1. Force of Friction

2. Introduction.
Friction is a force that appears whenever two surfaces, in contact, try to slide past one another.
Friction resists the slipping/sliding.

3. Friction
Friction is the result of small (or large) irregularities in surfaces hooking on to one another as the surfaces try to slide past one another,
the rougher the surface, the more the irregularities bump into one another.
Friction can also be the result of electrostatic attraction between two surfaces. Sometimes the smoother the surfaces, the greater the friction. For example two sheets of glass have a high amount of friction.
In both case, friction is an expression of electromagnetism…

4. Friction Depends on: The two surfaces sliding past one another.
How hard the two surfaces are “pushed together”
Another name for this is….. NORMAL FORCE!

5. 𝐹 𝑓 =𝜇 𝐹 𝑁 Friction The equation for Friction is: Where
Ff is the frictional force.
FN is the Normal force.
μ is the “coefficient of friction”
μ is the greek letter “mu” – pronounced Miew.

6. μ
That’s right, Miew

7. 𝐹 𝑓 =𝜇 𝐹 𝑁 Friction The equation for Friction is: Where
Notice, the only thing that friction depends on is Normal force and μ. It has nothing to do with anything else (such as surface area, or amount of contact).

8. Coefficient of friction
The coefficient of friction (𝜇) is the ratio of the frictional force to the normal force.
𝜇= 𝐹 𝑓 𝐹 𝑁
𝜇 is dimensionless (it has no units)
It is experimentally determined (this means that for any two surfaces we have to do an experiment to figure it out).
It is usually between 0 and 1, although can go above 1 for some sticky materials.
It depends on the two surfaces that are in contact
eg rubber and asphalt can have a μ of up to 0.9
Whereas two Teflon coated surfaces can have a μ as low as 0.04.

9. Problems
A 20 kg box is pulled across a surface at a constant velocity by a horizontal force of 75N, determine
A) Force of friction.
B) Coefficient of friction.
Work this out with your group before you go any further… but after struggling for a while, you are welcome to get some hints from the next slides.

10. Hints… Remember to draw FBD.
This is still a forces (F=ma problem) so try to frame this as a winners – losers question.
Is there some clue about the acceleration of the box?
Always consider vertical and horizontal planes totally separate…

11. Hints part 2 Is the box accelerating up or down?
So what does that tell you about the upwards forces and the downwards forces?
Is the box accelerating left or right?
So what does that tell you about the left and the right forces?

12. Solution
The object is not accelerating left or right (“Constant velocity”), so we can say
forces left = forces right
so 𝐹 𝑓 = 𝐹 𝐴
𝐹 𝑓 =75N

13. Solution
Now we know Ff lets use 𝐹 𝑓 =𝜇 𝐹 𝑁 to get coefficient of friction
The object is not accelerating up or down, so we can say
forces up = forces down
so 𝐹 𝑁 = 𝐹 𝑔
𝐹 𝑁 =mg
𝐹 𝑁 =20x9.8
so 𝜇= 𝐹 𝑓 𝐹 𝑁
so 𝜇= 75 (20)(9.8)
𝜇=0.38

14. Problems
An 85.0 kg object is pulled box is pulled across a surface by a force of 417 N. If the coefficient of friction between the object and the floor is 0.45, determine the acceleration of the object.
Work this out with your group before you go any further… but after struggling for a while, you are welcome to get some hints from the next slides.

15. Solution This is much more of a classic “push pull problem”
So you want to phrase it as a “winners – losers” type problem.
To know whether FA or Ff is the loser, you need to know the frictional force.
𝐹 𝑓 =𝜇 𝐹 𝑁 to get coefficient of friction.
Frist we need to get FN…
The object is not accelerating up or down, so we can say
forces up = forces down
so 𝐹 𝑁 = 𝐹 𝑔
𝐹 𝑁 =mg
𝐹 𝑁 =85x9.8
so 𝐹 𝑁 =85x9.8
𝐹 𝑁 =833N
𝐹 𝑓 =𝜇 𝐹 𝑁
𝐹 𝑓 =0.45x833N
𝐹 𝑓 = N

16. Solution Now turn this into a winners – losers problem. 𝐹 𝑛𝑒𝑡 =𝑚𝑎
𝐹 𝑛𝑒𝑡 =𝑊𝑖𝑛𝑛𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 −𝑙𝑜𝑠𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒
𝐹 𝑛𝑒𝑡 = 𝐹 𝐴 − 𝐹 𝑓
𝑚𝑎= 𝐹 𝐴 − 𝐹 𝑓
𝑎= 𝐹 𝐴 − 𝐹 𝑓 𝑚
𝑎= 417−
𝑎=0.50𝑚 𝑠 −1

17. Discuss…
Give 2 reasons why it is easier to push a box across a surface by exerting a horizontal force on it (Example A) than when you are pushing down on it at an angle (Example B)
(don’t ask me why the boxes or the people are different sizes, they assume the boxes in both cases are the same mass.)
Example A
Example B
Draw some FBDs to figure this out…

18. Problem 3 This one’s a bit trickier….
Now, a student is pulling a 25 kg box along a surface with a coefficient for friction of The student pulls with a cord that makes a 60º angle to the horizontal. The student applies a force of 210 N.
What’s the acceleration of the box?

