Friction (2) [compatibility mode]

Friction

Cause of dry friction
Contact between two surfaces.
Hence first task in a friction problem is correct
identification of contact surfaces
Identify the surface, the normal and the tangential
vectors.
Also important is to get an idea of probable direction
of relative motion
The contact force acts along the normal.
Gravity is the most common cause of normal force.
Friction acts along the tangent plane opposite to the
direction of relative motion

Normal

Relative velocity

Create PDF files without this message by purchasing novaPDF printer (http://www.novapdf.com)

Friction

Normal

Relative velocity

Friction problems are
essentially equilibrium
problems with one f the
forces being functions of
another

N
Fr=f(N,V)

Fr=f(N,V)
N


The correct way of writing the dry friction force

V ˆ
Fr    N   N V
V

N=Normal force vector

V=Relative velocity vector of the body
m= coefficient of dry friction or Coulomb
friction


Problem 1

Knowing that the coefficient of friction between the 13.5
kg block and the incline is ms = 0.25, determine

a) the smallest value of P required to maintain the block in
equilibrium,
b) the corresponding value of b.


Problem 1 y

N cos60  mg  f sin60  P sin   0
 N sin60  f cos 60  P cos   0
f  N x
mg sin60  f
P
sin  sin60  cos  cos 60
N cos60  mg   N sin60  P sin   0
mg  P sin 
N
cos 60   sin60
 N sin60   N cos60  P cos   0
mg  P sin 
  sin60   cos 60   P cos   0
cos 60   sin60
   mg  P sin   sin60   cos60   P cos   cos 60   sin60   0
  mg  sin60   cos60   P sin   sin60   cos 60   P cos   cos60   sin60   0
 P  sin60 sin    cos60 sin   cos60 cos    sin60 cos    mg  sin60   cos60 
 sin60   cos 60 
 P  mg
cos  60      sin  60   
1 cos  60      sin  60   
  mg
P  sin60   cos 60 
d 1 d
 P   0  d   cos  60      sin  60      0
d    
d
  sin60 sin    cos 60 sin   cos60 cos    sin60 cos    0
d
 sin60 cos    cos60 cos   cos 60 sin    sin60 sin   0
 sin60   cos 60  o
 tan      2.614    69
 cos60   sin60 

P  mg
 sin 60   cos60   mg
 0.866  0.125   0.72mg
cos  60      sin  60    cos  9    sin  9 


Problem 2

Knowing that P = 110 N, determine the range of values value of q
for which equilibrium of the 8 kg block is maintained.


Problem 2
y y

x x

mg mg

 P cos   N  0
 P cos   N  0
P sin  mg  f  0
P sin  mg  f  0
f  N
f  N
 N  P cos 
 N  P cos
P sin  mg   N  0
P sin  mg   N  0
 P sin  mg   P cos   0
 P sin  mg   P cos   0
 P  sin   cos    mg
 P  sin   cos    mg
mg
mg P
P sin   cos 
sin   cos 
Hence
Hence
upward movement will not start before
downward movement will not start before
mg
mg P
P sin   s cos 
sin  k cos 


Problem 3

The coefficients of friction are ms = 0.40 and mk = 0.30 between all
the surfaces of contact. Determine the force P for which
motion of the 27 kg block is impending if cable
a) is attached as shown,
b) is removed


Problem 3
v

T N1
f1
m1g f1
N1 T
m2g f2
N2

T   N1  0
N 1  m1 g  0
 T   N 1 , N 1  m1 g
 T   m1 g
T   N1   N 2  P  0
N 2  N 1  m2 g  0
N 2  N 1  m 2 g  0  N 2   m1  m 2  g
T   N1   N 2  P  0
  m 1 g   m1 g    m1  m 2  g  P  0
 P  3  m1 g   m 2 g


Problem 3
v

0 N1
f1
m1g f1
N1 0
m2g f2
N2

T   N 1  m1 a
Now T  0    N 1  m1 a
N 1  m1 g  0
 N 1  m1 g
 N1   N2  P  0
N 2  N 1  m2 g  0
N 2  N 1  m 2 g  0  N 2   m1  m2  g
 N1   N2  P  0
  m1 g    m 1  m 2  g  P  0
 P  2  m1 g   m2 g


Additional Problems

The 8 kg block A and the 16 kg block B are at rest on an
incline as shown. Knowing that the coefficient of static
friction is 0.25 between all surfaces of contact, determine the
value of q for which motion is impending.


