3/3 Equilibrium Conditions

Report
6 Friction
1
Objectives Students must be able to
• Utilize theory of dry friction
– Describe theory of dry friction
– Describe physical meanings of frictional effects
– Describe and differentiate between static and
kinetic coefficients of friction
– Describe the angles of frictions
– Add friction into the analyses of objects and
structures in equilibrium
2
Objectives Students must be able to
• Describe and analyze machines with frictions
– Wedges
– Threads, screws
– Belts
– Disks and clutches
– Collar, pivot, thrust and journal bearings
• Outline rolling resistance
– Describe the physical meanings of rolling
resistance
– Differentiate between frictions and rolling
resistance
3
Topic in textbook
• Section A: Frictional Phenomena
– Characteristics, theory, coefficient of
friction, angle of friction
• Section B: Applications
– Wedges
– Screws
– Journal Bearings
– Thrust Bearings; Disk friction
– Flexible Belts
– Rolling Resistance
We will
study this
Part first.
4
Dry Friction
Force of resistance acting on a body which prevents
or retards slipping of the body relative to a surface
with which it is in contact.
Friction
exists?
roughnesses of the contacting surfaces.
Magnitude: friction’s magnitude limitation will be discussed later
Direction:
tangent to the contacting surface and
opposed to the relative motion or tendency for motion
Line of Action (Point of application): contact surface
5
Equilibrium
Friction Model
[ F  0]
F  P N W
[ M  0] x  h
P
P
h
N
W
y
W
In equilibrium
FBD is correct?
a/2
a/2
x
P
modeling
h
F
The DN at right side is supporting
force more than its left side.
x
N
If x > a/2 ?
• Frictional force F
• The application
point (x) of N
increases
with force P
Slipping and/or
Tipping Effect
P
N
The object is toppling
(not in equilibrium)
toppling
Slipping
x-limit
F-limit
x
F
6
a/2
a/2
Motion
P
h
F
x
N
N
• Slipping / Sliding
– Relative sliding (translation motion) between two
surfaces
• Toppling / Tippling
– Fall over (rotation) about the edge
– Topple, tipping, rolling, tumble, trip
7
Experiment for determining Friction
mg
P
m
P
FBD
F
F
Static friction
(no motion)
Object at rest (no motion)
Fk = kN
Kinetic friction
(motion)
Fs:max = sN
k , s
N
“impending motion”’
(on the verge of motion)
P

Fs  S N

S : coefficient of static friction
Object with motion (steady state)


Fk  k N
k : coefficient of kinetic friction
: constant on 2 certain contacting surfaces
s  k (generally)8
no change in  k
Angle of Friction
however P, mg are
Not depend on N
Fk  k N
 k  tan 1
not depend on P,v,a
Fk
 tan 1 k
N
k
 k = arctan(k) = angle of kinetic friction
 k  k
Fs  S N
(object in motion)
? tan 1 (s )
Fs
 s  tan
N
1

S N
mg
s
F
1
1  s  max

tan
s
tan
N
 s = arctan(s) = angle of (max) static friction
s  s
(object at rest)
P
N

F
R
P  Fs 
s 
9
Dry Friction
Dry Friction Characteristics
• Frictional force acts tangentially to the contacting surfaces,
opposing the relative or tendency for motion.
• Fs is independent of the area of contact, provided that the
normal pressure is not very low nor great enough for
deformation of the surfaces.
• In equilibrium:
Impending slipping:
Slipping:
Very low velocity:
 = s
 = k
k s
10
Dry Friction
Impending Motion
Static Friction Typical Values
Contact Materials
Metal /
Wood /
Leather /
Leather /
Aluminum /
ice
wood
wood
metal
Aluminum
μs
0.03 – 0.05
0.30 – 0.70
0.20 – 0.50
0.30 – 0.60
1.10 – 1.70
How to obtain values of s ?
11
Sample 6/1 Determine the maximum angle 
which the adjustable incline may have before
the block of mass m begins to slip. The
coefficient of static friction between the block
and the inclined surface is s.
 Fx  0 :
 Fy  0 :
W=mg
y

