PHYS 1441 – Section 501 Lecture #5Newton’s Third Law (Law of Action and Reaction)Example of Newton’s 3rd LawSome Basic InformationFree Body Diagrams and Solving ProblemsSlide 6Example of Using Newton’s LawsExample w/o FrictionForces of FrictionExample w/ FrictionNewton’s Second Law & Uniform Circular MotionExample of Uniform Circular MotionExample of Banked HighwayWednesday, June 16, 2004 PHYS 1441-501, Summer 2004Dr. Jaehoon Yu1PHYS 1441 – Section 501Lecture #5Wednesday, June 16, 2004Dr. Jaehoon Yu•Chapter four: Newton’s Laws of Motion–Newton’s Third Law of Motion–Solving problems using Newton’s Laws•Free-body diagram•Uniform circular motionToday’s homework is homework #3, due 5pm, next Friday!!Wednesday, June 16, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu2Newton’s Third Law (Law of Action and Reaction)If two objects interact, the force, F21, exerted on object 1 by object 2 is equal in magnitude and opposite in direction to the force, F12, exerted on object 1 by object 2. F12F212112FF The action force is equal in magnitude to the reaction force but in opposite direction. These two forces always act on different objects. What is the reaction force to the force of a free fall object?The force exerted by the ground when it completed the motion. Stationary objects on top of a table has a reaction force (normal force) from table to balance the action force, the gravitational force.Wednesday, June 16, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu3Example of Newton’s 3rd Law A 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?2112FF 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.221tatvxbxbxfb21F =ur12F =urMxfv =bxfv =masses theof ratio by the if mMvvMxfbxfFFF 2112 12F =ur212b Mxf MxMx v t a tm� �= + =� �� �12 xF =12 yF =21xF =21 yF =12F =ur bxa =22tamMtvmMMxMxftamMMxMxfvmMMMxmbmarbxmabyma =0MM arMxMaMyMa =021F-ur21F- =urFFm=MxMamMxi Mxv a t+ =Mxa tbxi bxv a t+ =bxa t =Wednesday, June 16, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu4Some Basic InformationNormal Force, n: When Newton’s laws are applied, external forces are only of interest!!Why?Because, as described in Newton’s first law, an object will keep its current motion unless non-zero net external force is applied.Tension, T: Reaction force that reacts to gravitational force due to the surface structure of an object. Its direction is perpendicular to the surface.The reactionary force by a stringy object against an external force exerted on it. A graphical tool which is a diagram diagram of external forces on an objectof external forces on an object and is extremely useful analyzing forces and motion!! Drawn only on an object.Free-body diagramWednesday, June 16, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu5Free Body Diagrams and Solving Problems•Free-body diagram: A diagram of vector forces acting on an object A great tool to solve a problem using forces or using dynamics1. Select a point on an object and w/ information given2. Identify all the forces acting only on the selected object3. Define a reference frame with positive and negative axes specified4. Draw arrows to represent the force vectors on the selected point5. Write down net force vector equation6. Write down the forces in components to solve the problemsNo matter which one we choose to draw the diagram on, the results should be the same, as long as they are from the same motionMWhich one would you like to select to draw FBD?What do you think are the forces acting on this object?Gravitational forcegMFGA force supporting the object exerted by the floorMeWhich one would you like to select to draw FBD?What do you think are the forces acting on this elevator?NFgMFGGravitational force The force pulling the elevator (Tension)mWhat about the box in the elevator?Gravitational forceNormal forceNFTFgMFGgmFGBTFgMFGNFgmFBGNFWednesday, June 16, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu6Applications 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=TxFIf T is a constant force, ax, is constantxfvyFMTax0yaxn= -FgFg=MgTTxMa nFgyMa0 tavxxitMTvxiifxx221tMTtvxiWhat happened to the motion in y-direction?Wednesday, June 16, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu7Example of Using Newton’s LawsA traffic light weighing 125 N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37.0o and 53.0o with the horizontal. Find the tension in the three cables. F =urFree-bodyDiagram53o37oxyT137oT253oT3( ) ( )1 2sin 37 sin 53T T mg+ - =o o 053cos37cos21TT1T\ = NTT 12525.137sin754.053sin222100 ; T N=031iiixxTF031iiiyyTFNewton’s 2nd lawx-comp. of net forcey-comp. of net force0( )( )2cos 53cos 37T =oo20.754 T1 2 0.754 75.4T T N= =1 2 3T T T+ + =ur ur urma =r0Wednesday, June 16, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu8Example w/o FrictionA crate of mass M is placed on a frictionless inclined plane of angle . a) Determine the acceleration of the crate after it is released. FFree-bodyDiagramxyMdaFgnnF= -MgSupposed 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?dyFxFsingaxxysin2gdt xfvxy nFgamxMagxFsinMgyMagyFncosmgn0221tatvxix2 sin21tg tavxixsin2singdgsin2dgsin2dgvxfWednesday, June 16, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu9Forces of FrictionnfssResistive 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
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