1. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 1 PHYS 1443 – Section 001 Lecture #17 Thursday, July 7, 2011 Dr. Jaehoon Yu Equilibrium Problems Elastic Property of Solids Fluid and Pressure Pascal’s Principle Absolute and Relative Pressure Flow Rate and Equation of Continuity Bernoulli’s Principle

2. Thursday, July 7, 2011 2 Announcements Please bring your planetarium special credit sheet during the intermission Final Comprehensive Exam –8 – 10am, Monday, July 11 in SH103 –Covers CH1.1 through what we finish today –Mixture of multiple choice and free response problems PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu

3. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 3 More on Conditions for Equilibrium To simplify the problem, we will only deal with forces acting on x-y plane, giving torque only along z-axis. What do you think the conditions for equilibrium be in this case? The six possible equations from the two vector equations turns to three equations. What happens if there are many forces exerting on an object? O F1F1 F4F4 F3F3 F2F2 F5F5 r5r5 O’ r’ If an object is at its translational static equilibrium, and if the net torque acting on the object is 0 about one axis, the net torque must be 0 about any arbitrary axis. Why is this true? Because the object is not moving moving, no matter what the rotational axis is, there should not be any motion. It is simply a matter of mathematical manipulation. AND

4. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 4 How do we solve equilibrium problems? 1.Identify all the forces and their directions and locations 2.Draw a free-body diagram with forces indicated on it with their directions and locations properly noted 3.Write down force equation for each x and y component with proper signs 4.Select a rotational axis for torque calculations  Selecting the axis such that the torque of one of the unknown forces become 0 makes the problem easier to solve 5.Write down the torque equation with proper signs 6.Solve the equations for unknown quantities

5. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 5 Ex. 12.4 for Mechanical Equilibrium A person holds a 50.0N sphere in his hand. The forearm is horizontal. The biceps muscle is attached 3.00 cm from the joint, and the sphere is 35.0cm from the joint. Find the upward force exerted by the biceps on the forearm and the downward force exerted by the upper arm on the forearm and acting at the joint. Neglect the weight of forearm. Since the system is in equilibrium, from the translational equilibrium condition From the rotational equilibrium condition O FBFB FUFU mg d l Thus, the force exerted by the biceps muscle is Force exerted by the upper arm is

6. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 6 Example 12 – 6 A 5.0 m long ladder leans against a wall at a point 4.0m above the ground. The ladder is uniform and has mass 12.0kg. Assuming the wall is frictionless (but ground is not), determine the forces exerted on the ladder by the ground and the wall. FBD First the translational equilibrium, using components Thus, the y component of the force by the ground is mgmg FWFW F Gx F Gy O The length x0 x0 is, from Pythagorian theorem

7. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 7 Example 12 – 6 cont’d From the rotational equilibrium Thus the force exerted on the ladder by the wall is Thus the force exerted on the ladder by the ground is The x component of the force by the ground is Solve for F Gx The angle between the ground force to the floor

8. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 8 Elastic Properties of Solids We have been assuming that the objects do not change their shapes when external forces are exerting on it. It this realistic? No. In reality, objects get deformed as external forces act on it, though the internal forces resist the deformation as it takes place. Deformation of solids can be understood in terms of Stress and Strain Stress : The amount of the deformation force per unit area the object is subjected Strain : The measure of the degree of deformation It is empirically known that for small stresses, strain is proportional to stress The constants of proportionality are called Elastic Modulus Three types of Elastic Modulus 1.Young’s modulus : Measure of the elasticity in a length 2.Shear modulus : Measure of the elasticity in an area 3.Bulk modulus : Measure of the elasticity in a volume

9. Elastic Limit and Ultimate Strength Elastic limit: The limit of elasticity beyond which an object cannot recover its original shape or the maximum stress that can be applied to the substance before it becomes permanently deformed Ultimate strength: The maximum force that can be applied on the object before breaking it Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 9

10. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 10 Young’s Modulus Let’s consider a long bar with cross sectional area A and initial length Li.Li. F ex =F in Young’s Modulus is defined as What is the unit of Young’s Modulus? Experimental Observations 1.For a fixed external force, the change in length is proportional to the original length 2.The necessary force to produce the given strain is proportional to the cross sectional area LiLi A:cross sectional area Tensile stress L f =L i +ΔL F ex After the stretch F ex F in Tensile strain Force per unit area Used to characterize a rod or wire stressed under tension or compression

11. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 11 Bulk Modulus Bulk Modulus characterizes the response of a substance to uniform squeezing or reduction of pressure. Bulk Modulus is defined as Volume stress =pressure After the pressure change If the pressure on an object changes by ΔP=ΔF/A, the object will undergo a volume change ΔV. V V’ F F F F Compressibility is the reciprocal of Bulk Modulus Because the change of volume is reverse to change of pressure.

12. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 12 Elastic Moduli and Ultimate Strengths of Materials

13. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 13 Example for Solid’s Elastic Property A solid brass sphere is initially under normal atmospheric pressure of 1.0x10 5 N/m 2. The sphere is lowered into the ocean to a depth at which the pressures is 2.0x10 7 N/m 2. The volume of the sphere in air is 0.5m 3. By how much its volume change once the sphere is submerged? The pressure change Δ P is Since bulk modulus is The amount of volume change is From table 12.1, bulk modulus of brass is 8.0x10 10 N/m 2 Therefore the resulting volume change Δ V is The volume has decreased.

14. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 14 Fluid and Pressure What are the three states of matter?Solid, Liquid and Gas Fluid cannot exert shearing or tensile stress. Thus, the only force the fluid exerts on an object immersed in it is the force perpendicular to the surface of the object. How do you distinguish them? Using the time it takes for a particular substance to change its shape in reaction to external forces. What is a fluid? A collection of molecules that are randomly arranged and loosely bound by forces between them or by an external container. We will first learn about mechanics of fluid at rest, fluid statics. In what ways do you think fluid exerts stress on the object submerged in it? This force by the fluid on an object usually is expressed in the form of the force per unit area at the given depth, the pressure, defined as Note that pressure is a scalar quantity because it’s the magnitude of the force on a surface area A. What is the unit and the dimension of pressure? Expression of pressure for an infinitesimal area dA by the force dF is Unit:N/m 2 Dim.: [M][L -1 ][T -2 ] Special SI unit for pressure is Pascal

15. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 15 Example for Pressure The mattress of a water bed is 2.00m long by 2.00m wide and 30.0cm deep. a) Find the weight of the water in the mattress. The volume density of water at the normal condition (4 o C and 1 atm) is 1000kg/m 3. So the total mass of the water in the mattress is Since the surface area of the mattress is 4.00 m 2, the pressure exerted on the floor is Therefore the weight of the water in the mattress is b) Find the pressure exerted by the water on the floor when the bed rests in its normal position, assuming the entire lower surface of the mattress makes contact with the floor.

16. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 16 Variation of Pressure and Depth Water pressure increases as a function of depth, and the air pressure decreases as a function of altitude. Why? If the liquid in the cylinder is the same substance as the fluid, the mass of the liquid in the cylinder is It seems that the pressure has a lot to do with the total mass of the fluid above the object that puts weight on the object. Let’s imagine the liquid contained in a cylinder with height h and the cross sectional area A immersed in a fluid of density  at rest, as shown in the figure, and the system is in its equilibrium. The pressure at the depth h below the surface of the fluid open to the atmosphere is greater than the atmospheric pressure by  gh. Therefore, we obtain Atmospheric pressure P0 P0 is P0AP0A PA MgMg h Since the system is in its equilibrium

17. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 17 Pascal’s Principle and Hydraulics A change in the pressure applied to a fluid is transmitted undiminished to every point of the fluid and to the walls of the container. The resultant pressure P at any given depth h increases as much as the change in P0.P0. This is the principle behind hydraulic pressure. How? Therefore, the resultant force F2 F2 is What happens if P 0 is changed? Since the pressure change caused by the force F1 F1 applied onto the area A1 A1 is transmitted to the F2 F2 on an area A2.A2. This seems to violate some kind of conservation law, doesn’t it? d1d1 d2d2 F1F1 A1A1 A2A2 F2F2 In other words, the force gets multiplied by the ratio of the areas A 2 /A 1 and is transmitted to the force F2 F2 on the surface. No, the actual displaced volume of the fluid is the same. And the work done by the forces are still the same.

18. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 18 Example for Pascal’s Principle In a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section and a radius of 5.00cm. This pressure is transmitted by a liquid to a piston that has a radius of 15.0cm. What force must the compressed air exert to lift a car weighing 13,300N? What air pressure produces this force? Using the Pascal’s principle, one can deduce the relationship between the forces, the force exerted by the compressed air is Therefore the necessary pressure of the compressed air is

19. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 19 Example for Pascal’s Principle Estimate the force exerted on your eardrum due to the water above when you are swimming at the bottom of the pool with a depth 5.0 m. Assume the surface area of the eardrum is 1.0cm 2. We first need to find out the pressure difference that is being exerted on the eardrum. Then estimate the area of the eardrum to find out the force exerted on the eardrum. Since the outward pressure in the middle of the eardrum is the same as normal air pressure Estimating the surface area of the eardrum at 1.0cm 2 =1.0x10 -4 m 2, we obtain

20. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 20 H dy y h Example for Pascal’s Principle Water is filled to a height H behind a dam of width w. Determine the resultant force exerted by the water on the dam. Since the water pressure varies as a function of depth, we will have to do some calculus to figure out the total force. Therefore the total force exerted by the water on the dam is The pressure at the depth h is The infinitesimal force dF exerting on a small strip of dam dy is

21. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 21 Absolute and Relative Pressure How can one measure pressure? One can measure the pressure using an open-tube manometer, where one end is connected to the system with unknown pressure P and the other open to air with pressure P0.P0. This is called the absolute pressure, because it is the actual value of the system’s pressure. In many cases we measure the pressure difference with respect to the atmospheric pressure to avoid the effect of the changes in P 0 that depends on the environment. This is called gauge or relative pressure. The common barometer which consists of a mercury column with one end closed at vacuum and the other open to the atmosphere was invented by Evangelista Torricelli. Since the closed end is at vacuum, it does not exert any force. 1 atm of air pressure pushes mercury up 76cm. So 1 atm is The measured pressure of the system is h P P0P0 If one measures the tire pressure with a gauge at 220kPa the actual pressure is 101kPa+220kPa=303kPa.

22. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 22 Finger Holds Water in Straw You insert a straw of length L into a tall glass of your favorite beverage. You place your finger over the top of the straw so that no air can get in or out, and then lift the straw from the liquid. You find that the straw strains the liquid such that the distance from the bottom of your finger to the top of the liquid is h. Does the air in the space between your finger and the top of the liquid in the straw have a pressure P that is (a) greater than, (b) equal to, or (c) less than, the atmospheric pressure PA PA outside the straw? What are the forces in this problem? Gravitational force on the mass of the liquid mg Force exerted on the top surface of the liquid by inside air pressure p in A Force exerted on the bottom surface of the liquid by the outside air Since it is at equilibrium pAApAA Cancel A and solve for p in Less So p in is less than PA PA by ρg(L-h).

23. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 23 Flow Rate and the Equation of Continuity Study of fluid in motion: Fluid Dynamics If the fluid is water: Streamline or Laminar flow : Each particle of the fluid follows a smooth path, a streamline Turbulent flow : Erratic, small, whirlpool-like circles called eddy current or eddies which absorbs a lot of energy Two primary types of flows Water dynamics?? Hydro-dynamics Flow rate: the mass of fluid that passes the given point per unit time since the total flow must be conserved Equation of Continuity

24. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 24 Ex. 13 – 14 for Equation of Continuity How large must a heating duct be if air moving at 3.0m/s through it can replenish the air in a room of 300m 3 volume every 15 minutes? Assume the air’s density remains constant. Using equation of continuity Since the air density is constant Now let’s imagine the room as the large section of the duct

25. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 25 Bernoulli’s Principle Bernoulli’s Principle: Where the velocity of fluid is high, the pressure is low, and where the velocity is low, the pressure is high. Amount of the work done by the force, F 1, that exerts pressure, P 1, at point 1 Work done by the gravitational force to move the fluid mass, m, from y1 y1 to y2 y2 is Amount of the work done by the force in the other section of the fluid is

26. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 26 Bernoulli’s Equation cont’d The total amount of the work done on the fluid is From the work-energy principle Since the mass m is contained in the volume that flowed in the motion and Thus,

27. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 27 Bernoulli’s Equation cont’d We obtain Re- organize Bernoulli’s Equation Since Thus, for any two points in the flow For static fluid For the same heights The pressure at the faster section of the fluid is smaller than slower section. Pascal’s Law Result of Energy conservation!

28. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 28 Ex. 13 – 15 for Bernoulli’s Equation Water circulates throughout a house in a hot-water heating system. If the water is pumped at the speed of 0.5m/s through a 4.0cm diameter pipe in the basement under a pressure of 3.0atm, what will be the flow speed and pressure in a 2.6cm diameter pipe on the second 5.0m above? Assume the pipes do not divide into branches. Using the equation of continuity, flow speed on the second floor is Using Bernoulli’s equation, the pressure in the pipe on the second floor is

29. Congratulations!!!! I certainly had a lot of fun with ya’ll and am truly proud of you! You all are impressive and have done very well!!! Good luck with your exam!!! Have a safe and fruitful summer!! Thursday, July 7, 2011 29 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu

30. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 30 Vibration or Oscillation What are the things that vibrate/oscillate? A periodic motion that repeats over the same path. A simplest case is a block attached at the end of a coil spring. Tuning fork A pendulum A car going over a bump Buildings and bridges The spider web with a prey So what is a vibration or oscillation?

31. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 31 Simple Harmonic Motion Motion that occurs by the force that depends on displacement, and the force is always directed toward the system’s equilibrium position. When a spring is stretched from its equilibrium position by a length x, the force acting on the mass is What is a system that has such characteristics? A system consists of a mass and a spring This is a second order differential equation that can be solved but it is beyond the scope of this class. It’s negative, because the force resists against the change of length, directed toward the equilibrium position. From Newton’s second law we obtain What do you observe from this equation? Acceleration is proportional to displacement from the equilibrium. Acceleration is opposite direction to displacement. This system is doing a simple harmonic motion (SHM). Condition for simple harmonic motion

32. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 32 Equation of Simple Harmonic Motion The solution for the 2 nd order differential equation What happens when t=0 and  =0? Let’s think about the meaning of this equation of motion What are the maximum/minimum possible values of x? Amplitude Phase Angular Frequency Phase constant What is  if x is not A at t=0? A/-A An oscillation is fully characterized by its: Amplitude Period or frequency Phase constant Generalized expression of a simple harmonic motion

33. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 33 Vibration or Oscillation Properties The maximum displacement from the equilibrium is The complete to-and-fro motion from an initial point One cycle of the oscillation Amplitude Period of the motion, T The time it takes to complete one full cycle Frequency of the motion, f The number of complete cycles per second Unit? s -1 Unit? sec Relationship between period and frequency? or

34. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 34 More on Equation of Simple Harmonic Motion Let’s now think about the object’s speed and acceleration. Since after a full cycle the position must be the same Speed at any given time The period What is the time for full cycle of oscillation? One of the properties of an oscillatory motion Frequency How many full cycles of oscillation does this undergo per unit time? What is the unit? 1/s=Hz Max speed Max acceleration Acceleration at any given time What do we learn about acceleration? Acceleration is reverse direction to displacement Acceleration and speed are  /2 off phase of each other: When v is maximum, a is at its minimum

35. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 35 Simple Harmonic Motion continued Let’s determine phase constant and amplitude By taking the ratio, one can obtain the phase constant Phase constant determines the starting position of a simple harmonic motion. This constant is important when there are more than one harmonic oscillation involved in the motion and to determine the overall effect of the composite motion At t=0 At t=0 By squaring the two equations and adding them together, one can obtain the amplitude

36. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 36 More Simple Block-Spring System Special case #1 How do the period and frequency of this harmonic motion look? Since the angular frequency  is Let’s consider that the spring is stretched to a distance A, and the block is let go from rest, giving 0 initial speed; x i =A, v i =0, The period, T, becomes So the frequency is Frequency and period do not depend on amplitude Period is inversely proportional to spring constant and proportional to mass This equation of motion satisfies all the conditions. So it is the solution for this motion. What can we learn from these? Special case #2 Suppose a block is given non-zero initial velocity vi vi to positive x at the instant it is at the equilibrium, x i =0 Is this a good solution?

37. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 37 Example for Spring Block System A car with a mass of 1300kg is constructed so that its frame is supported by four springs. Each spring has a force constant of 20,000N/m. If two people riding in the car have a combined mass of 160kg, find the frequency of vibration of the car after it is driven over a pothole in the road. Let’s assume that mass is evenly distributed to all four springs. Thus the frequency for vibration of each spring is The total mass of the system is 1460kg. Therefore each spring supports 365kg each. From the frequency relationship based on Hook’s law How long does it take for the car to complete two full vibrations? The period isFor two cycles

38. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 38 Example for Spring Block System A block with a mass of 200g is connected to a light spring for which the force constant is 5.00 N/m and is free to oscillate on a horizontal, frictionless surface. The block is displaced 5.00 cm from equilibrium and released from reset. Find the period of its motion. From the Hook’s law, we obtain From the general expression of the simple harmonic motion, the speed is X=0 X=0.05 As we know, period does not depend on the amplitude or phase constant of the oscillation, therefore the period, T, is simply Determine the maximum speed of the block.

39. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 39 Energy of the Simple Harmonic Oscillator What do you think the mechanical energy of the harmonic oscillator look without friction? Kinetic energy of a harmonic oscillator is The elastic potential energy stored in the spring Therefore the total mechanical energy of the harmonic oscillator is Since Total mechanical energy of a simple harmonic oscillator is proportional to the square of the amplitude.

40. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 40 Energy of the Simple Harmonic Oscillator cont’d Maximum KE is when PE=0 The speed at any given point of the oscillation x A-A KE/PE E=KE+PE=kA 2 /2 Maximum speed

41. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 41 Oscillation Properties Amplitude?A When is the force greatest? When is the speed greatest? When is the acceleration greatest? When is the potential energy greatest? When is the kinetic energy greatest?

42. Thursday, July 7, 2011PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu 42 Example for Energy of Simple Harmonic Oscillator A 0.500kg cube connected to a light spring for which the force constant is 20.0 N/m oscillates on a horizontal, frictionless track. a) Calculate the total energy of the system and the maximum speed of the cube if the amplitude of the motion is 3.00 cm. The total energy of the cube is From the problem statement, A and k are Maximum speed occurs when kinetic energy is the same as the total energy b) What is the velocity of the cube when the displacement is 2.00 cm. velocity at any given displacement is c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. Kinetic energy, KE Potential energy, PE