1. Chapter 4B. Friction and Equilibrium
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
© 2007

2. Equilibrium: Until motion begins, all forces on the mower are balanced
Equilibrium: Until motion begins, all forces on the mower are balanced. Friction in wheel bearings and on the ground oppose the lateral motion.

3. Objectives: After completing this module, you should be able to:
Define and calculate the coefficients of kinetic and static friction, and give the relationship of friction to the normal force.
Apply the concepts of static and kinetic friction to problems involving constant motion or impending motion.

4. Friction Forces
When two surfaces are in contact, friction forces oppose relative motion or impending motion. P
Friction forces are parallel to the surfaces in contact and oppose motion or impending motion.
Static Friction: No relative motion.
Kinetic Friction: Relative motion.

5. Friction and the Normal Force
The force required to overcome static or kinetic friction is proportional to the normal force, n.
fs = msn
fk = mkn

6. Friction forces are independent of area.
If the total mass pulled is constant, the same force (4 N) is required to overcome friction even with twice the area of contact.
For this to be true, it is essential that ALL other variables be rigidly controlled.

7. Friction forces are independent of temperature, provided no chemical or structural variations occur.
Heat can sometimes cause surfaces to become deformed or sticky. In such cases, temperature can be a factor.

8. Friction forces are independent of speed.
5 m/s
20 m/s
2 N
2 N
The force of kinetic friction is the same at 5 m/s as it is for 20 m/s. Again, we must assume that there are no chemical or mechanical changes due to speed.

9. The Static Friction Force
When an attempt is made to move an object on a surface, static friction slowly increases to a MAXIMUM value. n fs P W
In this module, when we use the following equation, we refer only to the maximum value of static friction and simply write:
fs = msn

10. Constant or Impending Motion
For motion that is impending and for motion at constant speed, the resultant force is zero and SF = 0. (Equilibrium) P fs
P – fs = 0
Rest P fk
P – fk = 0
Constant Speed
Here the weight and normal forces are balanced and do not affect motion.

11. Friction and Acceleration
When P is greater than the maximum fs the resultant force produces acceleration. a
This case will be discussed in a later chapter. P fk
Constant Speed
fk = mkn
Note that the kinetic friction force remains constant even as the velocity increases.

12. EXAMPLE 1: If mk = 0.3 and ms = 0.5, what horizontal pull P is required to just start a 250-N block moving?
1. Draw sketch and free-body diagram as shown.
2. List givens and label what is to be found: n fs P W +
mk = 0.3; ms = 0.5; W = 250 N
Find: P = ? to just start
3. Recognize for impending motion: P – fs = 0

13. EXAMPLE 1(Cont. ): ms = 0. 5, W = 250 N. Find P to overcome fs (max)
EXAMPLE 1(Cont.): ms = 0.5, W = 250 N. Find P to overcome fs (max). Static friction applies. n fs P
250 N +
For this case: P – fs = 0
4. To find P we need to know fs , which is:
fs = msn
n = ?
n – W = 0
5. To find n:
SFy = 0
W = 250 N
n = 250 N
(Continued)

14. EXAMPLE 1(Cont. ): ms = 0. 5, W = 250 N. Find P to overcome fs (max)
EXAMPLE 1(Cont.): ms = 0.5, W = 250 N. Find P to overcome fs (max). Now we know n = 250 N. n fs P
250 N +
ms = 0.5
6. Next we find fs from:
fs = msn = 0.5 (250 N)
7. For this case: P – fs = 0
P = fs = 0.5 (250 N)
P = 125 N
This force (125 N) is needed to just start motion. Next we consider P needed for constant speed.

15. EXAMPLE 1(Cont. ): If mk = 0. 3 and ms = 0
EXAMPLE 1(Cont.): If mk = 0.3 and ms = 0.5, what horizontal pull P is required to move with constant speed? (Overcoming kinetic friction) fk n P mg +
SFy = may = 0
n - W = 0
n = W
mk = 0.3
Now: fk = mkn = mkW
SFx = 0; P - fk = 0
P = fk = mkW
P = 75.0 N
P = (0.3)(250 N)

16. The Normal Force and Weight
The normal force is NOT always equal to the weight. The following are examples:
300 P m n W
Here the normal force is less than weight due to upward component of P.  P n W
Here the normal force is equal to only the compo- nent of weight perpendi- cular to the plane.

