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UW-Madison PHYSICS 207 - Physics 207 – Lecture 10

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Page 1Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 1Lecture 10Goals:Goals: Exploit Newton’s 3rdLaw in problems with friction Employ Newton’s Laws in 2D problems with circular motionAssignment: HW5, (Chapter 7, due 2/24, Wednesday)For Tuesday: Finish reading Chapter 8, First four sections in Chapter 9Physics 207: Lecture 10, Pg 2Example: Friction and Motion A box of mass m1= 1 kg is being pulled by a horizontal string having tension T = 40 N. It slides with friction (µk= 0.5) on top of a second box having mass m2= 2 kg, which in turn slides on a smooth (frictionless) surface. What is the acceleration of the bottom box ?Key Question: What is the force on mass 2 from mass 1?mm22Tmm11slides with friction (µk=0.5 )slides without frictiona = ?vPage 2Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 3ExampleSolution First draw FBD of the top box:m1N1m1gTfk= µKN1= µKm1gvPhysics 207: Lecture 10, Pg 4 Newtons 3rdlaw says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2.m1ff1,2 = µKm1g = 5 Nm2f2,1= -f1,2 As we just saw, this force is due to friction:ExampleSolutionActionReactionPage 3Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 5 Now consider the FBD of box 2: m2 f2,1 = µkm1gm2gN2N1 = m1gExampleSolutionPhysics 207: Lecture 10, Pg 6 Finally, solve Fx= maxin the horizontal direction: m2 f2,1 = µKm1gµKm1g = m2axExampleSolution= 2.5 m/s2 = =kg 2N 5ga2 k1mmxµPage 4Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 7Home Exercise Friction and Motion, Replay A box of mass m1= 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m2= 2 kg. The static and kineticcoefficients of friction between the 2 boxes are µs=1.5 and µk= 0.5. The second box can slide freely (frictionless) on an smooth surface.Question:Compare the acceleration of box 1 to the acceleration of box 2?m2Tm1friction coefficients µs=1.5 and µk=0.5slides without frictiona2a1Physics 207: Lecture 10, Pg 8Home Exercise Friction and Motion, Replay in the static case A box of mass m1= 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m2= 2 kg. The static and kineticcoefficients of friction between the 2 boxes are µs=1.5 and µk= 0.5. The second box can slide freely on an smooth surface (frictionless).In the case of “no slippage” what is the maximum frictional force between boxes 1 & 2?m2Tm1friction coefficients µs=1.5 and µk=0.5slides without frictiona2a1Page 5Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 9Home ExerciseFriction and Motionfs= 10 N and the acceleration of box 1 isAcceleration of box 2 equals that of box 1, with |a| = |T| / (m1+m2) and the frictional force f is m2a(Notice that if T were in excess of 15 N then it would break free)m2Tm1friction coefficients µs=1.5 and µk=0.5slides without frictiona2a1TTm1 ggNNfSfS≤ µSN = µSm1g = 1.5 x 1 kg x 10 m/s2which is 15 N (so m2can’t break free)Physics 207: Lecture 10, Pg 10Exercise Tension exampleA. Ta= ½ TbB. Ta= 2 TbC. Ta= TbD. Correct answer is not givenCompare the strings below in settings (a) and (b) and their tensions.Page 6Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 11Problem 7.34 Hint (HW 6)Suggested Steps Two independent free body diagrams are necessary Draw in the forces on the top and bottom blocks Top Block  Forces: 1. normal to bottom block 2. weight 3. rope tension and 4. friction with bottom block (model with sliding) Bottom Block  Forces:1. normal to bottom surface 2. normal to top block interface3. rope tension (to the left)4. weight (2 kg) 5. friction with top block 6. friction with surface7. 20 NUse Newton's 3rd Law to deal with the force pairs (horizontal & vertical) between the top and bottom block.Physics 207: Lecture 10, Pg 12On to Chapter8 Reprisal of : Uniform Circular MotionFor an object moving along a curved trajectory with constant speeda = ar(radial only)|ar|= vt2rarvPage 7Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 13Non-uniform Circular MotionFor an object moving along a curved trajectory, with non-uniform speeda = ar+ at(radial and tangential)aratdtd| |v|at |= |ar|= vT2rPhysics 207: Lecture 10, Pg 14Key steps Identify forces (i.e., a FBD) Identify axis of rotation Apply conditions (position, velocity &acceleration)Page 8Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 15ExampleThe pendulumConsider a person on a swing:When is the tension on the rope largest? And at that point is it :(A) greater than(B) the same as(C) less thanthe force due to gravity acting on the person?axis of rotationPhysics 207: Lecture 10, Pg 16ExampleGravity, Normal Forces etc.vTmgTat bottom of swing vTis maxFr= m ac= m vT2 / r = T - mgT = mg + m vT2 / rT > mgmgTat top of swing vT= 0Fr= m 02 / r = 0 = T – mg cos θT = mg cos θT < mgθaxis of rotationyxPage 9Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 17Conical Pendulum (very different) Swinging a ball on a string of length L around your head(r = L sin θ)axis of rotationΣ Fr= mar= T sin θΣ Fz= 0 = T cos θ – mgsoT = mg / cos θ (> mg)mar= (mg / cos θ ) (sin θ )ar= g tan θ = vT2/r vT= (gr tan θ)½Period:t = 2π r / vT=2π (r cot θ /g)½= 2π (L cos θ /g)½rLPhysics 207: Lecture 10, Pg 18Conical Pendulum (very different) Swinging a ball on a string of length L around your headaxis of rotationPeriod:t = 2π r / vT=2π (r cot θ /g)½= 2π (L cos θ / g )½= 2π (5 cos 5 / 9.8 )½= 4.38 s = 2π (5 cos 10 / 9.8 )½= 4.36 s= 2π (5 cos 15 / 9.8 )½= 4.32 srLPage 10Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 19A match box car is going to do a loop-the-loop of radius r. What must be its minimum speed vtat the top so that it can manage the loop successfully ?Another example of circular motionLoop-the-loop 1 Physics 207: Lecture 10, Pg 20To navigate the top of the circle its tangential velocity vTmust be such that its centripetal acceleration at least equals the force due to gravity. At this point N, the normal force, goes to zero (just touching). Loop-the-loop 1Fr= mar= mg = mvT2/rvT= (gr)1/2mgvTPage 11Physics 207 – Lecture 10Physics 207: Lecture 10, Pg 21The match box car is going to do a loop-the-loop. If the speed at the bottom is vB, what is the normal force, N, at that point?Hint: The car is constrained to


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UW-Madison PHYSICS 207 - Physics 207 – Lecture 10

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