19. Hint… This one’s going to require you to do components…
Break the applied force into x and y components…

20. Solution First break FA into components 𝐹 𝐴𝑥 = 𝐹 𝐴 cos⁡(60)
𝐹 𝐴𝑦 = 𝐹 𝐴 sin⁡(60)
𝐹 𝐴𝑦 =210sin⁡(60)

21. Solution
Now it becomes a classic push/pull problem, we simply have to take our components into account.
𝐹 𝑛𝑒𝑡 =𝑚𝑎
𝐹 𝑛𝑒𝑡 = 𝐹 𝐴𝑥 − 𝐹 𝑓
𝑚𝑎= 𝐹 𝐴𝑥 − 𝐹 𝑓
And we can use 𝐹 𝑓 =𝜇 𝐹 𝑁 to get Ff
Except Now there is an upwards force so FN is not Fg

22. Solution
But since there is now an extra upwards force, FN has also changed. The object is not accelerating up or down, so we can say
forces up = forces down
so 𝐹 𝑁 + 𝐹 𝑎𝑦 = 𝐹 𝑔
𝐹 𝑁 = 𝐹 𝑔 − 𝐹 𝑎𝑦
𝐹 𝑁 =25×9.8−210sin⁡(60)
𝐹 𝑁 =63.13
𝐹 𝑓 =𝜇 𝐹 𝑁
𝐹 𝑓 =.44(63.13)
𝐹 𝑓 =27.78
In this example, Fay and FN are in the same direction… This is not always the case!

23. Solution 𝐹 𝑛𝑒𝑡 = 𝐹 𝐴𝑥 − 𝐹 𝑓 𝑚𝑎=210 cos 60 −27.78
𝑎=3.1 𝑚 𝑠 −2

24. Static vs Kinetic friction.
It turns out that in the “real world”, it takes a greater force to start something moving than it does to keep it moving at a constant velocity.
Materials have two different coefficients of friction, 𝜇s and 𝜇k.
𝜇s is the coefficient of static friction.
this governs the maximum friction that exists between surfaces when there is no slippage between surfaces.
𝜇k is the coefficient of kinetic friction.
this governs the friction that exists between surfaces when there is slippage between surfaces.
For example for a tire on a wet road 𝜇s = 𝜇k = 0.40

25. Static vs Kinetic Friction
The graph on the left shows the frictional force that exists on an object as an external force is applied to it.
As applied force increases, so does does frictional force (while the object remains stationary).
Frictional force reaches a maximum value 8.5 N.
This is the maximum “static friction” acting on the box and is governed by the coefficient of static friction.
𝐹 𝑓𝑠 = 𝜇 𝑠 𝐹 𝑁

26. Static vs Kinetic Friction
Once the object starts to move, the frictional force drops down to an average value of about 5.4 N. This value is governed by the coffecient of kinetic friction.
𝐹 𝑓𝑘 = 𝜇 𝑘 𝐹 𝑁
The mass of the object is 2.33 kg
Can you use these data to determine the coefficient of static and kinetic friction?

27. Static vs Kinetic Friction
Mass of object 2.33 kg,
maximum static friction = 8.5 N
𝜇= 𝐹 𝑓 𝐹 𝑁
So we need FN
Object is not accelerating up or down, so Forces up = Forces Down
FN=Fg or FN
2.33x9.8 =
𝜇 𝑠 = 𝐹 𝑠 𝐹 𝑁
𝜇 𝑠 =
𝜇 𝑠 = 0.37
Coefficient of static friction = 0.37

28. Static vs Kinetic Friction
Mass of object 2.33 kg,
Average friction when sliding= 5.4 N
𝜇= 𝐹 𝑓 𝐹 𝑁
So we need FN
Object is not accelerating up or down, so Forces up = Forces Down
FN=Fg or FN
2.33x9.8 =
𝜇 𝑘 = 𝐹 𝑠 𝐹 𝑁
𝜇 𝑘 =
𝜇 𝑘 = 0.24
Coefficient of kinetic friction = 0.24