Chapter 8 - Friction
8.1 Characteristics of dry friction
Friction - a resistive force which prevents or retards the slipping of one body with respect
to another.
2 types of friction
1). Dry (coulomb)- non-lubricated
2). fluid
Coefficients of friction
P
W
F
N
Static friction force - the force that keeps the block from moving.




As P increases, F increases until it reaches Fmax If P increases more, the friction can
no longer balance it and the block begins sliding.


When the block begins sliding the magnitude of F drops to Fk the kinetic friction force.
Experimental evidence shows
Fmax   s N and Fk   s N
Table on pg. 357, graph on pg. 358.
Angle of friction
 s  tan 1  S
P
W
Fs



Rs  Fs  N
s
Also:  k  tan 1  k
N
Rs
Thus:  s   k
Angle of repose
s
Lecture23.doc
 s  tan 1  s
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8.2 Problems involving dry friction
4 types of friction problems
1). Equilibrium (number of unknowns = # of equations of equilibrium)
P
Fx  0
F  Px
W
F  s N
F
Fy  0
N  Py  W

No evidence that F has reached its maximum value, thus F =  S N CANNOT be used.
N
2). Impending motion at all points (# of unks = # of EOE + # of friction equations)
Fx  0
P
Fmax  Px
W
Fmax
Fmax   s N
Fy  0
N  Py  W
N
Friction force is in opposite direction of impending motion.
3). Tipping or impending motion at some points (# unks < # E.O.E. + # F.E.)
 Sliding: must do computations for all the different situations
- assume condition 1, solve
- assume condition 2, solve
etc.
 Tipping: - block will slip F   s N ,   x  b / 2
- block will tip F   S N , x  b / 2
Pg. 361
4). Body is sliding
P
Fk  Px
Fk   k N
W
Fk
Fy  0
N  Py  W
N
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Examples
1). Given: A support block is acted upon by 2 forces as shown.  S  0.35 and
 k  0.25 .
P
25o
800 N
Find: Determine the force P required to:
a). Start the block moving up the incline
b). Keep it moving up
c). Prevent it from sliding down
a). FBD
y
800 N
25o
x
P
Fmax
N
Use F because motion is impending.
Fx  0
Fy  0
800 sin 25   s N  Px  0
N  800 cos 25  Py  0

N  800 cos 25  P sin 25


338.1   s N  P cos 25
338.1  0.35 N  P cos 25
338.1  0.35(800 cos 25  P sin 25 )  P cos 25
338.1  280 cos 25  P(cos 25  0.35 sin 25 )
P  780.4 N
B). Same as a). only use  k =0.25
P = 648.7 N
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c). FBD
y
800 N
x
P
Fmax
N
Fx  0
800 sin 25  F  Px  0
338.1   x N  P cos 25 
Fy  0
N  800 cos 25   Py  0
N  800 cos 25   P sin 25 
338.1  0.35(800 cos 25  P sin 25 )  P cos 25
338.1  280 cos 25  P(cos 25  .0.35 sin 25 )
P  80 N
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2). Given:  s  0.30,  k  0.25 between all surfaces.
A
B
C
100
kg
D
150 kg
P
Find: Determine the smallest force P required to start block D moving.
FBD. Block C
Wc = 981 N
y
x
TAB
F1
N1
F y  0
F1   S N1  0.30(981)
N 1  WC  0
F1  294.3 N
N 1  981 N
FBD Block D
N1
F1
WD  1471.5N
P
F2
N2
Fy  0
F2   S N 2  0.30(2452.5)
Fx  0
N 2  N1  WD  0
F1  294.3 N
P  F1  F2  0
N 2  981  1471.5
N 2  2452.5 N
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P  294.3  735.75
P  1030 N 
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3). Given: Three 4 kg packages A, B, and C are placed on a conveyor belt which is at rest.
Between the belt and both packages A and C,  s  0.30 and  k  0.20 . Between package B
and the belt,  s  0.10 and  k  0.08 . The packages are placed on the belt so that they are
in contact with each other and at rest.
B
C
A
15o
Find: Determine which, if any, of the packages will move and the friction force acting on
each package.
Consider C by itself
- assume C is in equilibrium.
y
15o
W
x
C
Fc
Nc
Fy  0
Fx  0
N C  W cos 15  0
FC  W sin 15  0
N C  4(9.81) cos 15
FC  4(9.81) sin 15
N C  37.9 N
FC  10.2 N
Fmax   s N C  0.3(37.9)
Fmax  11.37 N
FC  Fmax
Thus, package C does not move
Consider B by itself
- assume B is in equilibrium
y
W
x
B
FB
NB
Fy  0
Fx  0
N B  W cos 15  0
FB  W sin 15
N B  4(9.81) cos 
FB  4(981) sin 15
N B  37.9 N
FB  10.2 N
Fmax   s N  0.1(37.9)
Fmax  3.79 N
FB  Fmax Thus, package B moves
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Consider A and B together
-assume A and B are in equilibrium
y
W
W
x
B
FB
A
FA
NB
NA
Package A
Fy  0  N A  37.9 N
Fx  0  FA  10.2 N
Both A and B
y
WT
x
B
FT
A
NB
Fy  0
FX  0
N T  WT cos15   0
FT  WT sin 15   0
N T  8(9.81) cos15 
FT  8(9.81) sin 15 
N T  75.8 N
FT  20.3 N
Fmax   sA N A   sB N B
Fmax  0.3(37.9)  0.1(37.9)
Fmax  15.16 N
FT  Fmax , thus, packages A and B move
F   k N   0.2(37.9)
F   kB N B  0.08(37.9)
F  7.58 N
F  3.03 N
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