Friction

1. Friction

2. Friction: • Friction is the force that opposes motion. • Without friction a car wouldn’t be able to start or stop (without hitting something). • Ff • Vector- shows magnitude and direction.

3. The force of friction is always parallel to the 2 surfaces that are in contact. • The amount of friction depends on the material in contact. The friction between those surfaces is going to be different than friction between metal and ice. Here the material in contact is rubber (tires) on concrete.

4. Two kinds of friction: Kinetic Friction (Ffk) Static Friction (Ffs) moving friction Not moving friction Friction between objects when they are in motion. Opposes the start of motion

5. Ff is proportional to the FN: Ff = µFN Coefficient of friction = ratio of the Ff to the FN- doesn’t have a unit

7. The µk is less than the µs so the Ffk is less than the Ffs. Think about how much force it would take to initially get a car to start rolling. But after it starts it’s easier to keep it moving.

9. Rubber on Concrete example: µs = .90 With how much force does this guy need to pull to get this ball to start moving? FN = 10 N FP > 9 N to overcome the static force. m = 1 kg Ffs = µFN = (.90) (10 N) = 9 N FP < 9 N it will not move Fg = mg = 1 kg (10 m/s2) = 10 N

10. What if the ball was already moving? µk = .68 FN = 10 N Now to keep the ball rolling with a constant velocity the guy only needs FP of 6.8 N m = 1 kg Ffk = µFN = (.68) (10 N) = 6.8 N Fg = mg = 1 kg (10 m/s2) = 10 N

11. 1. If the static coefficient of a wood on wood surface interaction is .42 than what does the force of the pull have to be to overcome the static friction. FN Ffs = µFN = (.42) (10 N) = 4.2 N Ffs FP m = 1 kg Fg

12. Fg = mg = (20 kg)(10 m/s2) = 200 N 2. What does the force of the push need to be to overcome the static friction in this situation. Fg = FN = 200 N µs = .42 Ffs = µFN = (.42) (200 N) = 84 N FP > 84 N FN FP m = 20 kg Ffs Fg

13. 3. The car below is moving with a constant velocity of 5 m/s. What is F1? Constant velocity = ΣF = 0 ΣF = Ff – F1 = 0 0 = Ff – F1 10 N – F1 = 0 10 N = F1 10 N F1

14. 4. A constant horizontal force of 60 N is applied to a wooden block which causes it to slide at a constant speed across a wooden table. Determine the weight of the block? (The Ffk for wood on wood is .3) FN Ffk F1 = 60 N Fg

15. Constant velocity = ΣF = 0 FN ΣF = Ffk – F1 = 0 0 = Ffk – F1 Ffk - 60 N = 0 Ffk F1 = 60 N 60 N = Ffk Ffk = µFN Fg FN = Fg FN = Ffk µ What’s the mass of the block? Fg = 200 N F = mg m = F g = 60 N .3 m = 200 N 10 m/s2 FN = 200 N m = 20 kg

16. 5. Luke uses "the force" to magically lift a 4.0-kg rock 10 m off the surface of Dagobah. The magnitude of "the force" required is 20 N. If he now drops the rock, how long will it take to hit the ground? F = ma or F= mg 20N = 4.0kg (g) g = 5 m/s2 F lift = 20 N m = 4.0 kg dy = 10 m Vi = 0 a or g = ? t = ? d = vit + ½ a t 2 10 m = 0 + ½ (5m/s2) t2 t2= 2d a t = 2 s

17. How much force does the guy need to push with to overcome the force of friction? µ = .85 FA = m = 10 kg Ffs = What if the guy pushes with a force of 70 N? With what force is the box pushing back at him?

18. This car is moving with a constant velocity so the force applied equals the force of friction. Where’s the friction coming from? Air friction 100 N 100 N FA

19. During freefall terminal velocity (approx. 200km/h) is reached when the force of friction equals the force of gravity. Forces are balanced and you are no longer accelerating.