Transcript Document

Physics
Session Opener
A recent space shuttle
accident occurred because
of failure of heat
protecting devices.
How was this heat generated ?
Session Objectives
Session Objective
1. Friction & frictional force
2. Bodies connected in frictional
surface
3. Kinetic friction
4. Static friction
5. Angle of friction
6. Coefficient of friction
7. Laws of friction
Friction and Frictional Force
Whenever an object moves or tends to
move, contact surfaces of object resist
the force tending to generate motion.
This property of the
surface is friction
The resistive force is the
friction force
Bodies Connected in Frictional
Surface
Consider the experiment :
1. F applied.
‘A’ does not move.
An equal and opposite
force f acts
A
F
MA
f
MB
2. ‘A’ still at rest on
increased F.
f adjusts to match F
Bodies Connected in Frictional
Surface
r
3. F increased further
A has acceleration.
f has a maximum limiting value
4. F is reduced
A moves uniformly
f during the motion is lower than at rest.
FF F
F
F
ff
f
F
MB
f
F
f
F
f
f
f
Bodies Connected in Frictional
Surface
 Friction retards motion
 Friction force (f) is zero if no
external force ( F ) exists
 f adjusts to stay equal and
opposite to F
 f has a limiting value
 moving friction force is less
than friction force at rest.
Bodies Connected in Frictional
Surface
Friction force is a contact force,
independent of area of contact.
f
F
Friction force is a non-conservative force as mechanical
energy changes to heat during friction.
That is why the space shuttle overheated and exploded
Kinetic Friction
 appears when two objects in
contact have relative motion.
 Directed opposite the direction
of motion
 fk is a constant.
 fk on B may start the motion
of B.
F
F
fk
A
B
fk
Static Friction
 F has component along contact
surface, but there is no motion.
 f acts along the surface and
opposite the force component
 f = Fcos till limiting friction.
 At limiting friction f(=fs)
is constant
 fs > fk
F
N
Fcos
fs
mg
Class Exercise
Class Exercise - 1
Blocks A and B are pressed against
a smooth wall and are in equilibrium
by a horizontal force F as shown.
Then friction force due to A on B
(a) upward
(b) downward
(c) dependent on relative mass of A and B
(d) system cannot be in equilibrium
Solution
A
B
F
Wall
As wall is smooth, A and B will keep
slipping down with acceleration g and
cannot be in equilibrium.
Hence answer is (d).
Coefficients of Friction
Limiting static friction
fs
:cons tan t  fs  sN
N
Kinetic friction
fk
:cons tan t  fk  kN
N
 are coefficient of friction
s  k
Coefficients of Friction
Graphically
f
fs,max   sN
Limiting
friction
fs  F
fk k N
F
0
Static region
Kinetic region
Laws of Friction
1. Bodies slip over each other
fk  kN
2. Direction of kinetic friction
is opposite to the velocity
3. Bodies do not slip over each
other
fs  sN
s N is called the limiting friction.
4. fk or fs do not depend
upon the area of contact.
Class Exercise
Class Exercise - 4
An object of mass m is pressed
against a wall with a force F and is
in equilibrium. The coefficient of
friction between the wall and the
object is . Then F must be equal to
mg

(a)  mg
(b)
(c) mg
(d) cannot be found
m
W a ll
F
Solution
As equilibrium exists, wall presses
against the object by a force of F
(equal and opposite to force exerted
by object against the wall)
 f  F
At equilibrium
f
mg  f   F
F
F
mg
 F

mg
Class Exercise - 5
Three blocks of masses ma,mb and
mc are arranged on a smooth
horizontal surface as shown. Surface
between ma and mb is smooth and
the coefficient of friction between mb
and mc is  .Then the minimum force
F required to keep mb from sliding
down is:
 ma  mb  mc  mbg
ma mb g
a
b
mc
 mc
c
d
ma  mb  mc  g

ma  mb  mc  mbg
 mc
F
m
b m
c
m
a
S
m
o
o
ths
u
rfa
c
e
Solution
Let the force F give an
acceleration a to the system
F  ma  mb  mc  a
To give an acceleration a to mc, mb presses
against mc with force mca, so mc gives a
normal reaction mc a (to left) on mb.
 mbg   mca
for mb not slipping down
mbg
ma  mb  mc  mbg

