# Chapter 10. Solids & Liquids

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1 Chapter 10 Solids & Liquids

2 Next 6 chapters use all the concepts developed in the first 9 chapters, recasting them into a form ready to apply to specific physical systems.

3 10.1 Phases of Matter, Mass Density THREE PHASES OF MATTER Solids, Liquids, Gases Combination of Temperature and Pressure determine the phase. DEFINITION OF MASS DENSITY The mass density of a substance is the mass of a substance divided by its volume: ρ = m V SI Unit of Mass Density: kg/m 3

4 10.3 Fluids Example: Blood as a Fraction of Body Weight The body of a man whose weight is 690 N contains about 5.2x10-3 m 3 of blood. (a) Find the blood s weight and (b) express it as a percentage of the body weight. m = ρv (a) W = mg = ρvg ( )( m 3 )( 9.80m s 2 ) = 54 N = 1060kg/m 3 (b) % = 54 N 690 N 100% = 7.8%

5 10.2 Solids and Elastic Deformation Because of these atomic-level springs, a material tends to return to its initial shape once forces have been removed. FORCES ATOMS

6 10.2 Solids and Elastic Deformation STRETCHING, COMPRESSION, AND YOUNG S MODULUS F = Y ΔL L A Young s modulus has the units of pressure: N/m 2 Young s modulus is a characteristic of the material (see table 10.2) Y Steel = N/m 2

7 10.2 Solids and Elastic Deformation Spring Constants and Young s Modulus (x) Y : Young's Modulus A, L : Area and length of rod ΔL : Change in rod length (x) Young's Modulus & Spring Constants F = Y with k = = YA L ΔL L THEN A ΔL; let ΔL = x F = kx (Hooke's law) YA L (spring constant)

8 10.2 Solids and Elastic Deformation Note: 1 Pascal (Pa) = 1 N/m 2 1 GPa = N/m 2

9 10.2 Solids and Elastic Deformation In general the quantity F A is called the Stress. The change in the quantity divided by that quantity is called the Strain: ΔV V ΔL L HOOKE S LAW FOR STRESS AND STRAIN Δx L F A = Y ΔL L Stress is directly proportional to strain. Slope is Young s modulus Y. F A Strain is a unitless quantity, and SI Unit of Stress: N/m 2 ΔL L

10 10.2 Elastic Deformation Example: Bone Compression In a circus act, a performer supports the combined weight (1080 N) of a number of colleagues. Each thighbone of this performer has a length of 0.55 m and an effective cross sectional area of m 2. Determine the amount that each thighbone compresses under the extra weight. F = Y ΔL = FL YA = ΔL L A ( 540 N) ( 0.55 m) ( N m 2 ) m 2 = m = 0.041mm each leg = 1080 N 2 ( )

11 Clicker Question 10.1 A cylindrical, m rod has a diameter of 0.02 m. The rod is stretched to a length of m by a force of 3000 N. What is the Young s modulus of the material? F = Y ΔL L A a) N/m 2 b) N/m 2 c) N/m 2 d) N/m 2 e) N/m 2

12 10.2 Elastic Deformation SHEAR DEFORMATION AND THE SHEAR MODULUS L F = S Δx L A VOLUME DEFORMATION AND THE BULK MODULUS Pressure Change ΔP = B ΔV V B: Bulk modulus Table 10.2

13 Clicker Question 10.2 A cube made of brass (B = N/m 2 ) is taken by submarine from the surface where the pressure is N/m 2 to the deepest part of the ocean at a depth of m where it is exposed to a pressure is N/m 2. What is the percent change in volume as a result of this movement? ΔP = B ΔV V a) 0.413% b) 0.297% c) 0.187% d) 0.114% e) Need to know the initial size of the cube

14 10.3 Pressure P = F A Pressure = Force per unit Area The same pressure acts inward in every direction on a small volume. SI Unit of Pressure: 1 N/m 2 = 1Pa Pascal

15 10.3 Pressure Pressure is the amount of force acting on an area: Example 2 The Force on a Swimmer Suppose the pressure acting on the back of a swimmer s hand is 1.2x10 5 Pa. The surface area of the back of the hand is 8.4x10-3 m 2. (a) Determine the magnitude of the force that acts on it. (b) Discuss the direction of the force. P = F A SI unit: N/m 2 (1 Pa = 1 N/m 2 ) F = PA = ( N m )( m ) 2 = N Since the water pushes perpendicularly against the back of the hand, the force is directed downward in the drawing. Pressure on the underside of the hand is somewhat greater (greater depth). So force upward is somewhat greater - bouyancy