H/2
x
F
Three force member
F  Fmax  s N
x

“Impending
Slip”:
N
3 eq. , 3 unknowns
mg sin max  s (mg cosmax )
(for slipping)
Ans
Possibility of toppling?
[ M  0 ]
  x
Fmax  s N  s (mgcosmax )
max  tan1 s
H
x  tan 
2
H
mg cos  ( x )  mg sin  ( )  0
2
H
x  tan 
2
 F
F  mgsin   0
N  mgcos  0
x
W=mg


  H tan  


 2

L

2
 L/2 

 H /2
max,toppling  tan 1 
12
6/125 A uniform block of mass m is at rest on an
incline z. Determine the maximum force P>0 that can
be applied to the block in the direction shown before
slipping begins. The coefficient of static friction
between the block and the incline is s
Consider only when
z
f  s N
y
x
F
 0 : N  mg cos  0
F
 0 : mg sin   f sin   0
f sin   mg sin 
F
 0 : P  f cos   0
f cos   P
z
x

mg
y
tan  s 
y
f  P   mg sin  
2
P
N

At this max P, object is
about to move at which
*
direction?
(  ?)
P  mg cos  ( s  tan  )
2
N  mg cos
 s mg cos
Pmax  mg cos  ( s  tan  )
mg sin 
tan 
tan  

Pmax
( s  tan  )
*
13
Dry Friction
Example Friction 2 #1
Will this crate slide or topple
over?
14
 Fx  0 P  F
 Fy  0 W  N
 M  0 Ph  xN
Known : W , h, s
1st Eq.
4 Unknowns: P, F, N, x
2nd Eq.
3rd Eq.
Two possibilities
1) about to slip
F  s N
2) about to tip
4th Eq.
b
x
2
4th Eq.
use Pmin as answer
It’s time
consuming,
Better to know it
exactly
P   sW
*
N W
*
x*   s h
Check
condition:
s
bW
P 
2 h
N† W
†
F† 
b
x 
2
*
?
Check
condition:
bW
2 h
F † ? s N †
The block of mass m is homogeneous, moving at a constant
velocity. The coefficient of kinetic friction is k
-
Determine
a) the greatest value that h may have so
that the block will not tip over
b) The location of point C on the bottom
face of the block through with the
resultant of the friction and normal
force act if h = H/2.
k
16
known values: m, H , b, k
Moving
With no acceleration
- The block of mass m is homogeneous,
moving at a constant velocity. The
coefficient of kinetic friction is k
- Determine
=?
k
Solution 1
a) the greatest value that h may have
so that the block will not tip over
on the verge of tipping
F  k N
  tan1 k (certain value)
Moving
The box is in Equilibrium
P

mg
h=?
F F
 N N
on the verge of tipping
(no acceleration)
Three force member in Statics
Concurrent at one point OR
1 b
h
( )
tan  2

1 b
( )
k 2
parallel
Ans
17
known values: m, H , b, k
Moving
with constant velocity
=H/2
-
The block of mass m is homogeneous.
-
The block is moving at a constant velocity.
-
Determine
Solution 1
b) The location of point C on the bottom face
of the block through with the resultant of
the friction and normal force act if h = H/2.
C =?
N
k
F  k N
tan   k
Moving
The box is in Equilibrium
(no acceleration)
Three force member
P
 mg
F

