FRICTION

6.1 INTRODUCTION : irregularities will be sheared off and breaking of

Newton’s first law tells us that in the absence of welded joints takes place and the body starts
an external force, the state of a body remains unal- sliding on the surface.
tered. In practice, certain cases appear to contradict 6. Friction is regarded as electromagnetic force.
this. For instance, when a body is sliding on a rough 6.3 ADVANTAGES AND DISADVANTAGES OF
surface, it slows down and finally comes to rest. The FRICTION:
possible explanation for this is a force acting between 6.3.1 Advantages of Friction:
the surfaces in contact. If the surface on which the body The advantage of friction can be understood if
is sliding is also in motion, this force tries to oppose the we imagine the situation when friction is absent.
motion of the surface as well. One also notices that if The surfaces in contact will slip on each other with-
the direction of motion of the body is reversed, the out opposition in the absence of friction. This may
opposing force also gets reversed in direction exhibit- produce undersirable and dangerous motion. Fric-
ing its self adjusting nature. tion is very much necessary to produce a con-
Friction may be defined as the opposing trolled motion. The following are a few examples.
force which comes into play tangentially between 1. Safe walking on the floor is possible because of
two surfaces so as to destroy the relative motion friction between the floor and the feet.
between them. 2. Nails and screws are held in the walls or wooden
Friction is self adjusting force and is a non – surfaces due to friction.
conservative force. 3. Friction helps the fingers to hold a drinking wa-
6.2CAUSES OF FRICTION : ter tumbler or a pen.
1. A surface though appears smooth by visual in- 4. Vehicles move on the roads without slipping due
spection or by touch, when viewed under a pow- to friction and they can be stopped due to fric-
erful microscope, it consists of a large number of tion.
surface irregularities. 5. The mechanical power transmission of belt-drive
2. When two bodies are placed one above the is possible due to friction.
other, these surface irregularities interlock to- 6.3.2 Disadvantages of Friction
gether and oppose any attempt to bring a rela- Friction is disadvantageous because it opposes
tive motion between the bodies. motion and hence produces many undersirable
3. At the actual points of contact a sort of “weld- effects. Few examples are given below.
ing” takes place due to intermolecular forces. 1. Friction results in the large amount of power loss
4. Friction is due to cohesive or adhesive force in engines and hence brings down their efficiency.
among molecules at close proximity. 2. The wear and tear of the machines increases due
5. When a sufficiently large force is applied, these to friction reducing their life.
This area element is lifted up bringing the neighbouring
points on the body in contact with the surface and this
process is continued and the rolling is affected.
Note 6.1 :
Friction on the wheels of a bicycle :
a) While pedalling a bicycle, the force exerted
by the rear wheel on ground makes the force of friction Here we notice that until the static friction reaches
act on it in the forward direction. Front wheel moving its maximum value applied force is directly proportional
by itself experiences force of friction in backward di- to the frictional force. Thus the angle made by the
rection. straight line with X-axis is equal to 450.
b) If pedalling is stopped, or in uniform motion Slope of the line “m”= tan  = tan450 = 1.
or free wheeling, both wheels move by themselves and The static friction is always equal to the force
so experience force of friction in backward direction. or component of force parallel to the two surfaces
6.6 Variation of Friction with applied force : in contact in magnitude.
N The dynamic friction is equal to the applied
F1 force in magnitude only when it is sliding with con-
f B stant velocity.
While the static friction is always opposite to
W
the applied force in direction.
The kinetic frictional force is always oppo-
site to the direction of the velocity of sliding.
6.7 Normal Reaction :
Consider a block ‘B’ which is at rest on a hori- Whenever a body is in contact with a surface a
zontal table as shown in figure. A a small pan is at- force acts on it due to the contact. This contact force is
tached to the block by means of a thread passing over due to the pressure of contact and will be perpendicu-
a frictionless pulley. When the weight in the pan is in- lar to the tangent to the surface at the point of contact
creased the applied force also increases, the static fric- Normal reaction is the resultant contact
tion also increase in equal magnitude. force acting on a body placed on a rigid surface
Let the static friction reach its maximum value perpendicular to the plane of contact.
Normal reactiuon depends only on the nature
for an applied force F1. This value of F1 is numerically
of the contact. It does not depend on the area of con-
equal to limiting friction fL . Therefore the net force on
tact. Even in the event of the motion of the body on the
the body is zero and the body is in equilibrium. If the surface this force remains unaltered.
applied force is increased slightly more than fl , the N
body starts sliding on the surface. However, the mag- N