Toppling

The magnitude of the force P is slowly increased.
Does the homogeneous box of mass m slip or tip
first? State the value of P which would cause each
occurrence.


Slip or topple?

mg

f1 f2

N1 N2

P cos 30   N 1   N 2  0
P sin 30  N 1  N 2  mg  0
 P cos 30  d  mg  d   N 2  2d  0
P cos 30   N 1   N 2    N 1  N 2 
P sin 30    N 1  N 2   mg
P cos 30  mg  2  N 2
 P sin 30     N 1  N 2    mg   P cos 30   mg
 P sin 30  P cos 30   mg  P   sin 30  cos 30    mg
 mg
P
 sin 30  cos 30
0.5mg
P  0.448mg
0.25  0.866


Slip or topple?

mg

f1 f2

N1 N2

P cos 30  f 2  0
P sin 30  N 2  mg  0
 P cos 30  d  P sin 30  2d  mg  d  0

P cos 30  f 2
P sin 30   N 2  mg
P cos 30  2P sin 30  mg  0
 P  cos 30  2 sin 30   mg
mg
P
cos 30  2 sin 30
mg
P  0.536mg
0.866  1


Additional Problems


Wedge
f=mN
P

q

N mg

P  f cos   N sin  0
f sin  N cos   mg  0
f  N
mg
 N sin  N cos   mg  0  N 
 sin  cos 
P   N cos   N sin  0
 cos   sin
 P    cos   sin  N  mg
 sin  cos 
P   tan
 
mg  tan  1
P tan  tan
   tan    
mg tan tan  1


A screw thread is a wedge



M=Pa

Q=Pa/r
= equivalent force

W

Q
N
f=mN



W

Q
N
q f=mN

Q  f cos   N sin  0
 f sin  N cos   W  0
f  N
W
  N sin  N cos   W  0  N 
  sin  cos 
Q   N cos   N sin  0
 cos   sin
 Q    cos   sin  N  W
  sin  cos 
Q   tan
 
W 1   tan Pa   tan

Wr 1   tan
1 P
if tan   
 W
Screw locks



The
W

P
N
q f=mN

1 Pa
tan    Screw locks
 Wr
1 Wr
  0
tan Pa

There is a critical value of friction
coefficient beyond which the thread
does not move irrespective of the
force applied.
This happens when a screw is not
maintained properly. Because of dirt
and rust m becomes more than
critical.



W

N
f=mN



W

N
q f

For no movement
f cos   N sin  0
f sin  N cos   W  0
Self locking
f sin
 
N cos 
sin
  tan
cos 

Therefore after raising the load if we
let go of the screw the load will not
cause the screw to unscrew by itself.


Terminologies

Lead (L)
2pr

Pitch (p)

Lead  L  np
where
n=no. of parallely running threads = starts
L
tan =
2 r


Turnbuckle

T1 T2

Used to apply tension.
The sleeve is rotated to pull the
threads together.