x
N
h=H/2
Concurrent at one point OR parallel
H
x  tan 
2

H
k
2
Ans
18
Inequality:
hard to deal
Dry Friction’s Problem
F  s N
F  0
M
O
static friction
0
+
or
F  k N
Equilibrium eq.
kinetic friction
fricitional eq.
Friction
• Kinetic motion is known.
F  0 M
O
0
Equilibrium eq.
+
Moving at constant vel.
F  k N
Kinetic friction
20
F  0 M
O
0
F  s N
+
Static friction
not
assuming
In static equilibrium
No. of unknown
must =
No. of Equilibrium eq.
“impending
motion”
friction can
support
equilibrium
Static
equilibrium?
Assume:
static equilibrium.
F  s N (usually) : can't be used
F  0 M
O
0
Equilibrium eq.
Only equilibrium
eq. can be used to
determine
unknown values.
Solve for F (and also N)
using only equilibrium eq.
Assumption Checking
Friction is determined by equilibrium eq’s.
Static friction
If
F  s N
Static friction
However,
F must <= N
F  s N
F  Fmax   s N  O.K.
F  Fmax
Assumption is not true!
21
Motion occurs. F = kN
F  0 M
O
0
Equilibrium eq.
+
F  s N
Static friction
Find min 
to hold
equilibrium
Find min P
to initiate the
motion
Impending motions occur at
both point in the same time.
F  0 M
total unknown
<
equilibrium +
frictional eq.
O
0
Equilibrium eq.
F = s N
static friction
unknown :
unknown :  F N F N
A
A
B
B
P FA N A Fc N c Bx By
Both can be
used
total unknown
together to
=
determine
equilibrium +
unknown
frictional eq.
values.
impending motions do
not occur in both point at
the same time.
several possibilities to slip
Assumption needs to be
make and
that assumption should
be checked later.
22
500N and
100N
Equilibrium?
Sample 6/3 Determine the friction force acting
on the block shown if P = 500N and P = 100N.
The block is initially at rest.
s  0.2
k  0.17
Equilibrium state
is unknown
Assume: Body in equilibrium
(not assuming the impending motion)
2 Eq ,
2 unknown
 Fx  0 :
P cos  F  mgsin 
 Fy  0 :
N  P sin   mg cos  0
F ? s N
impending motion
P=500N
Assumption Checking:
P=100N
 Fy  0 :
Not valid!
still valid!
OK !
| F |  s N  0.2(1093)  219N
F  242 N
Object not in Equilibrium
 Fx  0 :
N  1093 N
F  134 .3N
N  956 N
Contradict!
 0
F  s N
Friction must be
enough to
maintain
Correct?
equilibrium
| F |  s N  0.20(956)  191.2 N
N  P sin   mgcos  0
N  956 N
F  k N  0.17(956)  162.5N 23
Dry Friction
Example Hibbeler Ex 8-1 #1
The uniform crate has a mass of 20 kg. If a force P = 80 N is
applied to the crate, determine whether it remains in equilibrium.
The coefficient of static friction = 0.3.
Equilibrium State
is not known
24
Example Hibbeler Ex 8-1 #2
Assume: Equilibrium
  Fx  0   
80cos30  F  0  F  69.3 N
F  sNC (70.8 N)
  Fy  0   
80 sin30  196.2  NC  0  NC  236 N
  MO  0  
80 sin30(0.4)  80cos30(0.2)  NC ( x )  0  x  0.0908 m
|x| < 0.4 m
(physical boundary
of toppling)
x  0.4 m
 The crate will not topple.
F  s NC (70.8 N)
 The crate, is close to but doesn't not slip.
25
Dry Friction
Example Bedford Ex 9.5 #1
Suppose that  = 10° and the coefficient of friction between the
surface of the wedge and the log are s = 0.22 and k = 0.20.
Neglect the weight of the wedge.
– If the wedge is driven into
the log at a constant rate
by vertical force F, what are
the magnitude of the
normal forces exerted on
the log by the wedge?
– Will the wedge remain in
place in the log when the
force is removed?
26
 = 10° s = 0.22 and k = 0.20.
Neglect the weight of the wedge.
Symmetry about the center of the wedge
Moving down at
constant rate
  Fy  0   


2N sin( )  2k N cos( )  F  0
2
2

2
N

2
s* N
symmetric/mirror
concept
F


2 sin( )  2k ( )
2
2
F
N
 1.75F #
10
10
2 sin(
)  2(0.20)(
)
2
2
27
 = 10° s = 0.22 and k = 0.20.
Neglect the weight of the wedge.
Will the wedge remain in place in
the log when the force is
removed?
Impending Motion
*

