Monday, Sept. 20, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu11. Newton’s Laws of Motion • Gravitational Force and Weight• Newton’s third law of motion2. Application of Newton’s Laws• Free-body diagrams• Application of Newton’s Laws• Motion without friction• Forces of friction• Motion with frictionPHYS 1443 – Section 003Lecture #8Monday, Sept. 20, 2004Dr. Jaehoon YuMonday, Sept. 20, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu2Announcements• e-mail distribution list: 37/42 of you have subscribed so far. – A test message was sent Friday for verification• 26 of you replied! Thank you!• Remember the 1stterm exam, Monday, Sept. 27, one week from today– Covers up to chapter 6.– No make-up exams• Miss an exam without pre-approval or a good reason: Your grade is F.– Mixture of multiple choice and free style problemsMonday, Sept. 20, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu3Acceleration Vector aF2F1Example for Newton’s 2ndLaw of MotionDetermine the magnitude and direction of acceleration of the puck whose mass is 0.30kg and is being pulled by two forces, F1 and F2, as shown in the picture, whose magnitudes of the forces are 8.0 N and 5.0 N, respectively.=xF1θ1=60οθ2=−20ο=xFComponents of F1Components of F2Components oftotal force Fxa =Magnitude and direction of acceleration a=yF1=xF2=yF2=yF ya=a=G θ=a=G1cosFθ1=JJG1sinFθ1=JJG22cosFθ=JJG22sinFθ=JJG12xxFF+=xma12yyFF+=ymaxFm=28.729 /0.3ms=yFm=25.217 /0.3ms=()()22xyaa+=() ()2229 17+234 /ms=1tanyxaa−⎛⎞=⎜⎟⎝⎠117tan 3029−⎛⎞=⎜⎟⎝⎠Dxyai a j∧∧+=229 17 /ijms∧∧⎛⎞+⎜⎟⎝⎠()8.0 cos 60 4.0N×=D()8.0 sin 60 6.9N×=D()5.0 cos 20 4.7N×−=D()5.0 sin 20 1.7N×−=−D4.0 4.7 8.7N+==6.9 1.7 5.2N−==Monday, Sept. 20, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu4Gravitational Force and WeightSince weight depends on the magnitude of gravitational acceleration, g, it varies depending on geographical location.The attractive force exerted on an object by the Earth Gravitational Force, Fg== amFGWeight of an object with mass M isW=By measuring the forces one can determine masses. This is why you can measure mass using the spring scale.gmGF=JGMg=JGMgMonday, Sept. 20, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu5Newton’s Third Law (Law of Action and Reaction)If two objects interact, the force F21exerted on object 1 by object 2 is equal in magnitude and opposite in direction to the force F12exerted on object 2 by object 1. 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 gravitational force exerted by the object to the Earth!Stationary objects on top of a table has a reaction force (normal force) from table to balance the action force, the gravitational force.Monday, Sept. 20, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu6Example of Newton’s 3rdLaw 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?12F=JGMmF12F21=-F1221F=JG2112FF −=12 F=JG12F=JG12 xF=12 yF=21xF=21yF=FFF =−=2112 bxma=bmaGbxmabyma=021F−JG21F=JGFMxMaMyMa=0SinceandEstablish the equation bxFam==Divide by mMMaG MxMamF=MxMaMonday, Sept. 20, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu7Example of Newton’s 3rdLaw b) Who moves farther while their hands are in contact?Given in the same time interval, since the boy has higher acceleration and thereby higher speed, he moves farther than the man.212bbxi bxxvt at=+ =bxfv=212bMxMxatm⎛⎞==⎜⎟⎝⎠22MxMatm=tamMMxMxfvmMMMxmbxi bxvat+=bxat=Mxfv=Mxi Mxvat+=MxatSo boy’s velocity if higher than man’s, if M>m, by the ratio of the masses.Man’s velocityBoy’s velocityBoy’s displacementMan’s displacementMonday, Sept. 20, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu8Some 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 keepits 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 of external diagram of external forces on an objectforces on an object and is extremely useful analyzing forces and motion!! Drawn only on an object.Free-body diagramMonday, Sept. 20, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu9Free 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 in the problem2. 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=NFMonday, Sept. 20, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu10Applications 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?Monday, Sept. 20, 2004 PHYS 1443-003, Fall 2004Dr. Jaehoon Yu11Example w/o FrictionA crate of mass M is placed on a frictionless
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