# Lecture 8 Equilibrium and Elasticity

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1 Lecture 8 Equilibrium and Elasticity

2 July 19 EQUILIBRIUM AND ELASTICITY CHAPTER 12 Give a sharp blow one end of a stick on the table. Find center of percussion. Baseball bat center of percussion Equilibrium demos? Pile of tiles Ladder with weight. Stretching and compressing metals. FLUIDS IN MOTION CHAPTER 14 Bathroom scale atmosphere crusher Pascal Vases Venturi tube Crush the soda can Magdeburg hemispheres Archimedes principle weigh cylinder in water Heat up water and watch floating object sink as density increases. Cartesian Diver Weigh thumb in water Buoyancy of air Weight of Air Surface tension and adhesive forces Blow a few bubbles on overhead projector. Note as bubble heats up, pressure increases inside bubble, and the radius of the bubble decreases. Air flow around objects

3 Equilibrium Equilibrium P com = constant, L com = constant Static Equilibrium constant =0 Stable static equilibrium Unstable static equilibrium

4 Balance of net Forces and net Torques F x = 0 F y = 0 F z = 0 τ x = 0 τ y = 0 τ z = 0 Gravity acts on a single point on a body called the center of gravity. If g is the same for every point on the body, then cog =com.

5 m= 1.8 kg, L=1 meter M=2.7 kg at L/4 from the left end. What do the scales read? Normal Force=F l and F r Free body diagram y-dir. F l + F r Mg mg = 0 Set net torque = 0 about right end (0)(F r ) + (3L / 4)Mg) + (L / 2)(mg) (L)(F l ) = 0 F l = 0.75(26.46) + 0.5(17.64) = 28.67N F r = Mg + mg F l F r = N

6 m= 1.8 kg, L=1 meter M=2.7 kg at L/4 from the left end. What do the scales read? y-dir. F l + F r Mg mg = 0 Set net torque = 0 about right end Free body diagram (0)(F r ) + (3L / 4)Mg) + (L / 2)(mg) (L)(F l ) = 0 F l = 0.75(26.46) + 0.5(17.64) = 28.67N mg = (1.8)9.8 = Left and right read = mg 2 Mg = (2.7)(9.8) = N = 8.82 N due to m. Left end reads.75(26.46) = N due to M F l =3/4 Mg +1/2 mg= N F r = Right end =mg+mg-f l = =15.43 N

7 Free body diagram Given L = 12m M = 72kg m = 45kg h = 9.3m Find F W F py F px

8 Now use F net,x =0 and F net,y =0 F w F px = 0 F px = 407N F py Mg mg = 0 F py = Mg + mg = ( )9.8 = N Take moments about which axis since net torque is 0 Choose O eliminate two variables (h)(f W ) + (a / 2)(Mg) + (a / 3)mg + (0)(F px ) + (0)(F py ) = 0 a = L 2 h 2 = = 7.58m F w = ga(m / 2 + m / 3) / h F w = 9.8(7.58)(72 / / 3) / 9.3 = 407N

9 Strain Elasticity Shear Compression

10 Strain Consider a long wire with a mass hanging off the end of it. A long wire stretches a distance ΔL. Divide by the length of the wire L. Call it the strain = ΔL/L. Imagine all the atoms connected by springs with a force F=k s s. Then the stretching is equivalent to all the springs stretching a little bit.

11 Strain = Stress = ΔL / L F T / A Young s Modulus is the ratio of stress/strain Y = F T / A ΔL / L

12

13 Various materials subjected to a stretching and compression test demo Material Youngs Modulus E 10 9 N/m 2 Steel Aluminum Ultimate Strength S 10 6 N/m 2 glass c concrete c wood c bone 9 c 170 c plastic 3 48

14 Sample problem Suppose we have a steel ball of mass 10 kg hanging from a steel wire 3 m long and 3mm in diameter. Young s modulus is 2 x N/m 2. How much does the wire stretch? ΔL / L = F T / A Y F T = 9.8(10) = 98N A = 3.14(0.003)(0.003) = m 2 ΔL = L Y F T / A = (98 / ) = m

15 Microscopic picture of Young s Modulus For one atom the atomic stress is k s s/d 2 and the strain for one atom is s/d where d is the size of the atom and s is the amount the atom is stretched. Y is the ratio of stress/ strain. Then Y = k s s / d 2 s / d = k s d for one atom. d d+s d

16 Chapter 14 Fluids (Liquids and Gases) What is a fluid Density, pressure, of gases and liquids Static fluids p =p 0 + gh Mercury barometer,open tube manometer Pascal s principle Archimedes principle Surface tension and adhesive forces Fluids in motion Equation of continuity Bernoulli s Equation about soap bubbbles

17 Fluids: liquids and gases Fluids conform to the container that holds them and we normally talk about their density and pressure Liquids are Incompressible Gases are very compressible Uniform Density =m/v kg/m 3 Pressure (Static) P=F/A N/m 2 or Pa Scalar- independent of orientation Atmospheric pressure 1atm 1.01 x 10 5 N/m mm of Hg or Torr 14.7 lb/in 2

18 What are all the forces acting on this book? Put a book on the table.

19 Static Pressure What is the force acting downward at level 1? at level 2? Level 1 F = Weight of Atmosphere Level 2 F = Weight of Atmosphere + Weight of water The pressure is just the weight of the water plus the weight of the air per square meter F 2 = F 1 + mg AP = AP 0 + ρhag P = F A P = P 0 + ρgh

20 What is the absolute and gauge pressure at a 2 meter depth in the water? First find the atmospheric pressure P 0. P = P 0 + ρgh h = 2m ρ = 1000kg / m 3 g = 9.8m / s 2 P 0 = Pa Or N/m 2 P = ( ) + (1000)(9.8)(2) P = Pa P P 0 = Pa (Gauge Pressure)

21 What is a Mercury Barometer? F = P 0 A This is the upward acting force due to the pressure at the bottom of the tube. This will support an amount of mercury equal to the weight haρg So P 0 A = haρg or P 0 = hρg Solving for h, We get or h = P 0 ρg h = = 0.760m = 760mm (13550)(9.8) How high would a column of water go?

