Physics 8 Friday, October 21, 2011

Transcription

1 Physics 8 Friday, October 21, 2011 Bill and Zoey are away next week at Medical Imaging Conference in warm, sunny Valencia, Spain. Simon Hastings (simonhas@sas.upenn.edu) will run the class meetings on October 24, 26, 28. I will still read all of and reply to some of your online reading responses, as usual. You can also feel free to me questions about the reading or the homework. I am also hoping to Skype into one or both of next week s evening homework sessions. HW6 has only 14 questions. But to make it a bit shorter, I give away my own answers to the two hardest questions (#4 and #12) at the end of these lecture notes! Think about Nov 30 (Wed) or Dec 1 (Thu) at 7pm as time to meet with Richard Farley about the connection between physics and architecture / structures. We ll vote once I m back in town.

2 Summary of Chapter 10 results (vector math) Various ways to write a vector: A = (A x, A y ) = A x (1, 0) + A y (0, 1) = A x î + A y ĵ Can separate into two vectors that add up to original, e.g. A x = A x î, Ay = A y ĵ A = A x + A y Scalar product ( dot product ) is a kind of multiplication that accounts for how well the two vectors are aligned with each other: A B = A x B x + A y B y = A B cos(θ AB )

3 Summary of Chapter 10 results (work) In one dimension, we learned W = F x x F x (x) dx Sometimes the force is not parallel to the displacement. For instance, what is the work done by gravity if you slide down a hill? W = F r = F x x + F y y which in the limit of many, many infinitessimal steps becomes W = F ( r) D = (F x (x, y) dx + F y (x, y) dy)

4 Suppose mass = 1 kg. x = +19 m y = 13 m ~ = (+19 m, 13 m) D ~ = (0, 9.8 N) F Sometimes the force is not parallel to the displacement. For instance, what is the work done by gravity if you slide down a hill? ~ D ~ = Fx x + Fy y W =F W = (0 N) (+19 m) + ( 9.8 N) ( 13 m) 130 J Same answer as W = mgh.

5 Summary of Chapter 10 results (friction) Static friction and kinetic (sometimes called sliding ) friction: F Static µ S F Normal F Kinetic = µ K F Normal normal & tangential components are to and to surface (Write these on board.)

6 Let s work through some examples using friction How hard do you have to push a 1000 kg car (with brakes on, all wheels, on level ground) to get it to start to slide? Let s take µ S 1.5 for rubber on dry pavement. F Normal = mg = 9800 N F Static µ S F N = (1.5)(9800 N) N So the static friction gives out (hence car starts to slide) when your push exceeds N. How hard do you then have to push to keep the car sliding at constant speed? Let s take µ K 0.8 for rubber on dry pavement. F Kinetic = µ K F N = (0.8)(9800 N) 8000 N

7 How far does your car slide on dry, level pavement if you jam on the brakes, from 60 mph (27 m/s)? F N = mg, F K = µ K mg a = F K /m = µ K g = (0.8)(9.8 m/s 2 ) 8 m/s 2 Constant force constant acceleration from 27 m/s down to zero: x = v i 2 2a = v 2 f = v 2 i + 2ax (27 m/s)2 2 (8 m/s 2 ) 45 m How much time elapses before you stop? v f = v i + at t = 27 m/s 8 m/s 2 = 3.4 s

8 How does this change if you have anti-lock brakes (or good reflexes) so that the tires never skid? Remember µ S > µ K. For rubber on dry pavement, µ S 1.5 (though there s a wide range) and µ K 0.8. The best you can do is maximum static friction: F S µ S mg a = F S /m = µ S g = (1.5)(9.8 m/s 2 ) 15 m/s 2 Constant force constant acceleration from 27 m/s down to zero: x = v i 2 2a = v 2 f = v 2 i + 2ax (27 m/s)2 2 (15 m/s 2 ) 24 m How much time elapses before you stop? v f = v i + at t = 27 m/s 15 m/s 2 = 1.8 s So you can stop in about half the time (and half the distance) if you don t let your tires skid. Or whatever µ K /µ S ratio is.

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10 Steel on steel µ K is about half that of rubber on concrete, and much less than that of µ S for rubber on concrete. So a train can take a while to skid to a stop! Even more so if the tracks are wet: µ K 0.1 At µ = 0.1 on level ground: 360 m to stop from 60 mph. At µ = 0.1 on 6 slope: not possible to stop.

11 Friction on inclined plane

12 Take x axis to be downhill, y axis to be upward from surface. F gravity = mg cos θ ĵ, F contact = +mg cos θ ĵ F gravity = +mg sin θ î If block is not sliding then friction balances downhill gravity: F static = mg sin θ î

13 Magnitude of normal force ( normal is a synonym for perpendicular ) between surfaces is F N = mg cos θ Magnitude of static friction must be less than maximum: F S µ S F N = µ S mg cos θ Block begins sliding when downhill component of gravity equals maximum magnitude of static friction...

14 Block begins sliding when downhill component of gravity equals maximum magnitude of static friction: µ S mg cos θ = mg sin θ µ S = mg sin θ mg cos θ µ S = tan θ

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16 A heavy crate has plastic skid plates beneath it and a tilted handle attached to one side. Which requires a smaller force to move the box? (A) Pushing the crate is easier than pulling. (B) Pulling the crate is easier than pushing. (C) There is no difference.