nitude of force required to keep the body sliding under 0

friction with constant velocity is slightly less than the mgSin 
mgCos
force required to start the motion. The friction in this  mg
mg
condition is called dynamic friction. Hence dynamic fric-
tion is less than the limiting friction for the same body Consider a body of mass ‘m’ lying on a horizon-
on the same surface. Even if the applied force is in- tal surface. The weight of the body acts vertically down
creased the dynamic friction remains constant. on the surface. The surface exerts a reaction “N” of
Let us consider a rectangular block of mass ‘m’ on a
F = s N ; F  s mg {  N = mg}
rough horizontal surface in limiting friction under the
applied force ‘F’The forces acting in the system are Consider a body of mass “m” placed on a rough hori-
weight of the body “mg” acting vertically downwards, zontal surface of coefficient of kinetic friction “  k ”. A
Normal reaction “N” vertically up and limiting friction horizontal force “F” greater than limiting friction is ap-
“fL” in a direction opposite to the direction in which the plied on it so that the body comes in to motion. Now
body tends to move. If a parallelogram is constructed kinetic friction acts on it in a direction opposite to its
taking limiting friction and normal reaction as the motion. If “N” is the normal reaction, then the kinetic
adjacent sides, its diagonal represents their resultant. friction is given by f k  k N  k mg
The angle made by this resultant with the normal If “a” is the acceleration produced in the body,
reaction is called “angle of friction (  )”. the resultant force acting on the body is given by
In fig. OA represent N, OB represents fL. Com- F  f k F   k mg
FR  F  fk  ma  F  fk  a  m

m
pleting the parallelogram OACB, OC gives the result-
ant of N and fL. N

AC OB fL
In  OAC, tan       s .  s  tan .
OA OA N F (Applied)
fk
The greater angle of friction the greater is the f k  k N
 k mg
value of coefficient of friction.
mg
Note : The forces exerted by the surface on the body
are friction “fL” and normal reaction “N”. As these two Note 6.3 : The distance travelled and velocity acquired
are mutually perpendicular, the net contact force acting in a given interval of time “t” can be obtained from the
1 2
on the body is given by kinematic equations s  ut  at and v  u  at .
2
FR = f L2 + N 2 2
= s N  +N 2 =N s2 +1 Note 6.4 : A force “F’ just enough to set the block into
motion is applied on the block. This by definition is
 FR =mg s2 +1 equal to maximum static friction  fL  s mg . If this
force is continued, even after the block starts moving,
FR =mg Tan 2+1  s=Tan the body now has to overcome kinetic fric-
tion  fk  k mg which is less than the applied force.
mg
FR  mg sec   Thus, there is a net resultant force which produces
cos  acceleration in the body.Then,
6.10 Motion of a body on Rough Horizontal F  f L  s mg
Surface : N f k  k mg,
Applying a Horizontal Force: a FR  f L  f k 
Consider a body of mass “m” lying on a hori- F FR  Smg  K mg 
fk
zontal surface of coefficient of static friction “  s ”. Let ma  S  K  mg
a horizontal force “F” is applied on the body as shown mg  a S  K  g
in the figure to bring it into motion. For the body to
come in to motion the applied force “F” must be at Problem 6.1 :
least equal to the limiting friction “fL”. A body of mass 60 kg is pushed up with just
enough force to start it moving on a rough surface
with µ s = 0.5 and µ k = 0.4 and the force contin-
ues to act afterwards. What is the acceleration of
Thus, when F = fL, FR = 0
Problem 6.7 :  friction = 8N
A pile driver of mass 200kg falls vertically Problem 6. 10 :
through 3.2m on to a pile of mass 600kg.
A block on table shown in figure is just on
Which it drives into the ground through a
the wedge of slipping. Find the co-efficient
distance of 0.08m. If the pile driver remains of static friction between the block and table
in contact with pile after collision. Find the top
resistance offered by the ground (g=10ms-2) T cos30
T
Sol : (M + m) V = m 2gh  800V = 200 (8) Sol:
30 0

40N T sin 30
 V = 2 ms–1
from work energy theorem T sin30 0  mg
 ;
1 T cos300 80
 f   M  m  g  S   M  m  V 2 80N