22 Pascal Vases Shows that the pressure in a liquid depends on the height of the liquid, not it's volume

23 Pascal s principle: A change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container Change in pressure = ΔP = f a = F A (1) D V f =V F V = AD = da, (2) D = d a A d a A = f F From (1) Work done FD = fd The work done by both systems is the same. The advantage is that a given force over a long distance can be transformed to a larger force over a shorter distance.

24 Cartesian Diver To illustrate the transmission of pressure in a liquid.

25 Archimedes principle The stone displaces a volume of water equal to its volume. The upwardly directed buoyant force is equal to the weight of the displaced fluid or water. This is the surprising thing. Floating: If the buoyant force is equal to the weight mg of the object, Then it is said to float. If it sinks then you can define its apparent weight. apparent weight = actual weight - buoyant force Archimedes Demo Example 26 chpt 14 ed 7

26 Weight of Thumb Stick thumb in the water and weigh it. Remove thumb and stick it in the water and weigh it again. Can you account for the readings.

27 Archimedes principle To show that the buoyant force exerted on an object is equal to the weight of the displaced water (or fluid).

28 Air effects Buoyancy of air Weight of Air

29 Laminar Flow Steady flow Incompressible flow Non-viscous flow Irrotational flow An example of viscous flow

30 Equation of continuity: A 1 v 1 = A 2 v 2 = Volume flow rate Mass flow rate= density x Volume flow rate. ρav = constant Show water stream narrowing in diameter as it falls.

31 Bernoulli s Equation Conservation of energy in a volume of fluid p ρv ρgy 1 = p ρv ρgy 2 p ρv2 + ρgy = a constant along a streamline Special case y =0 p ρv2 = a constant along a streamline

32 p ρv2 = a constant along a streamline

33 Venturi Tube and Qualitative Verification of Bernoulli's Equation To demonstrate operation of a venturi tube and to verify variation of gas velocity with area of tube. Show different pressures in flow tube by flowing air through system and noting pressure relationships by the level of colored liquid in the tubes.

34 Air Flow Around Different Objects Curve ball may be demonstrated by using the thrower and Styrofoam ball. The ball curves in the direction of the leading edge spin. Ping-pong ball may be supported on an air jet held vertical and also may be held in a funnel with an air jet. The cardboard cylinder is wrapped with a long length of rubber band and pulled back. It is released in such a manner that it spins out from the bottom. If it is released with enough energy, it will describe a loop in the air. The plexiglass plates can be held together with quite an impressive force by blowing air through the nozzle and bringing the plates together.

35 Water tank problem What is the speed of the water through the narrow hole of area a? The water tank as large area A. Apply Bernoulli s equation to the top of the water level where y=h and at the hole y= 0 where the water is ejected. p ρv 02 + ρgh = p ρv2 + ρg(0) Cancel out p 0 1 v 0 = a 2 ρ(v2 v 2 A v 0 ) = ρgh From the continuity eq. 1 2 v2 (1 a 2 A 2 ) = ρgh Since A/a >>1, neglect (a/a) 2 compared to ρv2 = ρgh v = 2gh

36 Chapter 14 Ed 7 Problem 26E Problem 36 Ejected ball from water Problem 20 Sucking up lemonade

37 ConcepTest 11.1 Balancing Rod A 1 kg ball is hung at the end of a rod 1 m long. If the system balances at a point on the rod 0.25 m from the end holding the mass, what is the mass of the rod? 1kg 1m

38 ConcepTest 11.2 Mobile A (static) mobile hangs as shown below. The rods are massless and have lengths as indicated. The mass of the ball at the bottom right is 1 kg. What is the total mass of the mobile?? 1 m 2 m? 1 kg 1 m 3 m

39 ConcepTest 11.3a Tipping Over I A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a blue dot in each case. In which case(s) does the box tip over? 1 2 3

40 ConcepTest 11.3b Tipping Over II Consider the two configurations of books shown below. Which of the following is true? 1 2 1/2 1/4 1/2 1/4

41 ConcepTest 11.1 Balancing Rod A 1 kg ball is hung at the end of a rod 1 m long. If the system balances at a point on the rod 0.25 m from the end holding the mass, what is the mass of the rod? 1 kg same distance m ROD = 1 kg X CM of rod

42 ConcepTest 11.2 Mobile A (static) mobile hangs as shown below. The rods are massless and have lengths as indicated. The mass of the ball at the bottom right is 1 kg. What is the total mass of the mobile?? 1 m 2 m? 1 kg 1 m 3 m

43 ConcepTest 11.3a Tipping Over I A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a blue dot in each case. In which case(s) does the box tip over? 1 2 3

44 ConcepTest 11.3b Tipping Over II Consider the two configurations of books shown below. Which of the following is true? 1 2 1/2 1/4 1/2 1/4

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