17. Review of Free-body Diagrams:
For Friction Problems:
Read problem; draw and label sketch.
Construct force diagram for each object, vectors at origin of x,y axes. Choose x or y axis along motion or impending motion.
Dot in rectangles and label x and y compo- nents opposite and adjacent to angles.
Label all components; choose positive direction.

18. For Friction in Equilibrium:
Read, draw and label problem.
Draw free-body diagram for each body.
Choose x or y-axis along motion or impending motion and choose direction of motion as positive.
Identify the normal force and write one of following:
fs = msn or fk = mkn
For equilibrium, we write for each axis:
SFx = SFy = 0
Solve for unknown quantities.

19. Example 2. A force of 60 N drags a 300-N block by a rope at an angle of 400 above the horizontal surface. If uk = 0.2, what force P will produce constant speed?
1. Draw and label a sketch of the problem.
400
P = ?
W = 300 N n fk
2. Draw free-body diagram. m W
400 P W n fk +
P sin 400 Py Py
The force P is to be replaced by its com- ponents Px and Py. Px
P cos 400

20. Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2.
400 P mg n fk +
P cos 400
P sin 400
3. Find components of P:
Px = P cos 400 = 0.766P
Py = P sin 400 = 0.643P
Px = 0.766P; Py = 0.643P
Note: Vertical forces are balanced, and for constant speed, horizontal forces are balanced.

21. Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2.
Px = P Py = P
400 P
300 N n fk +
0.766P
0.643P
4. Apply Equilibrium con- ditions to vertical axis.
SFy = 0
n P – 300 N= 0
[Py and n are up (+)]
n = 300 N – 0.643P;
Solve for n in terms of P
n = 300 N – 0.643P

22. Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2.
n = 300 N – 0.643P
400 P
300 N n fk +
0.766P
0.643P
5. Apply SFx = 0 to con- stant horizontal motion.
SFx = 0.766P – fk = 0
fk = mk n = (0.2)(300 N P)
fk = (0.2)(300 N P) = 60 N – 0.129P
0.766P – fk = 0;
0.766P – (60 N – 0.129P) = 0

23. Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2.
400 P
300 N n fk +
0.766P
0.643P
0.766P – (60 N – 0.129P )=0
6. Solve for unknown P.
0.766P – 60 N P =0
0.766P P = 60 N
If P = 67 N, the block will be dragged at a constant speed.
0.766P P = 60 N
0.895P = 60 N
P = 67.0 N
P = 67.0 N

24. Example 3: What push P up the incline is needed to move a 230-N block up the incline at constant speed if mk = 0.3? P
Step 1: Draw free-body including forces, angles and components.
600
W =230 N n x y P
Step 2: SFy = 0 fk
W cos 600
n – W cos 600 = 0
600
W sin 600
n = (230 N) cos 600
230 N
n = 115 N

25. Example 3 (Cont.): Find P to give move up the incline (W = 230 N).
600 x y P W fk n
600
W cos 600
W sin 600
n = 115 N
W = 230 N
Step 3. Apply SFx= 0
P - fk - W sin 600 = 0
fk = mkn = 0.2(115 N)
fk = 23 N, P = ?
P - 23 N - (230 N)sin 600 = 0
P = 222 N
P - 23 N N= 0

26. Summary: Important Points to Consider When Solving Friction Problems.
The maximum force of static friction is the force required to just start motion. n fs P W
Equilibrium exists at that instant:

27. Summary: Important Points (Cont.)
The force of kinetic friction is that force required to maintain constant motion. n fk P W
Equilibrium exists if speed is constant, but fk does not get larger as the speed is increased.

28. Summary: Important Points (Cont.)
Choose an x or y-axis along the direction of motion or impending motion.
The SF will be zero along the x-axis and along the y-axis. fk n P W +
mk = 0.3
In this figure, we have:

29. Summary: Important Points (Cont.)
Remember the normal force n is not always equal to the weight of an object.
300 P m n W 
It is necessary to draw the free-body diagram and sum forces to solve for the correct n value.

30. Summary Static Friction: No relative motion.
Kinetic Friction: Relative motion.
fs ≤ msn
fk = mkn
Procedure for solution of equilibrium problems is the same for each case:

31. CONCLUSION: Chapter 4B Friction and Equilibrium