17 HW6, Problem 12 There are two homework problems along the lines of previous question (but they re quantitative, not conceptual). 12. A woman applies a constant force to pull a 50 kg box across a floor at constant speed. She applies this force by pulling on a rope that makes an angle of 36.9 above the horizontal, and for the box-floor interface µ k = (a) Find the tension in the rope. (b) What is the work done by the woman as she moves the box 10 m? I ll give you my solution for the harder of those two problems, so that you can see how I solved it....

18 (a) The normal force this time is mg T sin θ because the pulling has an upward vertical component, while in Problem 10 the pushing had a downward vertical component. So then the force of friction is F K = µ K (mg T sin θ). To keep the box sliding at constant velocity, the horizontal component of the pull must balance the force of friction: T cos θ = µ K (mg T sin θ) T = µ K mg cos θ + µ K sin θ Notice that this reduces to the familiar T = µ K mg if θ = 0 (pulling horizontally) and even reduces to a sensible T = mg if θ = 90 (pulling vertically). Plugging in θ = 36.9, T = (0.10)(50 kg)(9.8 m/s2 ) (0.80) + (0.10)(0.60) = 57 N (b) Work is displacement times component of force along direction of displacement (which is horizontal in this case): W = (10 m)(t cos θ) = (10 m)(57 N)(0.80) = 456 J

19 OK, let s move on to another topic from next week s homework (which covers this week s reading): Projectile motion in two dimensions

20 Which path best represents the trajectory of a cantaloupe thrown horizontally off a bridge? (What s wrong with the other two?)

21 HW6, Problem 4 Here s a homework problem (for next week) along the lines of the previous discussion question. 4. An object is thrown at an angle θ (above the horizontal) from the edge of a cliff of height h. If the initial speed of the object is v 0, how far from the base of the cliff does it hit the ground? (Express your answer in terms of v 0, h, and θ.) Again, I ll give you my solution for this one, to save you some time on next week s homework....

22 Vertical and horizontal equations of motion are y f = y i + v iy t 1 2 gt2 x f = x i + v ix t Plug in y f = 0, y i = h, v iy = v 0 sin θ, v ix = v 0 cos θ: x f = (v 0 cos θ)t 0 = h + (v 0 sin θ)t 1 2 gt2 Solving At 2 + Bt + C = 0 with A = g/2, B = v 0 sin θ, C = h: t = 1 ( B ± ) B 2A 2 4AC = 1 g t = 1 g ( v 0 sin θ ± v0 2 sin2 θ 4( g/2)(h) ( ) v 0 sin θ v0 2 sin2 θ + 2gh )

23 Since the term is larger in magnitude than the term without the, the solution has t < 0, which is telling us when the object s parabolic trajectory would intersect the altitude y = 0 inside the cliff the equation doesn t know that the cliff is impenetrable. So we want the + solution. t = 1 ( ) v 0 sin θ + v0 2 g sin2 θ + 2gh x f = (v 0 cos θ)t = v 0 cos θ g ( ) v 0 sin θ + v0 2 sin2 θ + 2gh Whew! If you like, you can let s = v 0 sin θ/ 2gh and get 2h ( x f = v 0 cos θ s + ) s g which makes it clear that for θ = 0 the answer is the same as for Problem 3.

24 Now let s move on to two questions of much more practical significance: 1. Shall I shoot Nim Chimpsky, or Mr. Bill?

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26 Now let s move on to two questions of much more practical significance: 1. Shall I shoot Nim Chimpsky, or Mr. Bill? 2. It takes the pellet some time to travel across the width of the room. In that time interval, gravity will cause Nim/Bill to fall. So where should I aim the pea-shooter so that the pellet hits Nim/Bill as he drops?

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28 What shall I aim for? (A) Aim high, because the steel pellet is so much heavier than Mr. Bill, and will be pulled down more by gravity. (B) Aim low, because Mr. Bill will be falling while the pellet travels. (C) Aim directly for Mr. Bill. This is clearly what you would do if gravity were absent. The presence of gravity will affect Mr. Bill and the pellet in the same way, so aiming directly for Mr. Bill will result in a direct hit. (D) How much below Mr. Bill you need to aim depends on the speed with which you fire the pellet, because the time that it takes the pellet to reach Mr. Bill will depend on how fast the pellet is shot.

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31 Mr. Bill starts from rest at (x i, y i ). Pellet starts at (0, 0) with initial velocity (v i cos θ, v i sin θ). Equations of motion: x bill = x i y bill = y i 1 2 gt2 x pellet = v i cos θ t y pellet = v i sin θ t 1 2 gt2 When does pellet cross Mr. Bill s downward path? x pellet = x bill v i cos θ t = x i t = x i v i cos θ

32 ( ) Plugging in t = xi v i cos θ : x bill = x i ( ) xi x pellet = v i cos θ = x i v i cos θ y bill = y i 1 ( ) 2 g xi 2 v i cos θ ( ) xi y pellet = v i sin θ 1 ( v i cos θ 2 g xi v i cos θ What is vertical separation between Mr. Bill and the pellet at the instant when x pellet = x bill = x i? ( ) xi y bill y pellet = y i v i sin θ v i cos θ y bill y pellet = y i x i tan θ = y i y i = 0 ) 2

33 Have a good week! Be nice to Simon. I ll see you on 10/31.