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UT Arlington PHYS 1443 - Lecture Notes

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Slide 1AnnouncementsMore on Conditions for EquilibriumHow do we solve equilibrium problems?Example 12 – 6Example 12 – 6 cont’dElastic Properties of SolidsElastic Limit and Ultimate StrengthYoung’s ModulusBulk ModulusElastic Moduli and Ultimate Strengths of MaterialsExample for Solid’s Elastic PropertyDensity and Specific GravityFluid and PressureExample for PressureVariation of Pressure and DepthPascal’s Principle and HydraulicsExample for Pascal’s PrincipleAbsolute and Relative PressureThursday, July 7, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu1PHYS 1443 – Section 001Lecture #17Thursday, July 7, 2011Dr. Jaehoon Yu•Equilibrium Problems•Elastic Property of Solids•Fluid and Pressure•Pascal’s Principle•Absolute and Relative Pressure•Flow Rate and Equation of ContinuityThursday, July 7, 2011 2Announcements•Final Comprehensive Exam–8 – 10am, Monday, July 11 in SH103–Covers CH1.1 through what we finish today (CH13.6) plus Appendices A and B–Mixture of multiple choice and free response problems–Please do not miss the exam!! •Please bring your planetarium special credit sheet during the intermissionPHYS 1443-001, Summer 2011 Dr. Jaehoon YuThursday, July 7, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu3More on Conditions for EquilibriumTo 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 are 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?0F =�ur0t =�r0xF0zOr5O’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.0yFWhy is this true?Because the object is not moving in the first place, no matter where the rotational axis is, there should not be any motion. This simply is a matter of mathematical manipulation.ANDThursday, July 7, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu4How do we solve equilibrium problems?1. Identify all the forces and their directions and locations2. Draw a free-body diagram with forces indicated on it with their directions and locations properly noted3. Write down the force equation for each x and y component with proper signs4. 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 solve5. Write down the torque equation with proper signs6. Solve the equations for unknown quantitiesThursday, July 7, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu5Example 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. xFFBDFirst the translational equilibrium, using componentsThus, the y component of the force by the ground ismgFWFGxFGyOGyFGx WF F= -0yFGymg F=- +0mg12.0 9.8 118N N= � =The length x0 is, from Pythagorian theorem2 205.0 4.0 3.0x m= - =Thursday, July 7, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu6Example 12 – 6 cont’dOFrom the rotational equilibrium02 4.0Wmg x F=- +0Thus the force exerted on the ladder by the wall isWFThus the force exerted on the ladder by the ground isThe x component of the force by the ground is44Gx WF F N= =GF024.0mg x=118 1.5444.0N�= =0x Gx WF F F= - =�Solve for FGx2 2Gx GyF F= +2 244 118 130N= + �The angle between the ground force to the floorq1tanGyGxFF-� �=� �� �1118tan 7044-� �= =� �� �oThursday, July 7, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu7Elastic Properties of SolidsElastic Modulus �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 underStrain: The measure of the degree of deformationIt is empirically known that for small stresses, strain is proportional to stressThe constants of proportionality are called Elastic ModulusThree types of Elastic Modulus1. Young’s modulus: Measure of the elasticity in a length2. Shear modulus: Measure of the elasticity in an area3. Bulk modulus: Measure of the elasticity in a volumestressstrainElastic 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 itThursday, July 7, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu8Thursday, July 7, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu9Young’s ModulusAFexStress TensileLet’s consider a long bar with cross sectional area A and initial length Li. Fex=FinYoung’s Modulus is defined asWhat is the unit of Young’s Modulus?Experimental Observations1. For a fixed external force, the change in length is proportional to the original length2. The necessary force to produce the given strain is proportional to the cross sectional areaLiA:cross sectional areaTensile stressLf=Li+ΔLFexAfter the stretchFexFinTensile strainiLLStrain TensileYForce per unit areaUsed to characterize a rod or wire stressed under tension or compressionStrain TensileStress TensileiexLLAFThursday, July 7, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu10Strain VolumeStress VolumeB iVVAFiVVPBulk ModulusAFapplies force theArea SurfaceForce NormalPressureBulk Modulus characterizes the response of a substance to uniform squeezing or reduction of pressure.Bulk Modulus is defined asVolume stress =pressureAfter the pressure changeIf the pressure on an object changes by ΔP=ΔF/A, the object will undergo a volume change ΔV.VV’FFFFCompressibility is the reciprocal of Bulk Modulus Because the change of volume is reverse to change of pressure.Thursday, July 7, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu11Elastic Moduli and Ultimate Strengths of


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UT Arlington PHYS 1443 - Lecture Notes

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