1. CHAPTER 8
FRICTION

2. 8.1 Introduction
In preceding chapters, it was assumed that surfaces in contact were either frictionless (surfaces could move freely with respect to each other) or rough (tangential forces prevent relative motion between surfaces).
Actually, no perfectly frictionless surface exists. For two surfaces in contact, tangential forces, called friction forces, will develop if one attempts to move one relative to the other.
However, the friction forces are limited in magnitude and will not prevent motion if sufficiently large forces are applied.
The distinction between frictionless and rough is, therefore, a matter of degree.
There are two types of friction: dry or Coulomb friction and fluid friction. Fluid friction applies to lubricated mechanisms. The present discussion is limited to dry friction between nonlubricated surfaces.

3. Example of friction

4. Dry Friction
Friction is defined as a force of resistance acting on a body which prevents or retards slipping of the body relative to a second body.
Experiments show that frictional forces act tangent (parallel) to the contacting surface in a direction opposing the relative motion or tendency for motion.

5. For the body shown in the figure to be in equilibrium, the following must be true:
F = P, N = W, and Wx = Ph.

6. FS = msN Fk = mkN Impending motion Motion
coefficient of kinetic friction
coefficient of static friction

7. The maximum friction force is attained just before the block begins to move (a situation that is called “impending motion”). The value of the force is found using Fs = s N, where s is called the coefficient of static friction. The value of s depends on the materials in contact.

8. 8.2 The Law of Dry Friction, Coefficients of Friction
Block of weight W placed on horizontal surface. Forces acting on block are its weight and reaction of surface N.
Small horizontal force P applied to block. For block to remain stationary, in equilibrium, a horizontal component F of the surface reaction is required. F is a static-friction force.
As P increases, the static-friction force F increases as well until it reaches a maximum value Fm.
Further increase in P causes the block to begin to move as F drops to a smaller kinetic-friction force Fk.

9. Maximum static-friction force:
Kinetic-friction force:
Maximum static-friction force and kinetic-friction force are:
proportional to normal force
dependent on type and condition of contact surfaces
independent of contact area

10. Four situations can occur when a rigid body is in contact with a horizontal surface:
No friction, (Px = 0)
No motion, (Px < Fm)
Motion impending, (Px = Fm)
Motion, (Px > Fm)

11. 8.3 Angles of Friction
It is sometimes convenient to replace normal force N and friction force F by their resultant R:
No friction
Motion impending
No motion
Motion

12. No friction No motion Motion impending Motion
Another examples will show how the angle of friction can be used to advantage in the
analysis of certain types of problems.
No friction
No motion
Motion impending
Motion

13. 8.4 Problems Involving Dry Friction
Coefficient of static friction is known
Motion is impending
Determine magnitude or direction of one of the applied forces
All applied forces known
Motion is impending
Determine value of coefficient of static friction.
All applied forces known
Coefficient of static friction is known
Determine whether body will remain at rest or slide

14. Pushing the uniform crate that has a weight W and sits on the rough
Example
Pushing the uniform crate that has
a weight W and sits on the rough
surface.
Slip, tip over, or static?

15. Relation with moment
Verge of slipping
Tip over

16. Relation with moment + + + 200 N 200 N Fmax = (0.3) (200 N) = 60 [N]
μs = 0.3
200 N
0.2 m
Fmax = (0.3) (200 N)
= 60 [N]
80 N
80 N +
Fx = 0
0.4 m F F
80N – F = 0
F = 80 [N] O x +
Fy = 0
N = 200 N
N = 200 N
200 N + N = 0
N = 200 [N] +
x = 0.16 m

17. Relation with moment + + + 200 N 200 N Fmax = (0.3) (200 N) = 60 [N]
μs = 0.3
200 N
Fmax = (0.3) (200 N)
0.2 m
= 60 [N]
80 N
80 N +
Fx = 0
0.6 m F
80N – F = 0
F = 80 [N] F O x +
Fy = 0
N = 200 N
N = 200 N
200 N + N = 0
N = 200 [N] +
x = 0.24 m
tip over

18. Sample problem 8.1
1350N
450 N
A 450 N force acts as shown on a 1350N block placed on a inclined plane.
The coefficients of friction between the block and the plane are μs = 0.25 and
μk = 0.20.
Determine whether the block is in equilibrium, and find the value of the
frictiom force.

