1443-501 Spring 2002Lecture #5Dr. Jaehoon Yu1. Applications of Newton’s Laws2. Forces of Friction 3. Newton’s Second Law & Circular Motions4. Motion in Accelerated Frames5. Motion with Resistive Force1stterm exam on Monday Feb. 11, 2002, at 5:30pm, in the classroom!! Will cover chapters 1Will cover chapters 1--6!!6!!Feb. 4, 20021443-501 Spring 2002Dr. J. Yu, Lecture #52Newton’s LawsIn the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity. The acceleration of an object is directly proportional to the net force exerted on it and inversely proportional to the object’s mass. amFii=∑If two objects interact, the force, F12, exerted on object 1 by object 2 is equal magnitude to and opposite direction to the force, F21, exerted on object 1 by object 2. 1221FF −=1stLaw: Law of Inertia2ndLaw: Law of Forces3rdLaw: Law of Action and ReactionFeb. 4, 20021443-501 Spring 2002Dr. J. Yu, Lecture #53Example 5.3A large man and a small boy stand facing each other on frictionless ice. They put their hands together and push against each other so that they move apart. a) Who moves away with the higher speed and by how much?masses theof ratio by the if ; ;0 ; ;0 ; ; ;2112211221212112121221122112mMvvvmMtamMtatavvtatavvamMmFaFFFFFMaFMaFaMFmaFmaFamFFFFFFMxfbxfMxfbxbxbxibxfMxMxMxiMxfMxMxbxMyyMxxMbyybxxb〉〉∴===+==+====−=−===========−=b) Who moves farther while their hands are in contact?MmF12F21=-F12Given in the same time interval, since the boy has higher acceleration and thereby higher speed, he moves farther than the man. MMxMxfMxMxfbxbxfbxmMtatvmMtamMtvmMtatvx=+=+=+=22221221Feb. 4, 20021443-501 Spring 2002Dr. J. Yu, Lecture #54Some Basic InformationNormal Force, n: When Newton’s laws are applied, external forces are only of interestWhy?Because, as described in Newton’s first law, an object will keepits current motion unless non-zero net external forces are applied.Tension, T: Reaction force that balances gravitational force, keeping objects stationary.Magnitude of the force exerted on an object by a string or a rope.A graphical tool which is a diagram of external diagram of external forces on an objectforces on an object and is extremely useful analyzing forces and motion!! Drawn only on the object.Free-body diagramFeb. 4, 20021443-501 Spring 2002Dr. J. Yu, Lecture #55Applications of Newton’s LawsMSuppose you are pulling a box on frictionless ice, using a rope.TWhat are the forces being exerted on the box?Gravitational force: FgNormal force: nTension force: Tn= -FgTFree-body diagramFg=MgTotal force: F=Fg+n+T=T0 ;0 ;==+====∑∑ygyxxxanFFMTaMaTFIf T is a constant force, ax, is constant221tMTtvxxxtMTvtavvxiifxixxixf+=−=∆+=+=Feb. 4, 20021443-501 Spring 2002Dr. J. Yu, Lecture #56Example 5.4A traffic light weighing 125 N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37.0oand 53.0o with the horizontal. Find the tension in the three cables. ( ) ( )( ) ( )( )( )( ) ( )[ ]NTTNTNTTTTTTTmgTTTFTFTTTFiiiyyiiixx4.75754.0 ;10012525.137sin754.053sin 754.037cos53cos053cos37cos053sin37sin0 ;0 ;2122222121213131321=====×+==∴=+−=−+====++=∑∑====ooooooooFree-bodyDiagramT1T2T353o37o53o37oxyFeb. 4, 20021443-501 Spring 2002Dr. J. Yu, Lecture #57Example 5.6A crate of mass M is placed on a frictionless inclined plane of angle θ. a) Determine the acceleration of the crate after it is released.θθsinsin0 gaMgFMaFFnMaFnFFxgxxxgyyyg=====+==+=Free-bodyDiagramθxyMdaFgnnF=MaF= -MgθSupposed the crate was released at the top of the incline, and the length of the incline is d. How long does it take for the crate to reach the bottom and what is its speed at the bottom?θθθθθsin2sin2sinsin2sin212122dggdgtavvgdttgtatvdxixxfxix==+==∴=+=Feb. 4, 20021443-501 Spring 2002Dr. J. Yu, Lecture #58Forces of Frictionnfssµ≤Resistive force exerted on a moving object due to viscosity or other types frictional property of the medium in or surface on which the object moves.Force of static friction, fs:Force of kinetic friction, fkThe resistive force exerted on the object until just before the beginning of its movementThe resistive force exerted on the object during its movementnfkkµ=These forces are either proportional to velocity or normal forceEmpirical Formula What does this formula tell you? Frictional force increases till it reaches to the limit!!Beyond the limit, there is no more static frictional force but kinetic frictional force takes it over.Feb. 4, 20021443-501 Spring 2002Dr. J. Yu, Lecture #59Example 5.12Suppose a block is placed on a rough surface inclined relative to the horizontal. The inclination angle is increased till the block starts to move. Show that by measuring this critical angle, θc, one can determine coefficient of static friction, µs.ccccscsssgxxcgygyyysgMgMgnMgMgnffFFMgFnFnMaFfnFFθθθθµθµθtancossinsinsin ;0cos ;0======−==−==+==++=Free-bodyDiagramθxyMaFgnnF=MaF= -Mgθfs=µknEx05-12.ipFeb. 4, 20021443-501 Spring 2002Dr. J. Yu, Lecture #510Newton’s Second Law & Uniform Circular MotionFrFrrmrvar2=The centripetal acceleration is always perpendicular to velocity vector, v, for uniform circular motion.The force that causes the centripetal acceleration acts toward the center of the circular path and causes a change in the direction of the velocity vector. This force is called centripetal force.Is there force in this motion? If there is, what does it do?rvmamFrr2==∑What do you think will happen to the ball if the string that holds the ball breaks? Why?Based on Newton’s 1stlaw, since the external force no longer exist, the ball will continue its motion without change and will fly away following the tangential direction to the circle.Feb. 4, 20021443-501 Spring 2002Dr. J. Yu, Lecture #511Example 6.2A ball of mass 0.500kg is attached to the end of a cord 1.50m long. The ball is moving in a horizontal circle. If the string can withstand maximum tension of 50.0 N, what is the maximum speed the ball can attain before the cord breaks? Frmrvar2=Centripetal acceleration:TrvmamFrr===∑2When does the string break?When the centripetal force is greater than the sustainable tension.( )smmTrvTrvm/2.12500.05.10.502=×===Calculate the tension of the cord when speed of the ball is 5.00m/s.()( )NrvmT 33.85.100.5500.022=×==Ex06-02.ipFeb. 4, 20021443-501 Spring 2002Dr.
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