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Chapter 8 - Friction
8.1 Characteristics of dry friction
Friction - a resistive force that 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
Static friction force - the force that keeps the block from moving.
As P increases, F increases until it reaches m ax F 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 k F the kinetic friction force.
Experimental evidence shows N F s μ=max and N F s k μ=
Angle of friction
S s μφ1tan -= N F R s s +=
Also: k k μφ1tan -=
Thus: k s φφ≥
Angle of repose
F
F s
s
s
s μθ1tan -=
8.2 Problems involving dry friction
4 types of friction problems
1). Equilibrium (number of unknowns = # of equations of equilibrium) x
x P F F ==∑0
N F s μ<
W
P N F y y +==∑0
No evidence that F
has reached its maximum value, thus F = N S μ CANNOT be used.
2). Impending motion at all points (# of unks = # of EOE + # of friction equations)
x
x P F F ==∑max 0
N F s μ=max
W P N F y y +==∑0
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 2/,b x N F s ≤≤=θμ
- block will tip 2/,b x N F S =≤μ
4). Body is sliding
x k P F <
N F k k μ=
W
P N F y y +==∑0
F
F max
F k
P
Examples
1). Given: A support block is acted upon by 2 forces as shown. 35.0=S μ and
25.0=k μ.
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
Use max F because motion is impending.
b). Same as a). only use k μ=0.25 P = 648.7 N
25
sin 25cos 800025cos 8000
P N P N F y y +==--=∑
25cos 35.01.33825cos 1.338025sin 8000P N P N P N F s x s x =+=+=-+=∑μμN 4.780)25sin 35.025(cos 25cos 2801.33825cos )25sin 25cos 800(35.01.338=-=+=++P P P P
c). FBD
25cos 1.338025sin 8000
P N P F F x x x =-=--=∑μ
25sin 25cos 8000
25cos 8000
P N P N F y y +==--=∑
N 80)25sin 35.0.25(cos 25cos 2801.33825cos )25sin 25cos 800(35.01.338=+=-=+-P P P P
P
2). Given: 25.0,30.0==k s μμ between all surfaces.
P
Find: Determine the smallest force P required to start block D moving.
FBD Block C y x T AB 1F 1N
98100
11==-=∑N W N F C y N
3.294)
981(30.0111===F N F S μ
FBD Block D
1N F P
F
2N
N
3.294)5.2452(30.0122===F N F S μN
5.24525.147198100221
2
=+==--=∑N N W N N F D
y →
=+==--=∑N 103075.7353.294002
1P P F F P F x
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 , 20.0and 30.0==k s μμ. Between package B and the belt, 08.0and 10.0==k s μμ. The packages are placed on the belt so that they are
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.
N 37.11)
9.
37(3.0max max ===F N F C s μ
max F F
C < Thus, package C does not move.
Consider B by itself - assume B is in equilibrium
N
79.3)9.37(1.0max max ===F N F s μ
max F F B > Thus, package B moves
F c
F B
N
9.3715cos )81.9(4015cos 0
===-=∑C C C y N N W N F N
2.1015sin )81.9(4015sin 0===-=∑C C C x F F W F F N
9.37cos )81.9(4015cos 0
===-=∑B B B y N N W N F N
2.1015sin )981(415sin 0====∑B B B x F F W F F
Consider A and B together -assume A and B are in equilibrium
Package A
N 2.100N
9.370=⇒=∑=⇒=∑A x A y F F N F
Both A and B
N 8.7515
cos )81.9(8015cos 0
===-=∑T T T T y N N W N F
N
3.2015sin )81.9(8015sin 0
===-=∑T T T T X F F W F F
N 16.15)9.37(1.0)9.37(3.0max max max =+=+=F F N N F B
sB A sA μμ
max F F T >, thus, packages A and B move
N 58.7)9.37(2.0==-A A A F N F k μ N 03.3)
9.37(08.0===B B F N F B kB μ
x
B
A
F
x
T
B。