M   tan

 T2  T1  r 1   tan


An improved screw jack

W

q
q

W
T

T T
T

2T cos q


An improved screw jack

W

f
f

M=Pa
W=2T cos f

Pa  cos  sin

Wr   sin  cos 
M  cos  sin
 
2Tr cos    sin  cos 


Worm gear

MG

M0/R

R N
M f

MG
M G  WR  W 
R
Pa   tan

Wr 1   tan
M   tan
 
MG 1   tan
r
R
MR   tan
 
M G r 1   tan


Belt drives


Belt drives

 
F x
 0  T cos   F   T  T  cos 0
2 2
 
 Fy  0  T sin 2   N  T  T  sin 2  0
 F  s  N
 
T cos   F   T  T  cos 0
2 2
  
 T cos   F  T cos  T cos 0
2 2 2
 
   F  T cos  0  T cos   F  s  N
2 2
 
T sin   N   T  T  sin 0
2 2
  
 T sin   N  T sin  T sin 0
2 2 2
 
 2T sin   N  T sin 0
2 2
 
 2T sin   N  0  2T sin  N
2 2


Belt drives


T cos  s  N
2

2T sin  N
2
 
 T cos  2T  s sin
2 2

sin
1 T  2
 cos  s
2T  2 

sin
1 T  2
Lim cos  Lim 
  0 ,T  0 2T  2  0 ,T 0 s 
1 dT 
 1 s
2T d 2


dT   d
T s


Belt drives

Belt is just about to
slide to the right

dT   d
T s
T2
dT     
     s d T2
e s T e s T
T T 10 2 1
T2 
T1
  lnT 
  T1
   s 0
  Torque required to drive the pulley
 lnT  lnT2 1
 s     
T2  T1   e s  1 T
 ln T   
2 

 1

T 1
s

T
 2  es
T
1


Belt drives : Important points

T
 2  es
T
1

 Angle b must be expressed in
radians
 Larger tension occurs at the end of
the belt where motion is about to begin
or is already moving
 T2 is used to denote the larger
tension


Belt drives

A B

q

q

A B

q

q


Belt drives

T2
T2
 exp      2   ???
 
T1
A Always check
for mb value for
each pulley in a
system. The one
with the
T1
smallest mb
q
value will
determine the
Which expression is tensions.
correct???? Is T2>T1
Or T1>T2.
One pulley must slip. T2
Friction force is larger
for the larger pulley since
angle of wrap is larger.
Hence smaller pulley
B
slips and determines the
tension

T1 q
 exp      2   ???
  T1
T2


Band brakes

Important!!!
In pulleys the belt
is mobile and
tries to drive the
pulley. Here the
drum moves and
tries to move the
belt. Hence the
sense of slack is
opposite to that
for a pulley.

T1

T2


Band brakes
y
The drum rotates towards the right.
Ry If this were a pulley the slack side
would be on the left. But here as the
Rx
Mext drum is the moving part, the slack
x side is on the left.

T1  T2 exp(  k )

T1  M  0  M ext
 r  T1  T2   0

T2  M ext   T1  T2  r
F x
 0  Rx  T2 sin(    )
F y
 0  R y  T1  T2 cos(    )

a -p

F x
 0  Bx  T2 sin(    )  0
 Bx  T2 sin
F y
 0  B y  T2 cos(    )  T1  P  0
 B y  T2 cos   T1  P  0
M B
 0  cT2  bT1   a  b  P  0
bT1  cT2
P
ab


Band brakes

T1  T2 exp(  k ), M ext   T1  T2  r
bT1  cT2
Bx  T2 sin  , B y  T2 cos   T1  P  0 , P
ab
M ext   T1  T2  r  T2 exp(  k )  T2  r  exp( k  )  1T2 r
   
M ext
 T2 
exp(  k )  1 r
 
bT1  cT2 bT2 exp( k  )  cT2  bexp( k  )  c 
P    T2
ab ab  ab 
 bexp(  k )  c  M ext
P  exp(   )  1 r
 ab  k 
 c 
 1  exp(   k ) 
b M ext
P 
 1  exp(   k )   a
1  r

 
 b
c
Now 1  exp(    )  0  P  0 if 1  exp(    )  0
k b k
 k
 P  0 iff b > ce > ce  s   s  k 

 

Otherwise P is negative
negative.
Which means no force is required for braking!
Hence self locking occurs.
Since Mext is equal in magnitude to the torque required
for braking it is also called the braking torque.


Clutches

Assume pressure is uniform over the contact zone
P
 p=
 R2
Consider a small element of width dr at a radial distance r
and spanning an angle d
dA  r dr d
dF   pdA   prdrd
dM  rdF  r  prdrd   pr 2 drd
2
2 R
2
 r 3  R   p 3 2
 M     pr drd   p       2 R   pR 3
0 0  3 0 0
  3 3
2  P  3 2
M   R   PR
3   R2  3


Clutches

Assume pressure is uniform
over the contact zone
P
 p=
  r12  r02 
Consider a small element of spanning angle d at a radial distance r
dA  r dr d
2
2  r1  r 3 r 
1
p 3 2
  pr drd   p   3     3  r1  r0  2  3  p  r1  r0 
2 3 3 3
M 
0 r
0  r 0
 0
2  P  3
2  1
 r  r0 
3
M 
3    r1  r0  

