Chapter 8 - Friction
Sections 8.1 - 8.4
Friction



Frictional forces resist movement of a rigid
body over a rough surface.
It is assumed that the friction force acts
parallel to the surface.
A distinction is made between frictional
forces that act on an object that is not yet
moving and an object that is moving.
Friction cont.


Static-friction force is the term for the
friction force that exists before the object
begins to move. This force can be counted
as part of the equilibrium on a stationary
rigid body.
Kinetic-friction force is the term for the
friction force that exists on a rigid body that
started to move.
Friction cont.


Both the static and kinetic friction forces are
related to the normal force on the rigid
body.
A normal force is by definition the support
reaction force exerted by the supporting
surface on the rigid body. It always acts
perpendicular to the surface and up from the
surface. (See Fig 8.1)
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Friction cont.

In equation form, the forces are:
where: •Fm is the maximum static
Fm = sN
and
Fk = kN
friction force that exists just
before the body begins to
move.
•s is the coefficient of static
friction.
•N is the normal force.
Friction cont.



Fk is the kinetic friction force that exists
once the object begins to move. Its value
remains fairly constant as the velocity of the
object increases.
k is the coefficient of kinetic friction.
Some typical values for s (coefficient of
static friction) are given in Table 8.1.
Friction cont.


Four possible situations for frictional forces
are shown in Fig 8.2.
Of these, Case C (where motion is
impending) is probably the most important
to a study of statics. For this case, an
equilibrium free body diagram can be
drawn because the body is not in motion.
And the friction force is Fm which is equal
to sN.
Angle of Friction


The angle of static friction (s) is the angle
between a line perpendicular to the surface
and the vector resultant of the normal force
and the frictional force (see Fig’s 83(c) and
8.4(c)) when motion is impending on the
rigid body.
Note from the geometry of Fig’s 8.3(c) and
8.4(c) that:
tan s = Fm/N = sN/N = μs
Angle of Friction cont.


Looking at Fig 8.4(c) it can be seen that the
angle between the surface and the
horizontal (angle ) is equal to  s when
motion is impending.
This angle is called the angle of repose. If
this angle is increased any further, the block
will begin to slide down the board!
Angle of Friction cont.

Problems involving friction are solved by
setting up a free body diagram including all
weights, external loads, normal reaction
forces, and frictional forces. Break all
forces into x and y components and use
equilibrium equations to solve for
unknowns. (ΣF = 0 + ΣM = 0). As long as
there are no more than 3 unknown forces
the problem should be statically
determinate.
Angle of Friction cont.

