What do you Remember about Friction Friction opposes motion Friction is dependent on the texture of the surfaces Friction is dependent on normal force motion friction What do you Remember about Friction ID: 1001860
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1. Friction
2. What do you Remember about Friction?
3. What do you Remember about Friction?Friction opposes motionFriction is dependent on the texture of the surfacesFriction is dependent on normal forcemotionfriction
4. What do you Remember about Friction?Friction opposes motionFriction is dependent on the texture of the surfacesFriction is dependent on normal forceFfr = m Fn μ is called coefficient of friction m has no units depends on characteristics of both surfacesHigher m = rougher surface / more frictionmotionfrictionNote: Friction does NOT depend on the surface area of contact
5. surface-on-surfaceμs μkhook velcro-on-fuzzy velcro>6.0>5.9avg tire-on-dry pavement0.90.8grooved tire-on-wet pavement0.80.7smooth tire-on-wet pavement0.50.4metal-on-metal (lubricated)0.10.05steel-on-ice0.10.05steel-on-Teflon0.050.05The coefficient of friction is different for when an object is at rest and when it is moving. μs = coefficient of static friction (object at rest)μk = coefficient of kinetic friction (object moving)Static friction is greater than kinetic friction - its harder to start an object moving than it is to keep it moving.
6. Friction Problems – We DoA 28 kg crate initially at rest on a horizontal floor requires a 75 N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor.What do we do first?
7. Friction Problems – We DoA 28 kg crate initially at rest on a horizontal floor requires a 75 N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor.Remember our strategy:Draw a free body diagramIdentify all variablesIdentify relevant equationsSolve!Check your work!
8. Friction Problems – We DoA 28 kg crate initially at rest on a horizontal floor requires a 75 N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor.μs = 0.27
9. Friction Problems – We DoA force of 40.0 N accelerates a 5.0-kg block at 6.0 m/s2 along a horizontal surface.a. How large is the frictional force?b. What is the coefficient of friction?
10. Friction Problems – We DoA force of 40.0 N accelerates a 5.0-kg block at 6.0 m/s2 along a horizontal surface.a. How large is the frictional force?b. What is the coefficient of friction? Ff = 10 N μk = 0.2
11. Friction Problems – We Do2) A 12 kg suitcase is pushed with a force of 38 N to the left. If the coefficient of kinetic friction between the suitcase and the floor is 0.3, how far will the suitcase move after 5 sec? 2.8 m
12. Check your understanding
13. Check your understanding
14. Friction Problems – You DoFOLLOW THE STEPS! Make a plan before you plug in numbers! Be able to explain your reasoning!A 30 kg crate requires a 53 N force to keep it moving at 1 m/s. Find the coefficient of kinetic friction.You need to move a 105-kg sofa to a different location in the room. It takes a 403-N force to start the sofa moving. What is the coefficient of static friction between the sofa and the carpet?mk = 0.18
15. Elevator problemQuestion: How does the weight of a person in an elevator depend on the motion of that elevator?What will the scale show if the elevator isat rest or moving with constant speedspeeding upslowing downNewton’s 3. law: Force with which the person acts on the scale (reading of the scale) is equal to the normal force on the person. So, if we find normal force we know the reading of the scale, so called APPARENT WEIGHTEXAMPLE of FBD
16. Let’s assume that elevator is moving upward, and let this be positive direction. 1. draw free body diagram 2. apply Newton’s 2. law : Fnet = ma Fn – mg = ma = 0 → Fn = mg apparent weight = weight+1. elevator is at rest or moving with constant speed2. elevator is speeding up: a is positiveFn – mg = ma → Fn = mg + ma apparent weight > weightthe scale would show more, and you would feel heavier3. elevator is slowing down: a is negativeFn – mg = - ma → Fn = mg - ma apparent weight < weightthe scale would show less, and you would feel lighterFnFnFn
17. Check your understanding
18. You DoYou are riding in an elevator holding a spring scale with a 1kg mass suspended from it. You look at the scale and see that it reads 9.3 N. What, if anything, can you conclude about the elevator’s motion at that time?
19. Exit Ticket 1) Solve the elevator problem for an elevator traveling downward. YOU MUST EXPLAIN YOUR REASONING USING NEWTON’S 2nd LAW2) Write down 2 things you learned friction problems.