2

2  r13  r03 
 M  P 2
3  r1  r02 


Disc Brake

dq

Area of a whole circle (spanning 2 radians) =  r 2
 r2  r2
 Area of a sector =   =
2 2
2P
 p=
  R02  Ri2 
Consider a small element of spanning angle d at a radial distance r
dA  r dr d

 Ro  r 3 R  o
p 3  p
2
 M     pr drd   p        Ro  Ri3    3  Ro3  Ri3 
0 R i  3  R 0
  i
3
 p 3
M  Ro  Ri3     Ro3  Ri3   R2P R 2
3 3  02  i 
2   Ro  Ri 
3 3

M
3  R02  Ri2 


Thrust bearing
P

d
q2
q q

d1

P

q d1
q

d1
d2
q


Thrust bearing

y dl

df X

dN

O x


Thrust bearing y dl
r dr

df X

dN

O x
P
P=  R2 2  R1 2  p  p=
  R2 2  R1 2  What
R2 2 
about q
M    rdf 
R1 0
=0?
dN  pdA  prd dl , dl  drcosec
dN  prd dl  prd  dr cosec   pcosec  d rdr
df   dN   pcosec  d rdr
R2 2 

N   pcosec  d rdr
R1 0

R2 R2 R2
2  r2 
N   p cosec   0
rdr   2 pcosec  rdr   2 pcosec  
2 R
R1 R 1  1

 N   pcosec   R2 2  R1 2 
R2 2  R2 2 

M    rdf     r  rpcosec  d rdr
R1 0 R1 0

R2 R2 R2
2  r3 
   pcosec  
2 2
M 0
r dr   2 pcosec  r dr   2 pcosec  
R1 R 1  3 R 1

2 P  R2  R1 
3 3
2 p
M  R2  R1   3sin R 2  R 2
3 3

3sin  2 1


Torque wrench

r  T1  T2   aP
Smallest m for self locking ?
T2
T1 T1  T2 exp    2    
 

N

mN

T1  N cos    T2   N  sin  0
 P  N sin   T2   N  cos   0

mN
T1

N
T2


Torque wrench

Condition for self locking ?

T1  T2 exp    2    
 
r  T1  T2   aP
T1  N cos    T2   N  sin  0
 P  N sin   T2   N  cos  0
 T2  T1  sin   cos    P  cos    sin 
T1  T1  sin   cos    P  cos    sin   exp    2    
   
T1  T1  sin   cos   exp    2       P  cos    sin  exp    2    
   
T1   sin   cos  exp    2      1  P  cos   sin  exp    2    
   
P  cos    sin  exp    2    
  P  cos    sin 
T1  
  sin   cos  exp    2      1   sin   cos   exp     2     
   
aP
r  T1  T2   aP  T1 

r 1  exp     2    
  
aP P  cos   sin 
 

r 1  exp     2    
     sin   cos   exp     2     
 
r  cos    sin   a  sin   cos  
 exp     2     
   r  cos    sin   a 
 
r  cos    sin   a
 exp    2     
  r  cos   sin   a  sin   cos  

 a 
1   cos   sin   
 ln  r
a 
2   
 cos    sin    sin   cos  
 r 


Band brakes


Band brakes

TB T1 T3
B

T1 T3

T4
T2

 75   75 
T2  T1 exp      T1 exp  0.25 
 180   180 
T2  1.39T1
 135   135 
T3  T4 exp      T4 exp  0.25 
 180   180 
T4  0.55T3
M 0    T3  T4  T2  T1  R
Consider the pin B
F x
 0  T1 cos 45  T3 cos 45  T1  T3
F y
 0  T1 cos 45  T3 cos 45  TB  2T1 cos 45  TB  2T1  TB
 T2  1.39T1 ,T3  T1 ,T4  0.55T1
Thus the largest tension is T2  5.6  T1  T3  4.03,T4  2.22,TB  5.7
 M   5.6  2.22  R  3.38  0.16  0.54 KNm  540 Nm


Problems


Tip or Slide


Friction (2) [compatibility mode]

Recommended

Recommended

More Related Content

Recently uploaded

Recently uploaded (20)

Featured

Featured (20)

Friction (2) [compatibility mode]