You will encounter 3 common types of
friction problems.
(1) All applied forces are given and the coeff. of
friction s is also given. You must determine if
the body will remain at rest or begin to slide!
Solve the problem by setting up a FBD and
solving for the frictional force and the normal
force. Calculate Fm (Fm = μsN). If F is  Fm, the
body will not slide. If F is > Fm then the body
will begin to move.
Angle of Friction cont.
(2) All applied forces are given and it is stated that
motion is impending (F = Fm). You must find s.
Solve the problem by finding the maximum
friction force (Fm) and the normal force (N). Then
find s with Fm = sN.
(3) Given s and the fact that motion is impending
(F = Fm) find the magnitude and direction of one
applied force. Solve this problem with a FBD
showing Fm with a sence opposite to that for
impending motion.
Sample Problem 8.1
A 300 lb block is on the inclined plane.
s = .25 and k = .2.
Determine whether the block is in equilibrium and the
value of F.
y
F
100 lb
4 5
3
3
5
4
FBD
N
x
Sample Problem 8.1 cont.
ΣFx = 0 = 100(4/5) - F(4/5) - N(3/5)
N = 100(4/5 x 5/3) - F(4/5 x 5/3)
N = 133.3 - 1.33 F
ΣFy = 0 = 100(3/5) - 300 + N(4/5) - F(3/5)
F = -240(5/3) + N(4/5 x 5/3)
F = -400 + 1.33 N
F = -400 + 1.33(133.3 - 1.33F)
F + 1.72 F = -400 + 173
2.73 F = -227
F = -83.1 lb
Sample Problem 8.1 cont.
N = 133.3 - 1.33(-83.1)
N = 133.3 + 110.6
N = 243.8 lb
Fm = sN
= .25(2438)
= 62 lb
since F > Fm the block will slide and
Factual =kN = .2(243.8) = 48.8 lb
Sample Problem 8.2
If s = 0.35 and k = 0.25, determine the force P
needed to (a) start the block moving up the incline,
(b) to keep it moving, and (c) to prevent it from
sliding down.
Fm = sN
Fm = .35N
800 N
y
P
F
25°
25°
N
25°
x
Sample Problem 8.2 cont.
ΣFx = 0 = - P + F cos 25° + N sin 25°
P = .35 N(cos 25°) + N sin 25°
P = .317 N + .423 N
P = .74 N
ΣFy = 0 = -800 + N cos 25° - F sin 25°
800 LB = .906 N - (.35 N) x .42
N = 800/.758 = 1055 N
P = .74(1055) = 781 N
So when P exceeds 781 N the block will start to
move.
Sample Problem 8.2 cont.
Once the block starts moving:
Use the same equations, but F = kN = .25 N
ΣFx = 0 = - P + F cos 25° + N sin 25°
P = .25 N cos 25° + N sin 25°
P = .65 N
ΣFy = 0 = -800 N + N cos 25° - F sin 25°
800 = N cos 25° - .25 N sin 25°
800 = .8 N
N = 1000 N
P = .65 x 1000 = 650 N
Sample Problem 8.2 cont.
To prevent the block from sliding down:
w
ΣFx = 0 = - P - F cos 25° + N sin 25°
P
P = -(.35 N)cos 25° + N sin 25°
F
P = -.317 N + .423 N
25°
N
Fm = sN
P = .106 N
ΣFy = 0 = -800 + N cos 25° + F sin 25°
800 = .906 N + .423(.35 N)
800 = 1.05 N
P = .106(759)
N = 759 N
P = 80.5 N
Solution 8.1
s = 0.30
k = 0.20
Given:  = 30°, P = 50 lb
250 lb
P

Find: Friction Force acting on
block.
Solution 8.1 cont.
250 lb
y 30°
x
50 lb
30°
Assume Equilibrium
+
F
ΣFy = 0:
N - (250 lb)cos 30° - (50 lb)sin 30° = 0
N = +241.5 lb
+
N = 241.5 lb
ΣFx = 0:
F - (250 lb)sin 30° + (50 lb)cos 30° = 0
F = +81.7 lb
F = 81.7 lb
N
Solution 8.1 cont.
Maximum Friction Force:
Fm = sN = (0.30)(241.5 lb) = 72.5 lb
Since F > Fm, Block moves down
Friction Force: F = kN = (0.20)(241.5 lb) = 48.3 lb
F = 48.3 lb
Solution 8.2
Given:  = 35°, P = 100 lb
s = 0.30
k = 0.20
250 lb
P

Find: Friction Force acting on
block.
Solution 8.2 cont.
250 lb
y 35°
x
100 lb
35°
F
Assume Equilibrium
+
ΣFy = 0:
N - (250 lb)cos 35° - (100 lb)sin 35° = 0
N = +262.15 lb
+
N = 262.15 lb
ΣFx = 0:
F - (250 lb)sin 35° + (100 lb)cos 35° = 0
F = +61.48 lb
F = 61.48 lb
N
Solution 8.2 cont.
Maximum Friction Force:
Fm = sN = (0.30)(262.15 lb) = 78.64 lb
Since F < Fm, Block is in equilibrium
Friction Force: F = 61.5 lb
Solution 8.3
Given:  = 40°, P = 400 N
Find: Friction Force acting on block.
s = 0.20
k = 0.15
P
800 N

25°
Solution 8.3 cont.
y
40°
800 N
400 N
25°
15°
25°
x
Assume Equilibrium
+
F
N
ΣFy = 0:
N - (800 N)cos 25° + (400 N)sin 15° = 0
N = +621.5 N
+
N = 621.5 N
ΣFx = 0:
- F + (800 N)sin 25° - (400 N)cos 15° = 0
F = -48.28 N
F = 48.28 N
Solution 8.3 cont.
Maximum Friction Force:
Fm = sN = (0.20)(621.5 N) = 124.3 N
Since F < Fm, Block is in equilibrium
F = 48.3 N