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UT Arlington PHYS 1441 - Lecture Notes

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PHYS 1441 – Section 004 Lecture #9AnnouncementsNewton’s Third Law (Law of Action and Reaction)Example of Newton’s 3rd LawSome Basic InformationFree Body Diagrams and Solving ProblemsSlide 7Example 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, Feb. 18, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu1PHYS 1441 – Section 004Lecture #9Wednesday, Feb. 18, 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 #5, due 5pm, next Friday!!Wednesday, Feb. 18, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu2Announcements•1st term exam–1 – 2:20pm next Monday, Feb. 3, during the class–Chapters 1 – 5.3 –Mixture of multiple choices and numeric problems•Will drop the worst of the three term exams–Exams are not comprehensive•Rules for missing exams–No make-up exam–Missing any exam without a prior approval will results in an automatic failure•E-mail: –Received 30 verifications  Thank you!–I still have 13 of you not registered for e-mail•Will issue -3 point if not done by next Monday, Feb. 23•- 5 points if not done by Monday, Mar. 1Wednesday, Feb. 18, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu3Newton’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, Feb. 18, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu4Example 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 mMvvMxfbxfFFF 2112 12F =ur212b Mxf MxMx v t a tm� �= + =� �� �12 xF =12 yF =21xF =21 yF =12F =ur bxa =22tamMtvmMMxMxftamMMxMxfvmMMMxmbmarbxmabyma =0MM arMxMaMyMa =021F-ur21F- =urFFm=MxMamMxi Mxv a t+ =Mxa tbxi bxv a t+ =bxa t =Wednesday, Feb. 18, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu5Some 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, Feb. 18, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu6Free 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 problemsNo 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 forcegMFGA 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?NFgMFGGravitational force The force pulling the elevator (Tension)mWhat about the box in the elevator?Gravitational forceNormal forceNFTFgMFGgmFGBTFgMFGNFgmFBGNFWednesday, Feb. 18, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu7Applications 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=TxFIf T is a constant force, ax, is constantxfvyFMTax0yaxn= -FgFg=MgTTxMa nFgyMa0 tavxxitMTvxiifxx221tMTtvxiWhat happened to the motion in y-direction?Wednesday, Feb. 18, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu8Example 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   053cos37cos21TT1T\ =    NTT 12525.137sin754.053sin222100 ; T N=031iiixxTF031iiiyyTFNewton’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, Feb. 18, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu9Example 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-bodyDiagramxyMdaFgnnF= -MgSupposed the crate was released at the top of the incline, and the length of the


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UT Arlington PHYS 1441 - Lecture Notes

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