Think of minimum friction coefficient s
(On the verge of slipping)
that still “self-lock” the wedge
If s  s*
the wedge will remain in place
If s  s*
the wedge will move out.
Symmetry about the center of the wedge
s* N
s* N
symmetric/mirror
concept
  Fy  0   
2N sin( / 2)  2s*N cos( / 2)  0
s*  tan( / 2)  0.087
As s  s* , wedge will remain in place.
#
28
Dry Friction
Example Bedford 9.20 #1
The coefficient of static friction
between the two boxes and
between the lower box and the
inclined surface is s. What is
the largest angle  for which the
lower box will not slip.
29
Impending
Motion
at same time ?
The coefficient of static friction between the
two boxes and between the lower box and
the inclined surface is s. What is the largest
angle  for which the lower box will not slip.
FBD of upper block
  Fy  0 
N2  W cos   0
N2  W cos 
FBD of lower block
  Fy  0 
N2  W cos   N1  0
N1  2W cos 
  Fx  0 
s N2  s N1  W sin   0
s 3W cos   W sin 
s  tan  / 3 or   tan1(3 s )
Unknown:  N1 N2 T
#
Eqs : 2obj  2eq (don't think of tipping - no moment eq
30
Bedford 9.30 (ex)
Impending
Motion
at same time
The cylinder has weight W. The coefficient
of static friction between the cylinder and
the floor and between the cylinder and the
wall is s. What is the largest couple M
that can be applied to the stationary
cylinder without causing it to rotate.
  Fy  0   
s Nw  Nf  W  0
  Fx  0   
Nw  s Nf  0
  MO  0
M  s (Nw  Ns )r  0
 
sW
Nw 
,
2
1  s
Impending
rotation
W
Nf 
1  s2
(1  s )
M  sWr
1  s2
#
31
Bedford 9.33 (Ex)
The disk of weight W and radius R is
held in equilibrium on the circular
surface by a couple M. The coefficient
of static friction between the disk and
the surface is s. Find that the largest
value M can have without causing the
disk to slip.
  Fx  0  
Fs  W sin   0
  Fy  0  
N  W cos   0
  MO  0   M  Fs R  0
s  tan  , N  W cos 
cos  1  tan2   1, tan =s
(Impending slip)
M  s NR  sWR cos 
Fs  s N
M  sWR
1  s2
32
Dry Friction
Example Bedford 9.166 #1
Each of the uniform 1-m bars has a mass of 4 kg. The
coefficient of static friction between the bar and the surface at B
is 0.2. If the system is in equilibrium, what is the magnitude of
the friction force exerted on the bar at B.
33
Dry Friction
Example Bedford 9.166 #2
  Fx  0 
Ox  Ax  0
  Fy  0 
Oy  Ay  4g  0
  MO  0 
Ay (1cos 45)  Ax (1sin 45)  4g (0.5 cos 45)  0
34
Dry Friction
Example Bedford 9.166 #3
  Fx  0 
 Ax  F cos 30  N sin 30  0
  Fy  0 
 Ay  4g  F sin 30  N cos 30  0
  MB  0 
Ay (1cos 45)  Ax (1sin 45)  4g (0.5 cos 45)  0
Ay  0, Ax  2g, Oy  4g, Ox  2g
N  43.8 N, F  2.63 N #
35
Dry Friction
Example Bedford 9.166 #4
symmetry?
36
Dry Friction
Example Hibbeler Ex 8-3 #1
The rod with weight W is about to slip
on rough surfaces at A and B. Find
coefficient of static friction.
Direction of N?
38
Dry Friction
Example Hibbeler Ex 8-3 #2
Impending slip
FA  s N A
FB  s NB
3 equilibrium equations,
3 unknowns (N A , NB , s )
  Fx  0   
  Fy  0   
  M A  0  
s NA  s NB cos 30  NB sin30  0
(1)
N A  W  NB cos 30  s NB sin30  0
(2)
NB l  W cos 30(l / 2)  0
(3)
39
Dry Friction
Example
Hibbeler Ex 8-3 #3
From (3)
NB  0.4330W
From (1) & (2)
0.2165 s2  s  0.2165
s  0.228
#
40
6/9 For a 20 jaw opening, what is
the minimum coefficient of static
friction between the jaws and the
tube which will enable tongs to grip
the tube with out slipping Solution 1
y
M


