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## Lecture 9

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**Lecture 9**• Today: • Review session Assignment: For Monday, Read Chapter 8 Exam Thursday, Oct. 2nd from 7:15-8:45 PM Chapters 1-7 One 8 ½ X 11 note sheet and a calculator (for trig.) Room 2103: Sections 601 to 608 plus 614 Room 2223: Section 613 Room 2241: Sections 609 to 612**Textbook 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 1st and 2nd Laws • Chapter 7 Newton’s 3rd Law Exam will reflect most key points (but not all) ~30% of the exam will be more conceptual ~70% of the exam is problem solving**The 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?**Example with pulley**T4 T1 T3 T2 F T5 < M • A mass Mis 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 massless frictionless pulley is to change the direction of a tension. • This is an example of static equilibrium.**Example with pulley**T4 T1 T3 T2 F T5 < M • A mass Mis 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 massless and 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 SF = 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/2**ExampleWedge with friction**A mass m slides with friction down a wedge of angle q 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? FBD block N v fk m q mg**ExampleWedge with friction**FBD block fk N A mass m slides with friction down a wedge of mass M & angle q 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? mg 3rd Law FBD wedge -fk Fw -N v m Mg q FF**ExampleWedge with friction**FBD block N fk A mass m slides with friction down a wedge of mass M & angle q 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? y q mg x x-dir: S Fx = 0 = -fk+ mg sin q fk= mg sin q y-dir: S Fy = 0 = N - mg cos q N = mg cos q**ExampleWedge with friction**FBD wedge mg cos q sin q mg sin q mg cos q q q Fw q mg cos q sin q Mg FF A mass m slides with friction down a wedge of mass M & angle q 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 q sin q - mg cos q sin q = 0 ! Force wall = 0 But there are faster ways.**ExampleAnother setting**Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of mK=0.40, the masses are m1 = 4.0 kg, m2 = 1.0 kg and m3 = 2.0 kg. m2 T1 m1 m3 (A) What is the magnitude and direction of acceleration on the three blocks ? (B) What is the tension on the two cords ?**Another example with a pulley**Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of mK=0.40, the masses are m1 = 4.0 kg, m2 = 1.0 kg and m3 = 2.0 kg. N m2 T1 T1 T3 m1 m2g m1g m3 m3g (A) FBD (except for friction) (B) So what about friction ?**Problem recast as 1D motion**Three blocks are connected on the table as shown. The center table has a coefficient of kinetic friction of mK=0.40, the masses are m1 = 4.0 kg, m2 = 1.0 kg and m3 = 2.0 kg. N m3g m1g T1 T3 m3 m1 m2 ff frictionless frictionless m2g m1g > m3g and m1g > (mkm2g + m3g) and friction opposes motion (starting with v = 0) so ff is to the right and a is to the left (negative)**Problem recast as 1D motion**Three blocks are connected on the table as shown. The center table has a coefficient of kinetic friction of mK=0.40, the masses are m1 = 4.0 kg, m2 = 1.0 kg and m3 = 2.0 kg. N m3g m1g T1 T1 T3 T3 m3 m1 m2 ff frictionless frictionless m2g x-dir: 1.S Fx = m2a = mk m2g - T1 + T3 m3a = m3g - T3 m1a = - m1g + T1 Add all three: (m1 + m2 + m3) a = mk m2g+ m3g – m1g**Textbook 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 1st and 2nd Laws • Chapter 7 Newton’s 3rd Law Exam will reflect most key points (but not all) ~40% of the exam will be more conceptual ~60% of the exam is problem solving**Short word problems**• After breakfast, I weighed myself and the scale read 588 N. On my way out, I decide to take my bathroom scale in the elevator with me. What does the scale read as the elevator accelerates downwards with an acceleration of 1.5 m/s2 ? (500 N assuming g=10 m/s2) • A bear starts out and walks 1st with a velocity of 0.60 j m/s for 10 seconds and then walks at 0.40 i m/s for 20 seconds. What was the bear’s avg. velocity on the walk? (0.33 m/s) What was the bear’s avg. speed on the walk? (0.47 m/s)**Conceptual Problem**The pictures below depict cannonballs of identical mass which are launched upwards and forward. The cannonballs are launched at various angles above the horizontal, and with various velocities, but all have the same vertical component of velocity. Do not consider the effect of air resistance. Ans: d**Conceptual Problem**A bird sits in a birdfeeder suspended from a tree by a wire, as shown in the diagram at left. (Ans. f) Let WB and WF be the weight of the bird and the feeder respectively. Let T be the tension in the wire and N be the normal force of the feeder on the bird. Which of the following free-body diagrams best represents the birdfeeder? (The force vectors are not drawn to scale and are only meant to show the direction, not the magnitude, of each force.)**Graphing problem**The figure shows a plot of velocity vs. time for an object moving along the x-axis. Which of the following statements is true? (Ans. C) (A) The average acceleration over the 11.0 second interval is -0.36 m/s2 (B) The instantaneous acceleration at t = 5.0 s is -4.0 m/s2 (C) Both A and B are correct. (D) Neither A nor B are correct.**Conceptual Problem**A block is pushed up a 20º ramp by a 15 N force which may be applied either horizontally (P1) or parallel to the ramp (P2). How does the magnitude of the normal force N depend on the direction of P? Ans. B (A) N will be smaller if P is horizontal than if it is parallel the ramp. (B) N will be larger if P is horizontal than if it is parallel to the ramp. (C) N will be the same in both cases. (D) The answer will depend on the coefficient of friction. 20°**Conceptual Problem**A cart on a roller-coaster rolls down the track shown below. As the cart rolls beyond the point shown, what happens to its speed and acceleration in the direction of motion? Ans. D A. Both decrease. B. The speed decreases, but the acceleration increases. C. Both remain constant. D. The speed increases, but acceleration decreases. E. Both increase. F. Other**Conceptual Problem**• A person initially at point P in the illustration stays there a moment and then moves along the axis to Q and stays there a moment. She then runs quickly to R, stays there a moment, and then strolls slowly back to P. Which of the position vs. time graphs below correctly represents this motion? (Ans. 2)**Sample Problem**• A physics student on Planet Exidor throws a ball that follows the parabolic trajectory shown. The ball’s position is shown at one-second intervals until t = 3 s. At t = 1 s, the ball’s velocity is v= (2 i + 2 j) m/s. Determine the ball’s velocity at t = 0 s, 2 s, and 3 s. Ans.: 20½ m/s , 2 m/s , 8½ m/s b. What is the value of g on Planet Exidor? (2 m/s2 down)**Another question to ponder**How high will it go? • One day you are sitting somewhat pensively in an airplane seat and notice, looking out the window, one of the jet engines running at full throttle. From the pitch of the engine you estimate that the turbine is rotating at 3000 rpm and, give or take, the turbine blade has a radius of 1.00 m. If the tip of the blade were to suddenly break off (it occasionally does happen with negative consequences) and fly directly upwards, then how high would it go (assuming no air resistance and ignoring the fact that it would have to penetrate the metal cowling of the engine.)**Lecture 9**• Today: • Review session Assignment: For Monday, Read Chapter 8 Exam Thursday, Oct. 2nd from 7:15-8:45 PM Chapters 1-7 One 8 ½ X 11 note sheet and a calculator (for trig.) Room 2103: Sections 601 to 608 plus 614 Room 2223: Section 613 Room 2241: Sections 609 to 612