3. Friction Friction is the force that opposes the motion between two surfaces that touch. Friction forces are parallel to the contact surface and opposite the direction of motion It allows you to walk, turn a corner on your bike, warm your hands in the winter.

4. What Causes Friction? The surface of any object is rough. Even an object that feels smooth is covered with tiny hills and valleys. The contact between the hills and valleys of two surfaces causes them to stick, resulting in friction. The surface area in contact has negligible effect on friction!

5. The amount of friction depends on: –Roughness of the surfaces –Force pushing the surfaces together What Causes Friction?

6. Kinetic friction occurs when force is applied to an object and the object moves. Examples: Sliding Friction: pushing an object across a surface Rolling Friction: between wheels and a surface Fluid Friction: opposes the motion of objects traveling through a fluid (air or water) Types of Friction

7. Static friction occurs when force applied to an object does not cause the object to move. Types of Friction

8. To reduce the amount of friction, apply a lubricant between two surfaces. Motor oil, wax, and grease are examples. Friction can also be reduced by rolling, rather than pushing, an object. Affecting Friction

9. Coefficients of Friction Static coefficient …  s. Kinetic coefficient …  k. Both depend on the materials in contact. –Small for steel on ice or scrambled egg on Teflon frying pan –Large for rubber on concrete or cardboard box on carpeting The bigger the coefficient of friction, the bigger the frictional force.

10. Static Friction Force F fs   s F N static frictional force coefficient of static friction normal force f s, max =  s N maximum force of static friction f s, max is the force you must exceed in order to budge a resting object.

11. Static friction force varies f s, max is a constant in a given problem, but f s varies. f s matches F A until F A exceeds f s, max. Example: In the picture below, if  s for a wooden crate on a tile floor is 0.6, f s, max = 0.6 (10 ) (9.8) = 58.8 N. 10 kg F A = 27 Nf s = 27 N 10 kg F A = 43 Nf s = 43 N 10 kg F A = 66 N The box finally budges when F A surpasses f s, max. Then kinetic acts on the box. fkfk

12. Kinetic Friction Once object budges, forget about  s. Use  k instead. f k is a constant so long as the materials involved don’t change. There is no “maximum f k.” f k =  k N kinetic frictional force coefficient of kinetic friction normal force

13.  values Typically, 0 <  k <  s < 1. This is why it’s harder to budge an object than to keep it moving. If  k > 1, it would be easier to lift an object and carry it than to slide across the floor. Dimensionless (  ’s have no units, as is apparent from f =  N).

14. Friction Example 1 Barrel o’ Monkeys 14.7 kg You push a giant barrel o’ monkeys setting on a table with a constant force of 63 N. If  k = 0.35 and  s =0.58, when will the barrel have moved 15 m? answer: Never, since this force won’t even budge it! 63 < 0.58 (14.7) (9.8)  83.6 N

15. Friction Example 2 Barrel o’ Monkeys 14.7 kg Same as the last problem except with a bigger F A : You push the barrel o’ monkeys with a constant force of 281 N.  k = 0.35 and  s =0.58, same as before. When will the barrel have moved 15 m? step 1: f s, max = 0.58 (14.7) (9.8)  83.6 N step 2: F A = 281N > f s, max. Thus, it budges this time. step 3: Forget f s and calculate f k : f k = 0.35 (14.7) (9.8) = 50.421 N (continued on next slide)

16. Friction Example 2 (continued) step 4: Free body diagram while sliding: mg N F A fkfk step 5: F net = F A – f k = 281 - 50.421 = 230.579 N Note: To avoid compounding of error, do not round until the end of the problem. step 6: a = F net / m = 230.579 / 14.7 = 15.68564 m/s 2 step 7: Kinematics:  x = +15 m, v 0 = 0, a = +15.68564 m/s 2, t = ?  x = v 0 t + ½ a t 2  t = 2  x / a  1.38 s

17. Friction Conceptual Example Two cars, identical in mass and initial speed are traveling along two different roads. The road the first car is on is dry and the road the second car is on is wet. Why does it take longer for the 2 nd car to come to a stop? The wet road has a lower coefficient of friction and since the two cars have the same mass, they will have the same normal force so F f = µF N will be lower for the second car than for the first and since the force of friction is lower, there is less force slowing the car down and therefore less negative acceleration (by Newton’s second law) and it will take longer to stop

18. Friction as the net force example A runner is trying to steal second base. He’s running at a speed of 8 m/s; his mass is 70 kg. The coefficient of kinetic friction between his uniform and the base is 0.6. How far from second base should he begin his slide in order to come to a stop right at the base? mg N fkfk KEY: Once the slide begins, there is no applied force. Since N and mg cancel out, f k is the net force. So Newton’s 2 nd Law tells us: f k = m a. But the friction force is also given by f k =  N =  m g.

19. Friction as the net force (cont.) Therefore,  m g = m a. Mass cancels out, meaning the distance of his slide is completely independent of how big he is, and we have a =  g. (Note that the units work out since  is dimensionless.) This is just the magnitude of a. If the forward direction is positive, his acceleration (which is always in the direction of the net force) must be negative. So, a = -  g.

20. Friction as the net force (last) Since he comes to rest at 2 nd base, v f = 0. v f 2 - v 0 2 = 2 a  x  0 - 8 2 = -2  g  x   x = 8 2 / (2(.6)(10))   x = 8 2 / (2(.6)(10))=5.3m Note the slide distance is inversely proportional to the coefficient of friction, which makes sense, since the bigger  is, the bigger f is. Note also that here v and F net are in opposite directions, which is perfectly fine.

21. Example The man pushes/pulls with a force of 200 N. The child and sled combo has a mass of 30 kg and the coefficient of kinetic friction is 0.15. For each case: What is the frictional force opposing his efforts? What is the acceleration of the child? f=59 N, a=3.80 m/s 2 / f=29.1 N, a=4.8 m/s 2