M
F1
N1

x
F

N2


o
F2
A
x
 0:
 F1 (r)  F2 (r)  0
F1  F2
 0:
N1 (l )  N 2 (l )  0
N1  N 2
 0:
 N1 sin   N2 sin   F1 cos  F2 cos  0
2( N1 ) sin   2( F1 ) cos
F1  (tan  ) N1
tan   s
( s )min  tan   tan(20 / 2)  0.176
(...= ... F2  (tan  ) N2 )
F1  (tan  ) N1  ( F1 )max  s N1
41
6/9 For a 20 jaw opening, what is
the minimum coefficient of static
friction between the jaws and the
tube which will enable tongs to grip
the tube with out slipping
Solution 2
y
The object is in equilibrium
N1
F1

R1

N2
Two force systems


x
R1  F1  N1
R2  F2  N2
R2
F2
2  20
o
   10o
F2
F
 1  tan 
N2
N1
tan   s
F  s N
F
 s
N
(s )min  tan  tan10o
42
Sample 6/5 Find the maximum value which
P may have before any slipping takes place.
?
Possible ways to slip
1) Middle object is going to move lonely.
2) Middle + buttom object is going to move together.
y
T
N1
x
F1
F1
y
N2
x
y
x
F2
F3
F
F1  F2  P  m2 g sin   0
f 2  2 N 2
 0:
 Fx  0 :
 F2  F3  m3 g sin   0
f 3  3 N3
N1  m1g cos
 Fy  0 : N2  N1  m2 g cos  0
N2  N1  m2 g cos  (m1  m2 ) g cos
N3
N2
f1  1N1
 Fy  0 : N1  m1g cos  0
N1
m2 g
F2
T  F1  m1g sin   0
x
m1g
P
 Fx  0 :
m3 g
 Fy  0 : N3  N2  m3 g cos  0
N3  N2  m3g cos  (m1  m2  m3 ) g cos
44
Friction F3 can support the
equilibrium of 40-kg object
Check the
Assumption
Assume: 1) Middle object is going to move lonely.
Impending motion
y
T
N1
F2  ( F2 )max  2 N2  2 (m1  m2 ) g cos )
x
F1
m1g
F1
y
T  F1  m1g sin   0
F
 0:
F1  F2  P  m2 g sin   0
F
 0:
 F2  F3  m3 g sin   0
x
x
N1
P
y
x
F2
F3
2 friction eq
F3  F2  m3g sin   468N
( F3 )max  3 N3  3 (m1  m2  m3 ) g cos  459N
N3
N2
3 equilibrium eq
P  [1m1 2 (m1  m2 )]g cos  m2 g sin   103.074N
m2 g
F2
5 unknowns
 Fx  0 :
x
N2
F1  ( F1 )max  1N1  1 (m1g cos )
m3 g
F3  ( F3 )max
Impossible, the assumption is wrong 45
F2 can support the 50-kg and 40-kg
as one body
Assume:
2) Middle + bottom object is going to move together.
F1  ( F1 )max  1N1  1 (m1g cos )
F3  ( F3 )max  3 N3  3 (m1  m2  m3 ) g cos
Impending motion
y
T
N1
x
F1
F1
y
F
 0:
F1  F2  P  m2 g sin   0
x
F
 0:
 F2  F3  m3 g sin   0
x
N2
F2  3 (m1m2 m3) g cos m3g sin   263N
N1
P
m2 g
y
x
F2
