1. A person in an elevator begins going down. 2. The person reaches a constant velocity

Transcription

1 Friction.

2 Describe what this person is feeling and why: 1. A person in an elevator begins going down 2. The person reaches a constant velocity 3. The person reaches the bottom. 4. The person then realizes they forgot their phone, so they begin going upwards again 5. The person reaches a constant velocity. 6. The person is back on the floor that they started on.

3 Announcements HW5 is Due Thursday. Absent tomorrow - Testing Returning exams 1 and 2 on Thursday

4 Unit 1: Forces Objectives Part 1 - Dynamics 1 Define mass and inertia and explain the meaning of Newton's 1st Law. 2 Define a force and distinguish between contact forces and field forces. 3 Draw and label a free body diagram showing all forces acting on an object. 4 Determine the resultant of two or more vectors graphically and algebraically. 6 Resolve a vector into perpendicular components: both graphically and algebraically. [Moved to part 2 of forces unit] 7 Use vector diagrams to analyze mechanical systems (equilibrium and non-equilibrium + balanced and unbalanced). 8 Verify Newton's Second Law for linear motion. 9 Describe how mass and weight are related. 10 Define friction and distinguish between static and kinetic friction. 11 Determine the coefficient of friction for two surfaces. 12 Explain forces in terms of action-reaction pairs

5 Friction.

6 Two Types

7 Static Static = fixed, stable, not changing Static friction =

8 Kinetic moving! Kinetic Friction =

9 Which direction does friction act?

10 Which direction does friction act? It always opposes the motion.

11 Simulation: phet.colorado.edu/sims/html/ forces-and-motion-basics/latest/ forces-and-motion-basics_en.html

12 Static and Kinetic Friction Graph

13 Do all materials have the same amount of friction?

14 Coefficient of Friction (p1) 1. What is the coefficient of friction for a steel block sliding along a steel floor? 2. What is the coefficient of friction for a rubber tire at rest on a dry concrete surface?

15 Formula for Friction Ffriction= μ Fnormal Where μ represents the coefficient of friction (either kinetic or static)

16 Example 1 A N car with rubber tires is resting in a garage with a dry, concrete floor. What is the maximum force that can be applied to the car without it sliding across the floor?

17 A N car with rubber tires is resting in a garage with a dry, concrete floor. What is the maximum force that can be applied to the car without it sliding across the floor? Fapplied Ffriction

18 A N car with rubber tires is resting in a garage with a dry, concrete floor. What is the maximum force that can be applied to the car without it sliding across the floor? Fapplied Ffriction Fg = 12000N Fapplied =?

19 A N car with rubber tires is resting in a garage with a dry, concrete floor. What is the maximum force that can be applied to the car without it sliding across the floor? Fapplied Ffriction Fg = 12000N FN = 12000N Fapplied =?

20 A N car with rubber tires is resting in a garage with a dry, concrete floor. What is the maximum force that can be applied to the car without it sliding across the floor? Fapplied Ffriction Fg = 12000N FN = 12000N Fapplied =? Ffriction =?

21 A N car with rubber tires is resting in a garage with a dry, concrete floor. What is the maximum force that can be applied to the car without it sliding across the floor? Fapplied Ffriction Fg = 12000N FN = 12000N Fapplied =? Ffriction =? M = 0.9

22 A N car with rubber tires is resting in a garage with a dry, concrete floor. What is the maximum force that can be applied to the car without it sliding across the floor? Fapplied Ffriction Fg = 12000N FN = 12000N Fapplied =? Ffriction =? M = 0.9

23 A N car with rubber tires is resting in a garage with a dry, concrete floor. What is the maximum force that can be applied to the car without it sliding across the floor? Fapplied Ffriction Fg = 12000N FN = 12000N Fapplied =? Ffriction =? M = 0.9 Ff = (0.9)(12000N) Ff = 10800N

24 Example 2 A 400 kg copper block is sliding on a level, steel floor. Calculate the magnitude of the horizontal force required to continue to move the crate across the floor at a constant speed of 2 meters per second.

25 A 400 kg copper block is sliding on a level, steel floor. Calculate the magnitude of the horizontal force required to continue to move the crate across the floor at a constant speed of 2 meters per second.

26 A 400 kg copper block is sliding on a level, steel floor. Calculate the magnitude of the horizontal force required to continue to move the crate across the floor at a constant speed of 2 meters per second. m = 400kg Fapplied Ffriction

27 A 400 kg copper block is sliding on a level, steel floor. Calculate the magnitude of the horizontal force required to continue to move the crate across the floor at a constant speed of 2 meters per second. m = 400kg Fapplied Ffriction Fapplied =? Fapplied = Ffriction Fg =(400kg)(9.8m/s 2 ) = 3920N Fn = Fg M = 0.36

28 A 400 kg copper block is sliding on a level, steel floor. Calculate the magnitude of the horizontal force required to continue to move the crate across the floor at a constant speed of 2 meters per second. m = 400kg Fapplied Ffriction Ff = (0.36)(3920N) Fapplied =? Fapplied = Ffriction Fg =(400kg)(9.8m/s 2 ) = 3920N Ff = N Fn = Fg M = 0.36