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NJIT PHYS 111 - Chapter 12: Equilibrium and Elasticity

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Chapter 12: Equilibrium and ElasticityEquilibrium:Conditions of Equilibrium:The center or gravity:How is the center of gravity of an object determined?Problem-Solving Strategy:Slide 7ElasticityYoung’s Modulus: Elasticity in LengthSlide 10Shear Modulus: Elasticity in ShapeBulk Modulus: Elasticity in VolumeSlide 13Summary:Slide 15Sample Problem 12-3:Slide 17Slide 18Slide 19Chapter 12: Equilibrium and ElasticityConditions Under Which a Rigid Object is in EquilibriumProblem-Solving StrategyElasticityEquilibrium:An object at equilibrium is either ...• at rest and staying at rest (i.e., static equilibrium) , or • in motion and continuing in motion with the constant velocity and constant angular momentum. For the object in equilibrium, • the linear momentum ( ) of its center of mass is constant.• the angular momentum ( ) about its center of mass, or any other point, is constant.vmP)( vrmPrLConditions of Equilibrium:amrdtLdamdtPdFnetnet Net force: Net torque:Conditions of equilibrium: torques)of (balance 0 and forces) of (balance 0netnetF000,,,znetynetxnetFFF000,,,znetynetxnetAnother requirements for static equilibrium: 0PThe center or gravity:The gravitational force on a body effectively acts at a single point, called the center of gravity (cog) of the body.•the center of mass of an object depends on its shape and its density •the center of gravity of an object depends on its shape, density, and the external gravitational field. Does the center of gravity of the body always coincide with the center of mass (com)? Yes, if the body is in a uniform gravitational field.How is the center of gravity of an object determined?The center of gravity (cog) of a regularly shaped body of uniform composition lies at its geometric center.The (cog) of the body can be located by suspending it from several different points. The cog is always on the line-of-action of the force supporting the object.cogProblem-Solving Strategy:• Define the system to be analyzed• Identify the forces acting on the system• Draw a free-body diagram of the system and show all the forces acting on the system, labeling them and making sure that their points of application and lines of action are correctly shown.• Write down two equilibrium requirements in components and solve these for the unknownsSample Problem 12-1:• Define the system to be analyzed: beam & block• Identify the forces acting on the system:the gravitational forces: mg & Mg, the forces from the left and the right scales: Fl & Fr• Draw a force diagram• Write down the equilibrium requirements in components and solve these for the unknowns00)()(: torquesof balance0:forces of balance4121lrrlFMgLmgLLFmgMgFFOElasticitySome concepts:• Rigid Body: • Deformable Body:elastic body: rubber, steel, rock…plastic body: lead, moist clay, putty… • Stress: Deforming force per unit area (N/m2)• Strain: unit deformationStrainStressmodulus Elastic Young’s Modulus: Elasticity in LengthThe Young’s modulus, E, can be calculated by dividing the stress by the strain, i.e. where (in SI units)E is measured in newtons per square metre (N/m²). F is the force, measured in newtons (N)A is the cross-sectional area through which the force is applied, measured in square metres (m2)L is the extension, measured in metres (m)L is the natural length, measured in metres (m)LLAFLLAFstrainstressE//MaterialDensity (kg/m3)Young’s Modulus E(109N/m2)Ultimate Strength Su(106N/m2)Yield Strength Sy(106N/m2)Steel 7860 200 400 250Aluminum 2710 70 110 90Glass 2190 65 50Concrete 2320 30 40Wood 525 13 50Bone 1900 9 170Polystyrene 1050 3 48Table 12-1: Some elastic properties of selected material of engineering interestShear Modulus: Elasticity in ShapeThe shear modulus, G, can be calculated by dividing the shear stress by the strain, i.e. where (in SI units)G is measured in newtons per square metre (N/m²) F is the force, measured in newtons (N)A is the cross-sectional area through which the force is applied, measured in square metres (m2)x is the horizontal distance the sheared face moves, measured in metres (m)L is the height of the object, measured in metres (m)xLAFLxAFG//strainshear stressshearBulk Modulus: Elasticity in VolumeThe bulk modulus, B, can be calculated by dividing the hydraulic stress by the strain, i.e. where (in SI units)B is measured in newtons per square metre (N/m²) P is measured in in newtons per square metre (N/m²) V is the change in volume, measured in metres (m3)V is the original volume, measured in metres (m3)VVpVVpB/strain hydraulic pressure hydraulicYoung’s modulus Shear modulus Bulk modulusUnder tension and compressionUnder shearing Under hydraulic stressStrain is Strain is Strain isLLAFStrainStressExLAFStrainStressGVVpStrainStressBLL /Lx /VV /Summary:• Requirements for Equilibrium:• The cog of an object coincides with the com if the object is in a uniform gravitational field.• Solutions of Problems:• Elastic Moduli:tension and compressionshearinghydraulic stressstrain modulusstress LLEAF LxGAF VVBp 0 and 0netnetF•Define the system to be analyzed• Identify the forces acting on the system• Draw a force diagram• Write down the equilibrium requirements in components and solve these for the unknownsSample Problem 12-2:• Define the object to be analyzed: firefighter & ladder• Identify the forces acting on the system:the gravitational forces: mg & Mg, the force from the wall: Fwthe force from the pavement: Fpx & Fpy• Draw a force diagram• Write down the equilibrium requirements in components and solve these for the unknowns223121 where,0)()(: torquesof balance0 0:forces of balancehLahFmgaMgamgMgFFFwpypxwSample Problem 12-3:• Define the object to be analyzed: Beam• Identify the forces acting on the system:the gravitational force (mg), the force from the rope (Tr)the force from the cable (Tc), andthe force from the hinge (Fv and Fh)• Draw a force diagram• Write down the equilibrium requirements in components and solve these for the unknownsNamMgbmgbbTaTrcznet6093)(T 0))((21c21,NTFTFFchchxnet6093


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NJIT PHYS 111 - Chapter 12: Equilibrium and Elasticity

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