1Chapter 12 – Equilibrium and ElasticityI. Equilibrium- Definition- Requirements- Static equilibriumII. Center of gravityIII. Elasticity- Tension and compression- Shearing- Hydraulic stressI. Equilibrium- Definition:0,0 == LPExample: block resting on a table, hockey puck sliding across a frictionless surface with constant velocity, the rotating blades of a ceiling fan, the wheel of a bike traveling across a straight path at constant speed. An object is in equilibrium if:- The linear momentum of its center of mass is constant.- Its angular momentum about its center of mass is constant.- Static equilibrium:Objects that are not moving either in TRANSLATION or ROTATIONExample: block resting on a table.2Stable static equilibrium:If a body returns to a state of static equilibrium after having been displacedfrom it by a force marble at the bottom of a spherical bowl. Unstable static equilibrium:A small force can displace the body and end the equilibrium.(1) Torque about supporting edge by Fgis 0 because line of action of Fgpasses through rotation axis domino in equilibrium.(3) Not as unstable as (1) in order to topple it, one needs to rotate it beyond balance position in (1).(2) Slight force ends equilibrium line of action of Fgmoves to one side of supporting edge torque due to Fgincreases domino rotation.- Requirements of equilibrium:0==→=dtPdFctePnet0, ===dtLdcteLnetτBalance of forces translational equilibriumBalance of torques rotational equilibrium- Vector sum of all external forces that act on body must be zero.- Vector sum of all external torques that act on the body, measured about any possible point must be zero.Balance of forces Fnet,x= Fnet,y= Fnet,z=0Balance of torques τnet,x= τnet,y= τnet,z=03II. Center of gravitycog = Body’s point where the gravitational force “effectively” acts.Gravitational force on extended body vector sum of the gravitational forces acting on the individual body’s elements (atoms) .-This course initial assumption:The center of gravity is at the center of mass.If g is the same for all elements of a body, then the body’s Center Of Gravity (COG) is coincident with the body’s Center Of Mass (COM).Assumption valid for every day objects “g” varies only slightly along Earth’s surface and decreases in magnitude slightly with altitude.Proof:Each force Fgiproduces a torque τion the element of mass about the origin O, with moment arm xi.∑∑==→=→=⊥i igiiinetgiiiFxFxFrττττnetigicoggcogFxFxττ===∑∑∑∑∑∑∑∑==→=→=→=icomiicogiiiiicogiiiiiiicogigiiigicogxmxMxmxmxgmxgmxFxFx14III. ElasticityBranch of physics that describes how real bodies deform when forces are applied to them.Real rigid bodies are elastic we can slightly change their dimensions by pulling, pushing, twisting or compressing them.Stress: Deforming force per unit area. Strain: Unit deformationTensile stress: associated with stretchingShearing stressHydraulic stress(1) Stress = cte x Strain Recovers original dimensions when stress removed.(2) Stress > yield strength Sy specimen becomes permanently deformed.(3) Stress > ultimate strength Su specimen breaks.Tension and compression:AFStress =( F= force applied perpendicular to the area A of the object)Elastic modulus: describes the elastic behavior (deformations) of objects as they respond to forces that act on them.Stress = Elasticity Modulus x Strain5LLStrain∆=( fractional change in length of the specimen)Stress = (Young’s modulus) x StrainLLEAF∆=Shearing:AFStress =( F= force in the plane of the area A)LxStrain∆=( fractional change in length of the specimen)Units of Young modulus: F/m2LxGAF∆=Stress = (Shear modulus) x StrainHydraulic stress:AFppressureFluidStress ===VVStrain∆=Hydraulic Stress = (Bulk modulus) x Hydraulic
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