Physics 106P: Lecture 5 Notes - The Burns Home Page

Physics 106P: Lecture 5 Notes - The Burns Home Page

SPH4U: Lecture 1DynamicsDescribing motion – so far… (from last Year)How and why doobjects move• Linear motion with const acceleration:1x x0 v0t at2v v at0a const2v v 2a(x x )2 20 0v1 (v2v)av 0What about higher order rates of change?• If linear motion and circular motion are uniquelydetermined by acceleration, do we ever need higherderivatives? 3da d rJ Known as the “Jerk”3dt dt• Certainly acceleration changes, so does that mean we needto find some “action” that controls the third or higher timederivatives of position?• NO.Dynamics• Isaac Newton (1643 - 1727) published Principia Mathematicain 1687. In this work, he proposed three “laws” of motion:• principiaLaw 1: An object subject to no external forces is at rest or moveswith a constant velocity if viewed from an inertial referenceframe.Law 2: For any object, F NET = F = ma (not mv!)Law 3: Forces occur in pairs: F A ,B = - F B ,A(For every action there is an equal and opposite reaction.)These are the postulates of mechanicsThey are experimentally, not mathematically, justified.They work, and DEFINE what we mean by “forces”.Page 1

Newton’s Second Law...Example: Pushing a Box on Ice.• Components of F = ma :F X = ma XF Y = ma YF Z = ma Z• A skater is pushing a heavy box (mass m = 100 kg) across a sheetof ice (horizontal & frictionless). He applies a force of 50 N in the xdirection. If the box starts at rest, what is its speed v after beingpushed a distance d = 10 m?• Suppose we know m and F X , we can solve for a X and apply thethings we learned about kinematics over the last few lectures: (ifthe force is constant)v = 01x x0 v0xt axt2v v a tx 0x x2FmaxExample: Pushing a Box on Ice.Example: Pushing a Box on Ice...• A skater is pushing a heavy box (mass m = 100 kg) across a sheetof ice (horizontal & frictionless). He applies a force of 50 N in the xdirection. If the box starts at rest, what is its speed v after beingpushed a distance d = 10m ?v• Start with F = ma.• a = F / m.• Recall that v 2 - v 02= 2a(x - x 0 ) (Last Yeat)• So v 2 = 2Fd / m 2Fdv mvmFaFmadxdiPage 3

Example: Pushing a Box on Ice...v 2Fdm• Plug in F = 50 N, d = 10 m, m = 100 kg:• Find v = 3.2 m/s.QuestionForce and acceleration• A force F acting on a mass m 1 results in an acceleration a 1 .The same force acting on a different mass m 2 results in anacceleration a 2 = 2a 1 .m 1 m 2F a 1 F a 2 = 2a 1v• If m 1 and m 2 are glued together and the same force F actson this combination, what is the resulting acceleration?Fmam 1 m 2F a = ?di(a) 2/3 a 1 (b) 3/2 a 1 (c) 3/4 a 1SolutionForce and accelerationm 1 m 2F a = F / (m 1 + m 2 )• Since a 2 = 2a 1 for the same applied force, m 2 = (1/2)m 1 !• m 1 + m 2 = 3m 1 /2• So a = (2/3)F / m 1 but F/m 1 = a 1a = 2/3 a 1Forces• We will consider two kinds of forces:• Contact force:• This is the most familiar kind.• I push on the desk.• The ground pushes on the chair...• A spring pulls or pushes on a mass• A rocket engine provides some number of Newtons ofthrust (1 lb of thrust = mg = 2.205*9.81 = 21.62 Newtons)• Action at a distance:• Gravity• Electricity• Magnetism(a) 2/3 a 1 (b) 3/2 a 1 (c) 3/4 a 1Page 4

Contact forces:Action at a Distance• Objects in contact exert forces.• Convention: F a,b means acting on a due to b”.The Force• Gravity:• So F head,thumb means “the force onthe head due to the thumb”.Burp!F head,thumbGravitation(Courtesy of Newton)Understanding• Newton found that a moon / g = 0.000278• and noticed that R E2/ R 2 = 0.000273Hey, I’min UCM!a moon• This inspired him to propose theUniversal Law of Gravitation:gRR EFMmWe will discussthese conceptslaterGMm2RIf the distance between two point particles is doubled, thenthe gravitational force between them:A) Decreases by a factor of 4B) Decreases by a factor of 2C) Increases by a factor of 2D) Increases by a factor of 4E) Cannot be determined without knowing the masseswhere G = 6.67 x 10 -11 m 3 kg -1 s -2And the force is attractive along a line between the 2 objectsPage 5

Newton’s Third Law:Newton's Third Law...• Forces occur in pairs: F A ,B = - F B ,A .• For every “action” there is an equal and opposite “reaction”.• F A ,B = - F B ,A . is true for contact forces as well:F m,wForce on me from wall isequal and opposite to theforce on the wall from the me.• We have already seen this in the case of gravity:m 1m 2mmF G F R1 212 2 2112F 12 F 21We will discussF w,mF f,mF m,fForce on me from the floor isequal and opposite to the forceon the floor from the me.R 12these concepts inmore detail later.Example of Bad ThinkingExample of Good Thinking• Since F m,b = -F b,m , why isn’t F net = 0 and a = 0 ?• Consider only the box as the system!• F on box = ma box = F b,m• Free Body Diagram (next power point).a ??F m,bblockF b,ma boxF m,b F b,mblockiceiceNo iceFriction forcePage 6

Add a wall that stops the motion of the block• Now there are two forces acting (in the horizontal direction)on block and they cancel• F on box = ma box = F b,m + F b,w = 0• Free Body Diagram (next power point).iceNewton’s 3rd Law Understanding(a) 2a(b) 3F Fb,w F b,mw,bF m,bblocka box• Two blocks are stacked on the ground. How many action-reactionpairs of forces are present in this system?b(c) 4(d) 5F a,EaF E,ababF b,EF E,bSolution:gravitygravitygravity contactvery tinycontactaF b,aF a,bb5abF a,bF b,aF g,babF b,gUnderstandingA moon of mass m orbits a planet of mass 100m. Let thestrength of the gravitational force exerted by the planet on themoon be denoted by F 1 , and let the strength of thegravitational force exerted by the moon on the planet be F 2 .Which of the following is true?A) F 1 =100F 2B) F 1 =10F 2C) F 1 =F 2Newton’s Third LawE) F 2 =100F 1D) F 2 =10F 1Page 7

Flash: Newton’s 1 St LawFlash: Newton’s 2 nd LawFlash: Newton’s 3 rd LawFlash: Applications of NewtonPage 8

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