N2
( F2 )max  2 (m1m2 ) g cos  272N
F2  ( F2 )max
N3
F3
T  F1  m1g sin   0
x
m1g
F2
 Fx  0 :
m3 g
OK
P  1m1g cos  F2  m2 g sin   93.8N
Ans
46
Assume:
?
3) 40-kg object moves lonely --– possible?
F2  ( F2 )max  2 N2  2 (m1  m2 ) g cos
F3  ( F3 )max  3 N3  3 (m1  m2  m3 ) g cos
y
T
N1
x
F1
T  F1  m1g sin   0
F
 0:
F1  F2  P  m2 g sin   0
F
 0:
 F2  F3  m3 g sin   0
x
m1g
F1
y
x
x
N2
N1
P
Equation without unknown
(little chance to be true)
m2 g
F2
 Fx  0 :
y
x
F2
N3
F3
N2
m3 g
47
C  0.3
y
+
8-61 The uniform 50-N slender rod rests on the
top center of the 20-N block. Determine the
largest couple moment M which can be applied to
the rod without causing motion of the rod.
x
B  0.6
unknown: 8
N B  fC  W1  0
fc Nc f B N B
0.6 fC  0.8NC  M  0.3W1  0
 A  0.4
NC
W1  50
fC
NB
NB
fB
x
W2  20
NA
M fA NA x
N A  N B  W2  0
Motion Possibility?
f A  fB  0
1. Slip at C, B
0.3FB  x( N B  W2 )  0
M
fB
fA
f B  NC  0
2. Slip at C, A
fC  C NC
3. Slip at A, B ?
f B  B N B
4. other…
f A  A N A
48
C  0.3
y
+
8-61 The uniform 50-N slender rod rests on the
top center of the 20-N block. Determine the
largest couple moment M which can be applied to
the rod without causing motion of the rod.
x
B  0.6
Assume
f B  NC  0
2. Slip at C, A
N B  fC  W1  0
0.6 fC  0.8NC  M  0.3W1  0
 A  0.4
NC
W1  50
f A  fB  0
fC
0.3FB  x( N B  W2 )  0
M
fB
fC  C NC
NB
NB
fB
fA
x
N A  N B  W2  0
W2  20
NA
8 unknowns,
6+2 equation.
Don’t solve 8 eq
paralleling!
Solve only 6 eq.
excluding
fC  C NC
f B  B N B Assumption
f A  A N A
Checking
f A  A N A
49
C  0.3
y
+
8-56 The uniform 50-N slender rod rests on the
top center of the 20-N block. Determine the
largest couple moment M which can be applied to
the rod without causing motion of the rod.
x
B  0.6
Assume
f B  NC  0
N B  fC  W1  0
0.6 fC  0.8NC  M  0.3W1  0 Block is tipping
 A  0.4
NA 
W1  W2
= 62.5 N
1   A C
N A  N B  W2  0
x  0.3A = 0.12 m
f A  fB  0
M  (0.8  0.6C )
W1   A CW2 = 42.5 N 0.3FB  x( N B  W2 )  0
NB 
1   A C
f B  NC  f A  A
fC   AC
2. Slip at C, A
W1  W2
= 25 N
1   AC
W1  W2
1   AC
fC  C NC
f B  B N B Assumption
Checking
f A  A N A
= 7.5 N
 A (W1  W2 )
