P Bernoulli’s law: P + ½  v 2 = P 0 so P = P 0 - ½  v 2 So P < P 0 Because of the pressure difference left and right of each can, they move inward When air is blown in between the cans, the velocity is not equal to 0. (ignore the y part) case: 1: no blowing 2: blowing

46. MSU Physics 231 Fall 2015 46 P h = P 0 +  gh h x y If h=1m and y=3m what is x? Assume that the holes are small and the water level doesn’t drop noticeably. P0P0 A Hole in a Tank P h +  g y = P 0 +  g (y+h)

47. MSU Physics 231 Fall 2015 47 h x1x1 y If h=1 m and y=3 m what is x? Use Bernoulli’s law P A + ½  v A 2 +  gy A = P B + ½  v B 2 +  gy B at A: P A = P 0 v A =? y A = y at B: P B = P 0 v B =0 y B = y + h v A 2 = 2gh v A = 4.43 m/s A P0P0 B

48. MSU Physics 231 Fall 2015 48 x1x1 3m vAvA In the horizontal direction: x(t) = x 0 + v 0x t + ½at 2 = 0 + 4.43t + 0 In the vertical direction: y(t) = y 0 + v 0y t + ½at 2 = 3 - 0.5gt 2 = 0 when the water hits the ground, so t = 0.78 s so x(t=0.78) = (4.43)(0.78) = 3.45 m 0

49. MSU Physics 231 Fall 2015 49 PHY 231 49 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.

50. MSU Physics 231 Fall 2015 50 Viscosity Contact surface A fixed moving F=  Av/d  =coefficient of viscosity unit: Ns/m 2 or poise=0.1 Ns/m 2 F v d FluidT ( o C) Viscosity  (Ns/m 2 ) water201.0x10 -3 blood372.7x10 -3 oil30250x10 -3

51. MSU Physics 231 Fall 2015 51 Poiseuille’s Law (not covered in homework or exams) How fast does a fluid flow through a tube? Rate of flow Q=  v/  t=  R 4 (P 1 -P 2 ) 8L8L (unit: m 3 /s) R vP1P1 P2P2 L

52. MSU Physics 231 Fall 2015 52 Example P=10 6 Pa P=10 5 Pa Flow rate Q Tube length: L=3 m  =1.50 Ns/m 2 What should the radius of the tube be to allow for Q = 0.5 m 3 /s ? Rate of flow Q=  R 4 (P 1 -P 2 ) 8L8L R=[ (8Q  L) /  (P 1 -P 2 ) ] 1/4 = 0.05 m

53. MSU Physics 231 Fall 2015 53 Quiz A object with weight of W (in N) is resting on a table with K legs each having a contact surface S (m 2 ) 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)(W+V)/S b)W/S c)(W+V)/(KS) d)W/(KS) e)(W+V)g/S with g=9.81 m/s 2

54. MSU Physics 231 Fall 2015 54 Clicker Question A object with weight of W (in N) is resting on a table with K legs each having a contact surface S (m 2 ) 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)(W+V)/S b)W/S c)(W+V)/(KS) d)W/(KS) e)(W+V)g/S with g=9.81 m/s 2 P=F/A F=V+W A=KS

55. MSU Physics 231 Fall 2015 55 Measurement of Pressure The open-tube manometer. The pressure at A and B is the same: P A = P B P = P 0 +  gh so h = (P-P 0 )/(  g) If the pressure P = 1.1 atm, what is h? (the liquid is water) h = (1.10-1.00)x10 5 /(1.0x10 3 x9.81) = 1.0 m