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 amount of friction depends on the material in contact.
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. Friction between objects when
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
m = 1 kg
FP > 9 N to overcome
the static force.
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
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. 2. What does the force of the push need to be to overcome the static friction in this situation.
Fg = mg = (20 kg)(10 m/s2) = 200 N
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
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 = 0FN
ΣF = Ffk – F1 = 0
0 = Ffk – F1
Ffk N = 0
Ffk
F1 = 60 N
60 N = Ffk
Ffk = µFN Fg
FN = Fg
FN = Ffk
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
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.