1   A C
 0.3W1 = 9.5 N-m
fC  C NC
25 ? (0.6)(42.5) = 37.5
f A  A N A
Correct?
50
C  0.3
y
+
8-61 The uniform 50-N slender rod rests on the
top center of the 20-N block. Determine the
largest couple moment M which can be applied to
the rod without causing motion of the rod.
x
B  0.6
unknown:
N B  fC  W1  0
fc Nc f B N B
0.6 fC  0.8NC  M  0.3W1  0
 A  0.4
NC
W1  50
fC  C NC
fB
x
W2  20
NA
f B  B N B
M fA NA x
Possibility?
1. Slip at C, B
0.3FB  x( N B  W2 )  0
NB
NB
N A  N B  W2  0
f A  fB  0
fC
M
fB
fA
f B  NC  0
2. Slip at C, A
3. Slip at A, B ?
C2  6
4
4. C-Slip,
other…x-limit
f A  A N A
5. B-Slip, x-limit
x  0.1
6. A-Slip, x-limit
51
C  0.3
y
+
8-61 The uniform 50-N slender rod rests on the
top center of the 20-N block. Determine the
largest couple moment M which can be applied to
the rod without causing motion of the rod.
x
B  0.6
Assume
f B  NC  0
4. C-Slip, x-limit
N B  fC  W1  0
0.6 fC  0.8NC  M  0.3W1  0
 A  0.4
NC
W1  50
f A  fB  0
fC
0.3FB  x( N B  W2 )  0
M
fB
fC  C NC
NB
NB
fB
fA
x
N A  N B  W2  0
W2  20
NA
f B  B N B
f A  A N A
x  0.1
8 unknowns,
6+2 equation.
Don’t solve 8 eq
paralleling!
Solve only 6 eq.
excluding
fC  C NC
Assumption
Checking
x  0.1
52
C  0.3
y
+
8-61 The uniform 50-N slender rod rests on the
top center of the 20-N block. Determine the
largest couple moment M which can be applied to
the rod without causing motion of the rod.
x
B  0.6
Assume
f B  NC  0
4. C-Slip, x-limit
N B  fC  W1  0
0.6 fC  0.8NC  M  0.3W1  0 N  3(W1  W2 )
A
3  C
 A  0.4
N A  N B  W2  0
x  0.1
W1  W2
3  C
= 21.21 N
f A  f B  NC 
W  W2
N B  W1  C 1
3  C
= 43.64 N
fC 
C
(W1  W2 )
3  C
= 6.36 N
f A  fB  0
M
0.3FB  x( N B  W2 )  0
fC  C NC
f B  B N B
f A  A N A
x  0.1
= 63.64 N
(0.8  0.6C )(W1  W2 )
3  C
 0.3W1 = 5.78 N-m
fC  C NC
Assumption
Checking
Ans
21.21 <=0.6*43.64=26.18
21.21 <=0.4*63.64=25.46
x  0.1
53
54
s  0.05
6/28 Determine the minimum angle at which
slipping does not occur at either contact point A
or C. The coefficient of static friction at both A
and C is 0.05. (Note: you should assume the mass of each
bar – why?)
known : m s
2m g
Bx
Bx
N1
F1
mg
N2
By

