1. Friction
Pages in text
4/9/2019
Dr. Sasho MacKenzie - HK 376

2. What is Friction Friction is a force
A frictional force can exist when two substances contact each other.
The molecules of each surface interact according to Newton’s Laws of Motion.
Friction always opposes motion, i.e., it is opposite to the direction of velocity.
If there is no motion, then friction opposes the sum of all the other forces which are parallel to the surfaces in contact.
4/9/2019
Dr. Sasho MacKenzie - HK 376

3. > Types of Friction Dry Friction Fluid Friction Static Friction
Occurs with fluids,or lubricated surfaces
Dry Friction
Occurs between the non-lubricated surfaces of solid objects
Static Friction
When dry friction acts between two surfaces that are not moving relative to each other
Dynamic Friction
When dry friction acts between two surfaces that are moving relative to each other
>
4/9/2019
Dr. Sasho MacKenzie - HK 376

4. Contact Force
Force that occurs between objects that are in contact with each other.
Contact forces can be resolved into components that are perpendicular and parallel to the surfaces in contact.
The perpendicular component is called the normal force.
The parallel component is called friction.
4/9/2019
Dr. Sasho MacKenzie - HK 376

5. Contact Force in Running
Resultant force on runner
runner’s push
Normal Force
Friction Force
During the push off phase in running, the normal force acts upward on the runner, while the friction force acts forward on the runner. The friction force is the only force capable of moving the runner horizontally down the track. The normal force can only accelerate the runner upwards.
4/9/2019
Dr. Sasho MacKenzie - HK 376

6. Friction and the Normal Force
The maximum frictional force is proportional to the normal contact force.
An increase in the normal force results in an increase in the maximum friction.
This is because the molecules on the two surfaces are pushed together more, thus increasing their interactions.
4/9/2019
Dr. Sasho MacKenzie - HK 376

7.  Weight means  Normal Force, and therefore,  Maximum Friction
10 kg
5 kg
Surfaces are more compressed together and there are more interactions between molecules
4/9/2019
Dr. Sasho MacKenzie - HK 376

8. Friction and Surface Area
Friction is not affected by the size of the surface area in contact.
If the normal force remains constant, but the contacting surface area is increased, then the normal force is spread out over more molecules, thus the force on each molecule is reduced.
Amontons (1699)
What about race car tires?
4/9/2019
Dr. Sasho MacKenzie - HK 376

9. Calculating Friction Ff_max = FN
Ff_max is the maximum force of friction
 (Mu) is the coefficient of friction
FN is the normal force
Friction can range in value from -Ff_max to +Ff_max
 depends on the types of surfaces that are interacting. It would be low for rubber on ice, but high for rubber on asphalt. It also depends on whether the surfaces are moving relative to each other ( static or dynamic )
4/9/2019
Dr. Sasho MacKenzie - HK 376

10. Friction is not always = FF_max
Fapplied
Assume: FF_max = 50 N
10.2 kg
FFriction
50 N
50 N
10 N
25 N
0 N
50 N
40 N
25 N
10 N
0 N
Static Friction
Dynamic Friction 50
Ffriction
(magnitude)
What is the acceleration? 25
What is the normal force?
Calculate S? 10
Calculate D?
Fapplied
4/9/2019
Dr. Sasho MacKenzie - HK 376

11. 4/9/2019
Dr. Sasho MacKenzie - HK 376

12. Friction Example
A 5 kg block of wood rests on a ceramic counter. If the coefficient of static friction between the block and the counter is 0.4, what horizontal force is necessary to move the block. FN mg Ff Fh
Free body diagram
5 kg Fh
Fy = may
FN – mg = may = 0
FN = mg
Normal force = FN = mg = 5 x 9.81 = 49 N
Fx = max
Fh – Ff = max = 0
Fh = Ff
Fh= Friction force = FN = 0.4 x 49 = N
4/9/2019
Dr. Sasho MacKenzie - HK 376

13. Horse Pulling Cart
According to Newton’s 3rd Law, these forces are equal and opposite. So, if the horse pulls forward on the cart with the same force as the cart pulls back on the horse, how will the horse ever move the cart?
4/9/2019
Dr. Sasho MacKenzie - HK 376

14. Solution
Friction acts on the horse’s feet but very little acts on the wheels of the cart. Drawing a free body diagram reveals the answer. The horse and cart are one system so the forces in between them are internal and cannot produce a change in motion of the system. mg
Friction force resulting from the horse pulling back on the ground N
Force of friction on the wheel which opposes the motion of the horse-cart system
4/9/2019
Dr. Sasho MacKenzie - HK 376

15. Tug of War Fat Bastard vs. Phil Pfister
Pull Force = N
Mass = kg
Height = m
Pfister
Pull Force = N
Mass = kg
Height = m
Both competitors are wearing the same footwear which has a coefficient of friction of 1.5 with the rubber floor they are competing on. If both men employ the same technique, who wins?
4/9/2019
Dr. Sasho MacKenzie - HK 376

16. Two Free Body Diagrams Fat Bastard Pfister 3000 N 3000 N 2060 N 1180 N
Ff = FN
= 1.5 x 1180
= 1770 N
1180 N
2060 N
Ff = FN
= 1.5 x 2060
= 3090 N
4/9/2019
Dr. Sasho MacKenzie - HK 376

17. Fat Bastard Wins
Both competitors have a force of 3000 N pulling on them from the rope.
Fat Bastard’s extra mass gives him a potential friction force (3090 N) which is greater than the force of the rope, so he doesn’t move.
Pfister’s maximum friction force (1170 N) is less than the force of rope, so he is pulled toward Fat Bastard.
4/9/2019
Dr. Sasho MacKenzie - HK 376

18. Would it be better to pull up or down on the rope?
Suppose competitor A was taller than competitor B.
A would be pulling on an upward angle, while B would be pulling on a downward angle.
Who has the advantage?
4/9/2019
Dr. Sasho MacKenzie - HK 376

19. Pulling Up On The Rope Fy = may rope FN – Fy1 – mg = may = 0
FN = mg + Fy1
Friction force = Fx1 = FN
Fx2
Bigger N, means larger friction force
This component increases N mg
Fy1 FN
Fx1
4/9/2019
Dr. Sasho MacKenzie - HK 376

20. Pulling Down On The Rope
Fy = may
FN + Fy1 – mg = may = 0
FN = mg – Fy1
This component decreases N
Fy1
Friction force = Fx1 = FN
Fx2
Smaller N, means less friction force mg FN
Fx1
4/9/2019
Dr. Sasho MacKenzie - HK 376

21. Midterm Example Question
5 kg y
Fx1 x
40
A 5 kg box is being pushed up a 40 incline with an acceleration of 2 m/s/s. If the coefficient of dynamic friction between the incline and box is 0.2, then what is the value of Fx1? Remember that friction always opposes the direction of motion.
4/9/2019
Dr. Sasho MacKenzie - HK 376