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sy17_oct31_07hc

# sy17_oct31_07hc - Physics 207 Lecture 17 Physics 207...

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Page 1 Physics 207 – Lecture 17 Physics 207: Lecture 17, Pg 1 Physics 207, Physics 207, Lecture 17, Oct. 31 Lecture 17, Oct. 31 Agenda: circle6 Review for exam Review for exam circle6 Exam will be held in rooms B102 & Exam will be held in rooms B102 & B130 in Van Vleck at 7:15 PM Assignment: circle6 MP Homework 7, Ch. 11, 5 problems, NOTE: Due Wednesday at 4 PM NOTE: Due Wednesday at 4 PM circle6 MP Homework 7A, Ch. 13, 4 problems, available today Physics 207: Lecture 17, Pg 2 Example Gravity, Normal Forces etc. Consider a women on a swing: When is the tension on the rope largest ? And is it : (A) greater than (B) the same as (C) less than the force due to gravity acting on the woman Physics 207: Lecture 17, Pg 3 Gravity, Normal Forces etc. Gravity, Normal Forces etc. At the bottom of the swings and is it (A) greater than the force due to gravity acting on the woman v m g T F c = m a c = m v 2 / r = T - mg T = mg + m v 2 / r m g T F c = m 0 2 / r = 0 = T mg cos θ F T = m aT = mg sin θ θ Physics 207: Lecture 17, Pg 4 Newton Newton’ s Laws Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of 0.350, the masses are m 1 = 4.00 kg, m 2 = 1.00kg and m 3 = 2.00kg. a) What is the magnitude and direction of acceleration on the three blocks ? b) What is the tension in the two cords ? m 1 T 1 m 2 m 3 Physics 207: Lecture 17, Pg 5 Problem Three blocks are connected on the table as shown. The table has a coefficient of kinetic friction of μ K =0.40, the masses are m 1 = 4.0 kg, m 2 = 1.0 kg and m 3 = 2.0 kg. (A) FBD (except for friction) (B) So what about friction ? m 1 T 1 m 2 m 3 m 2 g N m 3 g m 1 g T 3 T 1 Physics 207: Lecture 17, Pg 6 Problem recast as 1D motion Problem recast as 1D motion Three blocks are connected on the table as shown. The center table has a coefficient of kinetic friction of μ K =0.40, the masses are m 1 = 4.0 kg, m 2 = 1.0 kg and m 3 = 2.0 kg. m 1 m 2 m 3 m 2 g N m 3 g m 1 g T 3 T 1 frictionless frictionless m 1 g > m 3 g and m 1 g > ( μ k m 2 g + m 3 g) and friction opposes motion (starting with v = 0) so f f is to the right and a is to the left (negative) f f

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Page 2 Physics 207 – Lecture 17 Physics 207: Lecture 17, Pg 7 Problem recast as 1D motion Problem recast as 1D motion Three blocks are connected on the table as shown. The center table has a coefficient of kinetic friction of μ K =0.40, the masses are m 1 = 4.0 kg, m 2 = 1.0 kg and m 3 = 2.0 kg. m 1 m 2 m 3 m 2 g N m 3 g m 1 g T 3 T 1 frictionless frictionless x-dir: 1. Σ F x = m 2 a = μ k m 2 g - T 1 + T 3 m 3 a = m 3 g - T 3 m 1 a = - m 1 g + T 1 Add all three: ( m 1 + m 2 + m 3 ) a = μ k m 2 g + m 3 g – m 1 g f f T 3 T 1 Physics 207: Lecture 17, Pg 8 Lecture 17, Exercise Exercise Work/Energy for Non Work/Energy for Non- Conservative Forces A. 2.5 J B. 5.0 J C. 10. J D. -2.5 J E. -5.0 J F. -10. J 1 meter 30° circle6 The air track is once again at an angle of 30°with resp ect to horizontal. The cart (with mass 1.0 kg) is released 1.0 meter from the bottom and hits the bumper at a speed, v 1 . This time the vacuum/ air generator breaks half-way through and the air stops. The cart only bounces up half as high as where it started.
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sy17_oct31_07hc - Physics 207 Lecture 17 Physics 207...

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