19. Solution 450 N 1350N 450N Force required for equilibrium.
We first determine the value of the friction force required to
maintain equilibrium. Assuming that F is directed down and to the left,
we draw the free-body diagram of the block and write
1350N
450N
The force F required to maintan equilibrium is an 360N force
directed up and to the right ; the tendency of the block is thus to
move down the plane.

20. 450 N 1350N 450N Maximum Friction force. 1350N
The magnitude of the maximum friction force can be developed is
Since the value of the force required to maintain equilbrium (360N)
is larger than the maximum value which cn be obtained (270 N),
equilibrium will not be maintained and the block will slide down the plane.
1350N
450N

21. 450 N Factual = 216 N Actual Value of Friction Force. 1350N
The magnitude of the actual friction force is obtained as follows :
The sence of this force is opposite to the sense of motion ;
the force is thus directed up and to the right :
1350 N
450 N
N = 1080 N
F = 216 N
Factual = 216 N
It should be noted that the forcee is opposite to the sense of motion ;
the force

22. Sample problem 8.2
A support block is acted upon by two forces as shown. Knowing that the coefficient of friction
between the block and the incline are μs = 0.35 and μk = 0.25, determine
(a) the force P for which motion of the block up the incline is impending,
(b) the friction force when the block is moving up
(c) the smallest force P required to prevent the block from sliding down

23. Solution Free Body Diagram.
For each part of the problem we draw a free-body diagram of the block and a force triangle including
the 800N vertical force, the horizontal force P, and the force R exerted on the block by the incline.
The direction of R must be determine in each separate case. We note that since P is perpendicular
to the 800N force, the force triangle is a right triangle, which can easily be solved for P. In most other
problems, however, the force triangle will be an oblique triangle and should be solved by applying the
law of sines.

24. a. Force P for impending Motion of the Block Up the Incline
P = (800 N) tan 44.29º
= 780 N

25. b. Friction Force F when the Block Moves Up the incline
R = (800 N) / cos 39.04º = N
F = R sin Φk = ( N) sin 14.04º
F = 250 N

26. c. Force P for impending Motion of the Block Down the Incline
P = (800 N) tan 5.71º
= 80 N

27. If the coefficient of friction between the pipe and bracket is 0.25,
Sample problem 8.3
150mm
75mm
The moveable bracket shown may be placed at any height on the 75mm diameter pipe.
If the coefficient of friction between the pipe and bracket is 0.25,
determine the minimum distance x at which the load can be supported.
Neglect the weight of the bracket.

28. Apply conditions for static equilibrium to find minimum x.
SOLUTION:
When W is placed at minimum x, the bracket is about to slip and friction forces in upper and lower collars are at maximum value.
150 mm
75 mm
37.5 mm
Apply conditions for static equilibrium to find minimum x.
NA (150 mm) – FA (75mm) – W (x mm) = 0
150 NA – 75 ( 0.25NA ) – Wx W = 0
150 (2W) – (2W) – Wx W = 0
x = 300 mm

29. PROBLEMS

30. Problem 8.1
Determine whether the block shown is in equilibrium and find the magnitude
and direction of the friction force when θ= 30º and P = 200 N

31. Solution
Free Body Diagram

33. Problem 8.4
Determine whether the 9 kg block shown is in equilibrium, and find
the magnitude and direction of the friction force when P = 60 N and θ = 15º

34. Solution
Free Body Diagram

36. Problem 8.11 and 8.12
The coefficients of friction are μs = 0.40 and μk = 0.30
between all surface of contact.
Determine the forces P for which motion of the 27kg block is impending if cable AB
is attached as shown
is removed
18kg
27kg
18kg
27kg

37. 18kg
27kg

39. 18kg
27kg

41. THE END