 a
 F
 mc
 mc
m
b m
c
a
F m
S
m
o
o
thsu
rfa
ce
a
Class Exercise - 6
A block of mass M is kept in a lift.
Coefficient of friction between the
block and the lift is . The force
required to initiate horizontal motion
of the block is maximum when
(a) Lift is moving up with constant acceleration
(b) Lift is moving down with constant acceleration
(c) Lift is stationary
(d) Lift is in free fall
Solution
Condition of motion of M: F   N
F is largest when N is maximum
N is maximum when lift moves up with
constant acceleration a [N = m(g + a)].
So F is maximum at that condition.
N
F
f
W
Class Exercise - 8
A 70 Kg box is pulled along a
horizontal surface by a 400 N force
at an angle of 30° above horizontal.
If the coefficient of friction is 0.50,
what is the acceleration of the box?
(g = 10 m/s2)
Solution
Vertical: N = mg – F sin30
Horizontal: F cos 30 – f = ma

F cos30   N  ma
F  cos30   sin30   mg  ma
Fsin 
 3 0.5 
 400 

  0.5  70  10  70 a
 2
2 

F
N
30°
f
Solving for a: a = 1.37 m/s2
mg
Fcos 
Angle of Friction
f along surface
(surface property)
N normal to surface
Angle of friction
f
 tan 
N
R
N

fs,max

Static
> kinetic
Class Exercise
Class Exercise - 2
The angle between the resultant contact force and the normal reaction force
exerted by a body on the other when
they are one on top of the other is f.
Then, what is the coefficient of friction,
(a)  = tan f
(b)  > tan f
(c)  < tan f
(d) f and  are not related
Solution :
f
   tan f
N
Hence answer is (a)
Class Exercise - 3
A pushing force F is applied to a
body of weight W, placed on a
horizontal table, at an angle  . The
angle of friction is f . The magnitude
of F to initiate motion in the body is
W sinf
a.
cos   f
W cos f
b.
cos   f
F
W tan f
c.
cos   f 
W sin f
d.
sin   f 

w
Solution
Vertical: N  W  Fsin
Horizontal: Fcos  f
(a = 0 at initiation of motion)
f  N tan f   W  F sin   tan f
F cos    W  F sin   tan f




sin f 
F  cos   sin .
 W tan f

cos f 

cos  cos f  sin  sin f  W sin f

F

cos f
W sin f
F
cos    f 
cos f

F sin 
N
F cos 
f
F cos    f    W sin f
W
Angle of Repose
Object at rest
f
increased .
Object slips
N
fs
f is angle of repose
f
 tan f
N
s
f f
mg
Angle of Repose
f
reduced
At
f
f
: Angle of repose (kinetic friction)
k
k
, object moves uniformly.
N
fk
fk
 tan fk
N
fk  fs
f f
mg
Class Exercise
Class Exercise - 9
A block ‘A’ slides from rest down an
incline with a 30° inclination with
horizontal. It covers 3 m in 5 seconds.
What is the value of coefficient of
kinetic friction? (g = 10 m/s2)
A
30°
Solution
Along y: N = mg cos
Along x : mg sin – f = ma
 g sin    cos   a
1 2 1 2
x  ut  at  at
2
2
N y
2x 
1

   gsin   2  .
t  cos .g

2
10 2  3 
   0.53


.
25  3  10
2
f
m
gsin
x

m
g
m
gcos
Class Exercise - 10
A block of mass 2kg lies on a rough
inclined plane of inclination 30° to
the horizontal. Coefficient of friction
between the block and the plane is
0.75. What minimum force will
make the block move up the
incline? (g = 10 m/s2)
M
30°
Solution
Along y: N = mg cos
F is up the incline (along x)
F – (mg sin  + f) = ma
To start the motion a is put equal
to zero
 Fmin  mg sin   f 
y
 mg sin    mg cos 
 mg  sin    cos 
1 3 3 
 2  10   .
  22.98 N
2 4 2 
F
x
N
mg cos 
mg sin 
mg

Thank you