Lecture 9. Friction in a viscous medium Drag Force Quantified

Transcription

1 Lecture 9 Goals Describe Friction in Air (Ch. 6) Differentiate between Newton s 1 st, 2 nd and 3 rd Laws Use Newton s 3 rd Law in problem solving Assignment: HW4, (Chap. 6 & 7, due 10/5) 1 st Exam Thurs., Oct. 7 th from 7:15-8:45 PM Chapters 1-7 in rooms 2103, 2141, & 2223 Chamberlin Hall Physics 207: Lecture 8, Pg 1 Friction in a viscous medium Drag Force Quantified With a cross sectional area, A (in m 2 ), coefficient of drag of 1.0 (most objects), ρ sea-level density of air, and velocity, v (m/s), the drag force is: D = ½ C ρ A v 2 c A v 2 c = ¼ kg/m 3 in Newtons In falling, when D = mg, then at terminal velocity Example: Bicycling at 10 m/s (22 m.p.h.), with projected area of 0.5 m 2 exerts a force of ~30 Newtons At low speeds air drag is proportional to v but at high speeds it is v 2 Minimizing drag is often important Physics 207: Lecture 8, Pg 2 Page 1

2 Newton s Third Law: If object 1 exerts a force on object 2 (F 2,1 ) then object 2 exerts an equal and opposite force on object 1 (F 1,2 ) F 1,2 = -F 2,1 For every action there is an equal and opposite reaction IMPORTANT: Newton s 3 rd law concerns force pairs which act on two different objects (not on the same object)! Physics 207: Lecture 8, Pg 3 Force Pairs vs. Free Body Diagrams Consider the following two cases (a falling ball and ball on table), Compare and contrast Free Body Diagram and Action-Reaction Force Pair sketch Physics 207: Lecture 8, Pg 4 Page 2

3 Free body diagrams mg F B,T = N mg Ball Falls For Static Situation N = mg Physics 207: Lecture 8, Pg 5 Force Pairs 1 st and 2 nd Laws Free-body diagram Relates force to acceleration 3 rd Law Action/reaction pairs Shows how forces act between objects F B,E = -mg F E,B = mg F B,T = N F T,B = -N F B,E = -mg F E,B = mg Physics 207: Lecture 8, Pg 6 Page 3

4 Example (non-contact) Consider the forces on an object undergoing projectile motion EARTH F B,E = - m B g F E,B = m B g F B,E = - m B g F E,B = m B g Question: By how much does g change at an altitude of 40 miles? (Radius of the Earth ~4000 mi) Physics 207: Lecture 8, Pg 7 Note on Gravitational Forces Newton also recognized that gravity is an attractive, long-range force between any two objects. When two objects with masses m 1 and m 2 are separated by distance r, each object pulls on the other with a force given by Newton s law of gravity, as follows: Physics 207: Lecture 8, Pg 8 Page 4

5 Cavendish s Experiment F = m 1 g = G m 1 m 2 / r 2 g = G m 2 / r 2 If we know big G, little g and r then will can find m 2 the mass of the Earth!!! Physics 207: Lecture 8, Pg 9 Example (non-contact) Consider the force on a satellite undergoing projectile motion 40 km above the surface of the earth: EARTH F B,E = - m B g F E,B = m B g F B,E = - m B g F E,B = m B g Compare: g = G m 2 / g = G m 2 / ( ) 2 g / g = / ( ) 2 / = 0.98 Physics 207: Lecture 8, Pg 10 Page 5

6 A conceptual question: A flying bird in a 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? Follow up question: So, what is holding the airplane up in the sky? Physics 207: Lecture 8, Pg 11 Static Friction with a bicycle wheel You are pedaling hard and the bicycle is speeding up. What is the direction of the frictional force? You are breaking and the bicycle is slowing down What is the direction of the frictional force? Physics 207: Lecture 8, Pg 12 Page 6

7 Exercise Newton s Third Law A fly is deformed by hitting the windshield of a speeding bus. v The force exerted by the bus on the fly is, A. greater than B. equal to C. less than that exerted by the fly on the bus. Physics 207: Lecture 8, Pg 13 Exercise 2 Newton s Third Law Same scenario but now we examine the accelerations A fly is deformed by hitting the windshield of a speeding bus. v The magnitude of the acceleration, due to this collision, of the bus is A. greater than B. equal to C. less than that of the fly. Physics 207: Lecture 8, Pg 14 Page 7

