Midterm Exam #1. First midterm is next Wednesday, September 23 (80 minutes!!!)

Transcription

1 Midterm Exam #1 First midterm is next Wednesday, September 23 (80 minutes!!!) Ø Ø Units 1-5: 1D kinematics Forces and FBDs (todays lecture) Thins to brin: Pencil(s), calculator UofU ID card Ø Ø Ø Brin your U of U ID card: you will not receive a rade unless you show it upon turnin in your exam No makeups!!! (Unless you are unavailable due to University/military business or documented medical emerency come talk to me!) Cheatin: You will receive a zero and referred to the University administration may result in an E for the course and expulsion You must contact Prof. Gerton by Wednesday, September 14 if you have a conflict wit the exam times/days!!! Exam Location: SFEBB 1110 (lare lecture hall on first floor of Spencer Fox Eccles Business Buildin) Mechanics Lecture 6, Slide 1

2 Classical Mechanics Lecture 6: Friction Today's Concept: Fric%on Mechanics Lecture 6, Slide 2

3 Time spent on Prelecture #6 Averae for quiz 1 = 8.4/15 (class averae = 11.6) Averae for quiz 2 = 8.9/15 (class averae = 10.4) Mechanics Lecture 6, Slide 3

4 Unit 6 Learnin Objectives If you master this unit, you should: be able to compute the force of static friction between two objects by solvin Newton s 2 nd Law. be able to determine the maximum force that can be applied to an object before it starts to slide. be able to compute the acceleration of one object slidin over another in the presence of friction. Mechanics Lecture 6, Slide 4

5 Friction Mechanics Lecture 6, Slide 5

6 Example problem A car of mass m drives around a circular track of radius R. The coefficient of static friction between the car s tires and the road is µ s. The coefficient of kinetic friction is µ k. What is the maximum speed of the car before it starts to skid? Mechanics Lecture 6, Slide 6

7 1&2: Understand Problem & Describe Physics A car of mass m drives around a circular track of radius R. The coefficient of static friction between the car s tires and the road is µ s. The coefficient of kinetic friction is µ k. What is the maximum speed of the car before it starts to skid? 1. What kind of friction are we dealin with? 2. Between which objects does this friction act? 3. As the car oes around the track, in which direction does the friction force point? Ø Which object are we talkin about? Ø Is the car maintainin constant speed? Ø How does the situation chane if the car speeds up or slows down? 4. How is the maximum friction force related to the car s weiht? 5. Should the maximum speed depend on the car s mass? Mechanics Lecture 6, Slide 7

8 Step 2: Describe the Physics A car of mass m drives around a circular track of radius R. The coefficient of static friction between the car s tires and the road is µ s. The coefficient of kinetic friction is µ k. What is the maximum speed of the car before it starts to skid? a = v2 R f s,max = µ s N = µ s m Mechanics Lecture 6, Slide 8

9 Steps 3 & 4: Plan and Execute the Solution A car of mass m drives around a circular track of radius R. The coefficient of static friction between the car s tires and the road is µ s. The coefficient of kinetic friction is µ k. What is the maximum speed of the car before it starts to skid? The maximum possible centripetal acceleration occurs at the maximum speed, v max This is when the static friction force is maximized. a max = v2 max R A) B) v max = µ s p R v max = µ s p R f s,max = µ s N = µ s m C) v max = µ s R p D) v max = p µ s R Mechanics Lecture 6, Slide 9

10 Follow up question A car of mass m drives around a circular track of radius R. The coefficient of static friction between the car s tires and the road is µ s. The coefficient of kinetic friction is µ k. What is the force of static friction on the car when it travels at half the maximum speed? Mechanics Lecture 6, Slide 10

11 Stuff you asked about: I struled with the trionometry on the box slidin down the ramp. i thouht that this was a ood lecture, very informa%ve, but could we have an example, just one li>le example, that doesn't involve a box? this lecture was very clear thouh. y 90 θ 90-θ 90 x θ m Mechanics Lecture 6, Slide 11

12 CheckPoint: Box in Acceleratin Truck A box sits on the horizontal bed of a movin truck. Sta%c fric%on between the box and the truck keeps the box from slidin around as the truck drives. a µ S If the truck moves with constant accelera%on to the lee as shown, which of the followin diarams best describes the sta%c fric%onal force ac%n on the box: A B C Mechanics Lecture 6, Slide 12

13 Revote: Box in acceleratin truck A C B D a µ S If the truck moves with constant acceleration to the left as shown, which of the followin diarams best describes the static frictional force actin on the box: A B C ~a = ~ F net M A) The friction is causin the box to accelerate in the same direction as the truck. B) The force of friction always acts in the opposite direction of acceleration. Mechanics Lecture 6, Slide 13