16 10.3 Pressure Atmospheric Pressure at Sea Level: 1.013x10 5 Pa = 1 atmosphere

17 10.3 Pressure and Depth in a Static Fluid Fluid density is ρ Equilibrium of a volume of fluid F 2 = F 1 + mg with F = PA, m = ρv P 2 A = P 1 A + ρvg with V = Ah P 2 = P 1 + ρ gh Pressure grows linearly with depth (h)

18 10.3 Pressure and Depth in a Static Fluid Example: The Swimming Hole Points A and B are located a distance of 5.50 m beneath the surface of the water. Find the pressure at each of these two locations. Atmospheric pressure P 1 = N/m 2 P 2 = P 1 + ρ gh P 2 = P 1 + ρ gh ( ) = ( Pa) + ( kg m 3 )( 9.80m s 2 ) 5.50 m = Pa

19 Clicker Question 10.3 The density of mercury is 13.6 x 10 3 kg/m 3. The pressure 100 cm below the surface of a pool of mercury is how much higher than at the surface? a)13 N/m 2 b)130 Pa c) N/m 2 d) Pa e) N/m 2 P 2 = P 1 + ρ gh

20 10.3 Pressure Gauges ρ Hg = kg m 3 P 2 = P 1 + ρ gh P 2 = ρ gh P 1 = 0 (vacuum) P atm = ρ gh h = P atm ρ g ( Pa) = ( kg m 3 ) 9.80m s 2 ( ) = m = 760 mm of Mercury

21 Clicker Question 10.4 What is the force that causes a liquid to move upward in a drinking straw a a person takes a drink? a) the force due to a low pressure generated by sucking b) the force due to the pressure within the liquid c) the force due to the atmospheric pressure d) the force due to the low pressure in the lungs e) the force due to friction on the surface of the straw

22 10.3 Pascal s Principle PASCAL S PRINCIPLE Any change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and enclosing walls. P 2 = F 2 A 2 ; P 1 = F 1 A 1 P 2 = P 1 P 2 = P 1 + ρ gh Assume weight of fluid in the tube is negligible ρ gh << P Small ratio F 2 A 2 = F 1 A 1 A F 1 = F 1 2 A 2

23 10.3 Pascal s Principle Example: A Car Lift The input piston has a radius of m and the output plunger has a radius of m. The combined weight of the car and the plunger is N. Suppose that the input piston has a negligible weight and the bottom surfaces of the piston and plunger are at the same level. What is the required input force? A F 1 = F 1 2 A 2 ( ) π ( m ) 2 = 131 N π ( m) 2 = N r 1 = m r 2 = 0.150m

24 10.4 Archimedes Principle Buoyant Force F B = F 2 + ( F 1 ) F 1 F 2 = P 2 A P 1 A = ( P 2 P ) 1 A = ρ gha = ρv g mass of displaced fluid since P2 = P1 + ρ gh and V = ha Buoyant force = Weight of displaced fluid

25 10.4 Archimedes Principle ARCHIMEDES PRINCIPLE Any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the buoyant force equals the weight of the fluid that the object displaces: CORROLARY If an object is floating then the magnitude of the buoyant force is equal to the magnitude of its weight.

26 10.4 Archimedes Principle Example: A Swimming Raft The raft is made of solid square pinewood. Determine whether the raft floats in water and if so, how much of the raft is beneath the surface.

27 10.4 Archimedes Principle W raft = m raft g = ρ pine V raft g = ( 550kg m 3 )( 4.8m 3 )( 9.80m s 2 ) = N Raft properties V raft = ( 4.0) ( 4.0) ( 0.30)m 3 = 4.8 m 3 ρ pine = 550 kg/m 3 If W raft < F B max, raft floats F B max = W fluid (full volume) F B max = ρvg = ρ water V water g = ( 1000kg m 3 )( 4.8m 3 )( 9.80m s 2 ) = N W raft < F B max Raft floats Part of the raft is above water

28 10.4 Archimedes Principle How much of raft below water? Floating object F B = W raft F B = ρ water gv water = ρ water g( A water h) h = = W raft ρ water ga water 26000N ( 1000kg m 3 ) 9.80m s 2 = 0.17 m W raft = N ( )( 16.0 m 2 )

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