8 Exercise 3 Newton s 3rd Law Two blocks are being pushed by a finger on a horizontal frictionless floor. How many action-reaction force pairs are present in this exercise? a b A. 2 B. 4 C. 6 D. Something else Physics 207: Lecture 8, Pg 15 Force pairs on an Inclined plane Forces on the block Normal means perpendicular Normal Force f Friction Force mg sin θ mg cos θ θ y θ θ x Block weight is mg Physics 207: Lecture 8, Pg 16 Page 8

9 Force pairs on an Inclined plane Forces on the block (equilibrium case) Normal Force f= µn Friction Force Forces on the plane θ y x Physics 207: Lecture 8, Pg 17 Force pairs on an Inclined plane Forces on the block (non equilibrium case) Friction Force f Normal Force Forces on the plane θ y x Physics 207: Lecture 8, Pg 18 Page 9

10 Example: Friction and Motion A box of mass m 1 = 1 kg is being pulled by a horizontal string having tension T = 40 N. It slides with friction (µ k = 0.5) on top of a second box having mass m 2 = 2 kg, which in turn slides on a smooth (frictionless) surface. What is the acceleration of the second box? 1 st Question: What is the force on mass 2 from mass 1? (A) a = 0 N (B) a = 5 N (C) a = 20 N (D) can t tell T m 1 v slides with friction (µ k =0.5 ) a =? m 2 slides without friction Physics 207: Lecture 8, Pg 19 Example Solution First draw FBD of the top box: v N 1 T m 1 f k = µ K N 1 = µ K m 1 g m 1 g Physics 207: Lecture 8, Pg 20 Page 10

11 Example Solution Newtons 3 rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2. As we just saw, this force is due to friction: Reaction f 2,1 = -f 1,2 m 1 Action f 1,2 = µ K m 1 g = 5 N m 2 (A) a = 0 N (B) a = 5 N (C) a = 20 N (D) can t tell Physics 207: Lecture 8, Pg 21 Example Solution Now consider the FBD of box 2: N 2 f 2,1 = µ k m 1 g m 2 m 2 g m 1 g Physics 207: Lecture 8, Pg 22 Page 11

12 Example Solution Finally, solve F x = ma in the horizontal direction: µ K m 1 g = m 2 a a = = 2.5 m/s 2 m 1µ k m 2 g = 5 N 2 kg f 2,1 = µ K m 1 g m 2 Physics 207: Lecture 8, Pg 23 Home Exercise Friction and Motion, Replay A box of mass m 1 = 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 m 2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are µ s =1.5 and µ k = 0.5. The second box can slide freely (frictionless) on an smooth surface. Compare the acceleration of box 1 to the acceleration of box 2? a 1 T m 1 friction coefficients µ s =1.5 and µ k =0.5 a 2 m 2 slides without friction Physics 207: Lecture 8, Pg 24 Page 12

13 Home Exercise Friction and Motion, Replay in the static case A box of mass m 1 = 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 m 2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are µ s =1.5 and µ k = 0.5. The second box can slide freely on an smooth surface (frictionless). If no slippage then the maximum frictional force between 1 & 2 is (A) 20 N (B) 15 N (C) 5 N (D) depends on T a 1 T m 1 friction coefficients µ s =1.5 and µ k =0.5 a 2 m 2 slides without friction Physics 207: Lecture 8, Pg 25 Home Exercise Friction and Motion f S µ S N = µ S m 1 g = 1.5 x 1 kg x 10 m/s 2 which is 15 N (so m 2 can t break free) f s = 10 N and the acceleration of box 1 is Acceleration of box 2 equals that of box 1, with a = T / (m 1 +m 2 ) and the frictional force f is m 2 a (Notice that if T were in excess of 15 N then it would break free) T N m 1 g f S a 1 T m 1 friction coefficients µ s =1.5 and µ k =0.5 a 2 m 2 slides without friction Physics 207: Lecture 8, Pg 26 Page 13

14 Recap Wednesday: Review for exam Physics 207: Lecture 8, Pg 27 Page 14