14 ACT! A flower pot of mass M sits on a horizontal table. A horizontal strin havin tension T applies a force on the pot, but static friction between the pot and the table keeps the pot from movin. A C B D What is the manitude of the net force actin on the pot? A) M B) μ s M C) T D) 0 f N T W = M X Fy = Ma =0) N = M X Fx = Ma x =0) T = f Since acceleration is zero, the net force MUST be zero also! Mechanics Lecture 6, Slide 14

15 CheckPoint: Box on Table A box of mass M sits on a horizontal table. A horizontal strin havin tension T applies a force on the box, but static friction between the box and the table keeps the box from movin. What is the manitude of the static frictional force actin on the box? A C B D A) M B) µ s M f M T C) T D) 0 Apply Newton s 2 nd Law: X =0 Fx = Ma x =0) T = f Mechanics Lecture 6, Slide 15

16 CheckPoint: Box on Table A box of mass M sits on a horizontal table. A horizontal strin havin tension T applies a force on the box, but static friction between the box and the table keeps the box from movin. What is the manitude of the static frictional force actin on the box? F f T M X Fx = Ma x =0) T = F f F f depends on weiht µ s does not!!! What happens if I pull harder? As lon as v = 0: The static friction force will continue to match T until F f = µ s N. After that, F f can t increase any more: Ø T > F f and the box will accelerate! Mechanics Lecture 6, Slide 16

17 Example problem A block ( ) slides on a table pulled by a strin attached to a mass ( ) hanin over the side. The coefficient of kinetic friction between the slidin block and the table is µ k. What is the tension in the strin? Mechanics Lecture 6, Slide 17

18 CheckPoint: Blocks and Pulley A block slides on a table pulled by a strin a>ached to a hanin weiht. In Case 1 the block slides without fric%on and in Case 2 there is kine%c fric%on between the slidin block and the table. Case 1 (No Friction) Case 2 (With Friction) In which case is the tension in the strin larer? A) Case 1 B) Case 2 C) Same 64% of you ot this riht Mechanics Lecture 6, Slide 18

19 ACT: Blocks and Pulley, I A block slides on a table pulled by a strin a>ached to a hanin weiht. In Case 1 the block slides without fric%on and in Case 2 there is kine%c fric%on between the slidin block and the table. + a a is positive when accelerates downward! A C B D Case 1 (No Friction) a T + T = a 1 = a =) T = ( a) What is the tension in the strin for Case 1? A) T = 0 B) T = C) 0 < T < Mechanics Lecture 6, Slide 19

20 ACT: Blocks and Pulley, II A block slides on a table pulled by a strin a>ached to a hanin weiht. In Case 1 the block slides without fric%on and in Case 2 there is kine%c fric%on between the slidin block and the table. A C B D Case 1 (No Friction) Case 2 (With Friction) In which case is the acceleration of the blocks larer? A) Case 1 B) Case 2 C) Same Mechanics Lecture 6, Slide 20

21 CheckPoint revisited: Blocks and Pulley A C B D Case 1 (No Friction) Case 2 (With Friction) In which case is the tension in the strin larer? A) Case 1 B) Case 2 C) Same One of your answers: B) M1 will not be accelera%n as fast in case two than in case one because there is the extra force of fric%on ac%n aainst mass 1. Since the accelera%on is smaller in case two, there has to be more of a force ac%n aainst ravity, the only other possible force is Tension. Mechanics Lecture 6, Slide 21

22 ACT! A block ( ) slides on a table pulled by a strin attached to a mass ( ) hanin over the side. The coefficient of kinetic friction between the slidin block and the table is µ k. What is the tension in the strin? T 1 A C B D What is the relationship between the manitude of the tension of the strin at block 2 and the manitude of the tension in the strin at block 1? T 2 A) T 1 > T 2 B) T 1 = T 2 C) T 1 < T 2 Mechanics Lecture 6, Slide 22

23 ACT: A block ( ) slides on a table pulled by a strin attached to a mass ( ) hanin over the side. The coefficient of kinetic friction between the slidin block and the table is µ k. What is the tension in the strin? A C B D What is the relationship between the manitudes of the acceleration of the two blocks? A) a 1 = a 2 B) a 1 < a 2 C) a 1 > a 2 Mechanics Lecture 6, Slide 23

24 Back to example problem A block ( ) slides on a table pulled by a strin attached to a mass ( ) hanin over the side. The coefficient of kinetic friction between the slidin block and the table is µ k. What is the tension in the strin? Mechanics Lecture 6, Slide 24

25 1) FBD N f T T Mechanics Lecture 6, Slide 25

26 1) FBD 2) ΣF=ma f N T T N = T µ = a T = a add µ = a + a a = µ + Mechanics Lecture 6, Slide 26

27 1) FBD 2) ΣF=ma f N T T T = a a = µ + T = a T is smaller when a is bier Mechanics Lecture 6, Slide 27