Page 1Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 1Lecture 9Today:Today: Review sessionAssignment: For Monday, Read Chapter 8Exam Thursday, Oct. 2ndfrom 7:15-8:45 PM Chapters 1-7One 8 ½ X 11 note sheet and a calculator (for trig.)1. Room 2103: Sections 601 to 608 plus 6142. Room 2223: Section 6133. Room 2241: Sections 609 to 612 Physics 207: Lecture 9, Pg 2Textbook Chapters Chapter 1 Concept of Motion Chapter 2 1D Kinematics Chapter 3 Vector and Coordinate Systems Chapter 4 Dynamics I, Two-dimensional motion Chapter 5 Forces and Free Body Diagrams Chapter 6 Force and Newton’s 1stand 2ndLaws Chapter 7 Newton’s 3rdLaw Exam will reflect most key points (but not all)~30% of the exam will be more conceptual~70% of the exam is problem solvingPage 2Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 3The flying bird in the cage You have a bird in a cage that is resting on your upward turned palm. The cage is completely sealed to the outside (at least while we run the experiment!). The bird is initially sitting at rest on the perch. It decides it needs a bit of exercise and starts to fly. Question: How does the weight of the cage plus bird vary when the bird is flying up, when the bird is flying sideways, when the bird is flying down? So, what is holding the airplane up in the sky?Physics 207: Lecture 9, Pg 4Example with pulley A mass M is held in place by a force F. Find the tension in each segment of the massless ropes and the magnitude of F. Assume the pulleys are massless and frictionless.• The action of a masslessfrictionless pulley is to change the direction of a tension.• This is an example of static equilibrium.MT5T4T3T2T1FPage 3Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 5Example with pulley A mass M is held in place by a force F. Find the tension in each segment of the rope and the magnitude of F. Assume the pulleys are masslessand frictionless. Assume the rope is massless.• The action of a massless frictionless pulley is to change the direction of a tension.• Here F = T1 = T2 = T3 = T• Equilibrium means Σ F = 0 for x, y & z• For example: y-dir ma = 0 = T2 + T3 – T5 and ma = 0 = T5– Mg • So T5= Mg = T2 + T3 = 2 F T = Mg/2MT5T4T3T2T1FPhysics 207: Lecture 9, Pg 6ExampleWedge with friction A mass m slides with friction down a wedge of angle θ at constant velocity. The wedge sits at rest on a frictionless surface and abuts a wall. What is the magnitude of the force of the wall on the block?mvθθθθmgNfkFBD blockPage 4Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 7Example Wedge with friction A mass m slides with friction down a wedge of mass M & angle θ at constant velocity. The wedge sits at rest on a frictionless surface and abuts a wall. What is the magnitude of the force of the wall on the block?mvθθθθmgNfkFBD blockFBD wedge-N-fkMgFwFF3rdLawPhysics 207: Lecture 9, Pg 8Example Wedge with friction A mass m slides with friction down a wedge of mass M & angle θ at constant velocity. The wedge sits at rest on a frictionless surface and abuts a wall. What is the magnitude of the force of the wall on the block?mgNfkFBD blockx-dir: ΣFx= 0 = -fk+ mg sin θfk= mg sin θy-dir: ΣFy= 0 = N - mg cosθN = mg cosθyxθPage 5Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 9Example Wedge with friction A mass m slides with friction down a wedge of mass M & angle θ at constant velocity. The wedge sits at rest on a frictionless surface and abuts a wall. What is the magnitude of the force of the wall on the block?Notice that mg cos θ sin θ − mg cos θ sin θ = 0 !Force wall = 0 But there are faster ways.FBD wedgemg cos θMgFwFFmg sin θθθθθθθθθmg cos θ sin θθθθθmg cos θ sin θPhysics 207: Lecture 9, Pg 10ExampleAnother setting Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of µK=0.40, the masses are m1 = 4.0 kg, m2 = 1.0 kg and m3 = 2.0 kg. (A) What is the magnitude and direction of acceleration on the three blocks ?(B) What is the tension on the two cords ?m1T1m2m3Page 6Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 11Another example with a pulleyThree blocks are connected on the table as shown. The table has a coefficient of kinetic friction of µK=0.40, the masses are m1 = 4.0 kg, m2 = 1.0 kg and m3 = 2.0 kg. (A) FBD (except for friction) (B) So what about friction ?m1T1m2m3m2gNm3gm1gT3T1Physics 207: Lecture 9, Pg 12Problem recast as 1D motionThree blocks are connected on the table as shown. The center table has a coefficient of kinetic friction of µK=0.40, the masses are m1 = 4.0 kg, m2 = 1.0 kg and m3 = 2.0 kg. m1m2m3m2gNm3gm1gT3T1frictionlessfrictionlessm1g > m3g and m1g > (µkm2g + m3g) and friction opposes motion (starting with v = 0)so ffis to the right and a is to the left (negative)ffPage 7Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 13Problem recast as 1D motionThree blocks are connected on the table as shown. The center table has a coefficient of kinetic friction of µK=0.40, the masses are m1 = 4.0 kg, m2 = 1.0 kg and m3 = 2.0 kg. m1m2m3m2gNm3gm1gT3T1frictionlessfrictionlessx-dir: 1.ΣFx= m2a = µkm2g - T1+ T3m3a = m3g - T3m1a = −m1g + T1Add all three: (m1 + m2 + m3) a = µkm2g+ m3g – m1g ffT3T1Physics 207: Lecture 9, Pg 14Chapter 2Page 8Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 15Chapter 2Physics 207: Lecture 9, Pg 16Chapter 3Page 9Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 17Chapter 3Physics 207: Lecture 9, Pg 18Chapter 4Page 10Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 19Chapter 4Physics 207: Lecture 9, Pg 20Chapter 5Page 11Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 21Chapter 5 & 6Physics 207: Lecture 9, Pg 22Chapter 6Page 12Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 23Chapter 7Physics 207: Lecture 9, Pg 24Chapter 7Page 13Physics 207 – Lecture 9Physics 207: Lecture 9, Pg 25Textbook Chapters Chapter 1 Concept of Motion Chapter 2 1D Kinematics Chapter 3 Vector and Coordinate Systems Chapter 4 Dynamics I, Two-dimensional motion Chapter 5 Forces and Free Body Diagrams Chapter 6 Force and Newton’s 1stand 2ndLaws Chapter 7 Newton’s 3rdLaw Exam will reflect most key points (but not all)~40% of the exam will be more conceptual~60% of the exam is problem solvingPhysics 207: Lecture 9, Pg 26Short word problems After breakfast, I weighed myself and the scale
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