F2
By
unknown :
F1 F2 N1 N 2
 Bx B y
2m g
Equilibrium :
mg
N1
F1
N2

3 2  6
Friction : 2
F2
55
s  0.05
F
F
2m g
mg
N1
N2

F1
2 force
F2
2m g
Bx
N1
F1

6/28 Determine the minimum angle at which
slipping does not occur at either contact point A
or C. The coefficient of static friction at both A
and C is 0.05
F1  F2
x
 0:
F1  F2  0
y
 0:
N1  N2  3mg  0
N1  N 2  3mg
MB  0 :  N1 (l cos )  N 2 (l cos )
 N1  N 2 
l
 F1 (l sin  )  F2 (l sin  )  m g( cos )  0
2
M
B
 0 :  N1 (l cos )  F1 (l sin  )  0
F1 
By
1
( N1 )
tan 

N1 
5
mg
4
mg
2
N2 
7
mg
4
1 5
( mg )
tan  4
( )  (tan )  ( F1  F2 )  ... A or C is going to slip first?
s ( N1 )  ( F1 )max
s ( N2 )  ( F2 ) max

F2  F1 
1 5
( mg )
tan  4
s ( N1 )  ( F1 )max 
min  tan1 (
1
s
1
( N1 )
tan min
)  tan1 (1 / 0.05)  63.435
56
P applies at
pin B, AB or BC?
8-61 The end C of the two-bar linkage rests on the
top center of the 50-kg cylinder. Determine the largest
vertical force P which can be applied at B without
causing motion. Neglect the mass of the bars.
P applies at pin B
C  0.6
P
From solving
equilibrium eqs.
NC  P
P
fc 
tan 
fC  C NC ?
fC  (
E  0.3
2 force
P
tan 
1
) P  0.577 P
tan 
 C NC  0.6P
NC
fc
fC always fC ,max
However value P are, BC will not slip at C.
57
8-61 The end C of the two-bar linkage rests on the top center of the
50-kg cylinder. Determine the largest vertical force P which can be
applied at B without causing motion. Neglect the mass of the bars.
Max P without causing motion
NC  P
fC 
P
tan 
Slipping or tipping?
C  0.6
f E  fC  P cot 
N E  NC  W  P  W
fE
 x( P  mg )  0.2 fC  0
E  0.3
Slipping?
f E  E N E
P cot   E ( P  mg )
P
E mg
 531.227
cot   E
x
W  mg
 50(9.81)
NE
Tipping (x=0.05)?
0.05( P  mg )  0.2 P cot   0
P
5mg
 374.6
20cot   5
Ans
minimum value
58
6/23 The 10-kg cylinder is initially placed at V-Block. If s  0.5 , find
(1) friction force F acting at each side when P=0
(2) The P to start sliding the cylinder up y
F1  F2 ?
Assume: body in equilibrium
mg
(incomplete)
[  M y ,origin  0]
 F1  F1
30 y
y
F2
F1
2F
m gcos30
[ Fx  0 ]
x
P
z
 2F  10(9.81) sin 30  0
F  24.5N

2N cos45
2F
45
N1
45
N2
No tendency to move
in z-axis direction !!
Fz  0  N1  N2 =N
F
y
 0  2N cos45  mg cos30
N = 601
Assumption Checking
Fmax  s N  0.5(60.1)  30.0N OK
(2) Find P min to move
F
x
 0:
impending motion
 P  2F  10(9.81)sin 30  0
60
P  2(0.5)(60.1)  10(9.81)sin 30  109.1N
A
C
B
D
Determine the minimum coefficient of static
friction at each point of contact so that the pile
does not collapse.
weight W, radius r
Think of Equilibrium to find out the relation of F and N
FA  FB
N A  NB
F
N
2 N cos30  2 F sin 30 - W  0
FC  FA  F
( s , A ) min 
N
W
2
sin 30
1  cos 30
 N sin 30  F cos 30  F  0
F
sin 30 W
sin 30
N 
1  cos 30 2
1  cos 30
NC  W  N cos30  F sin30  0
1  sin 30

( s ,C ) min  

3  1  cos30 
NC 
3W
2
FC
61
6/108 Describe what should happen
Posibility?
1) A B and C don’t move
2) A move, B and C don’t move
3) B move (so do A), C don’t move
Many possibilities?
4) C move (so do A and B)
 ? tan1 C
C  0.35
B  0.20
 A  0.30
  15o
Yes: C not moving
No, C is moving.
C may or may not moving depends
on friction at A and B.
15 ? tan 1 (0.35)  19.29
C not moving
15  tan 1 (0.2)  11.30
B may or may not moving
depends on friction at A.
15  tan 1 (0.3)  16.69
A may or may not moving
depends on friction at A.
62
6/108 Describe what should happen
Assume 1) A and B don’t move, C don’t move
f B  s N B = 75.806
NB  mB g cos = 379.03
mC g
y
mA g
N2
mB g
x
fC
N1
= 58.403
f A  mA g sin   N1
= 125.507
NB
fB
fA
N1  mB g sin   f B
NC
Assumption Checking
NA
fC ? A N A
mB g
125.507
mA g
 A (mA g cos  N1 )
Assumption is wrong !
fB
fA
NA
NB
So, A and B move and C don’t move
f A  k , A N A
f B  k , B NB
63
Recommend Problem
6/16 6/125 6/23 6/41 6/40 6/37 6/108
64

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