2
1 2  40 1   
2
 1.15
 f  8000 0.08  2 800  4 Tan300 = ; 
80 3 2 3
Problem 6. 11 :
Solving we get f = 28, 000N Two blocks A and B attached to
each other by a massless spring or kept on a
Problem 6. 8 : rough horizontal surface    0.1 and pulled
Two blocks A and B of masses 2kg and by a force of 200N as shown in figure. If at
3kg are connected by a light string as shown some instant, the 10 kg mass has an
in the figure and placed on a horizontal sur- acceleration of 12 ms –2 , what is the
face.  between all surfaces is 0.1 and g = acceleration of 20 kg mass (in ms2).
10 ms–2. The acceleration of the system is, B
A
when the froce applied F = 45N F=200N
10 kg 20 kg
Sol: F    m1  m2  g   m1  m2  a
45 – 0.1 (5) 10 = 5a Sol: For A: Kx –  mAg = mA a
kx – 0.1 (10) (10) = 10 (12)
45 – 5 = 5a For B: 200 – kx –  mBg = mBa
2kg 3kg 45N
a = 8ms–2 200 – 130 – 0.1 (20) 10 = 20a
a = 2.5 ms–2
Problem 6. 9 :
A block of mass 4 kg is placed on a rough Problem 6.12 :
horizontal plane. A time dependent horizontal A block of mass M slides along the
sides of bowl as shown in the figure. The walls
force F = kt acts on the block (k = 2 N/s).
of the bowl are frictionless and the base has
Find sthe frictional force between the block coefficient of friction 0.1, and length 0.5m. The
and the plane at t = 2 second and t = 5 second block is released from the point A which is
   0.2 0.2 m high as shown in figure. Then the block
comes to rest
Sol: When t = 2sec F = 4N
fms = s mg = 0.2  4  10 = 8N A
M
 F < fms  friction = 4N
When t = 5 sec F = 2 (5) = 10N 0.2m
Sol:
F>f P Q
0.5m
1 2
From law of conservation of energy mgh = mv
2
 v 2  2  0.2  g  0.4g
pull it on the horizontal surface as shown in the figure. s mg
The forces acting in the system are; Fmin 
s 2 1
1. Weight of the body “mg” acting vertically down.
This is the least force required to move the
2. Normal reaction “N” exerted by the surface
on the body vertically up. body on a horizontal force.
3. Frictional force “fL” opposing the motion. Application–6.2 :
4. The applied pulling force “F” acting at an angle Applying an Inclined Pushing Force :
“ θ ” with the horizontal. Consider a body of mass “m” lying on a hori-
The applied force “F” can be resolved in to two zontal surface of coefficient of static friction “mS”. Let
an inclined force “F” is applied on the body so as to
components “F Cos θ ” and “ F sin  ”. The body is in
push it on the horizontal surface as shown in the figure.
contact with the surface, there by
N + F Sin θ = mg  N = mg – F Sin θ ——— (1) N
For the body to be pulled F Cos θ = fL
F Cos θ =  S N  F Cos θ =  S (mg – F Sin θ ) 
F Cos
F Sin 
fL
N F
mg F

F Sin
fL F Cos

The forces acting in the system are ;

1. Weight of the body “mg” acting vertically down.
mg
2. Normal reaction “N” exerted by the surface
F Cos θ =  S mg –  S F Sin θ on the body vertically up.
F Cos θ +  S F Sin θ =  S mg 3. Frictional force “fL” opposing the motion.
4. The applied pushing force “F” acting at angle
F (Cos θ +  S Sin θ ) =  S mg
“  ” with the horizontal.
 s mg
F The applied force “F” can be resolved in to two compo-
cos   sSin
nents “F Cos  ” and “F Sin  ”. The body is in contact
Sin
mg  Tan   s  with the surface, there by N = mg + F Sin  –––– (1)
cos   
F sin  For the body to be pulled F Cos  = fL
   
cos   Sin Sin
 s
 cos  
 cos   
   angleof friction

 F Cos  =  S N  F Cos  =  S (mg + F Sin  )
F Cos  =  S mg +  S F Sin 
mgSin mgSin 
F F F Cos  –  S F Sin  =  S mg
cos  cos  SinSin  cos    
F (Cos  –  S Sin  ) =  S mg
For F to be minimum cos     should be s mg
F
(co s    S in  ) 
maximum  cos     1      0 ,    s
 
F
mgSin    = Sin 
 Fmin  mg sin  or  mg sin  (cos  Cos  Sin Sin)  s Cos 
 s2  1
s
s
From the figure, Sin  mgSin 
 s2  1 F
cos (  )
1
 cos  cos   sin  sin   cos(   ) 
travels before coming to rest, mv2
s mg  v   s rg
W  KE r
1  1 This is the safe maximum speed with which the
f k S   mv 2  0  k mgS  mv 2
 2  2 vehicle can negotiate the turn, to avoid skidding of the
S g
N vehicle. The maximum angular velocity is  
r
Problem 6.15 :
u V v 0
A car is driven round a curved path of radius
V2 fk 18 m without the danger of skidding. The coeffi-
S cient of friction between the tyres of the car and
2 k g S
the surface of the curved path is 0.2. What is the
mg maximum speed in kmph of the car for safe
One can say that the safe maximum speed of the driving ? (g = 10 ms-2 )
vehicle for which it comes to rest within a distance of So1. Maximum speed. v = s gr

“S” is given by V  2kgS v = 0.21018  36  6ms 1

Application–6.7 :
If “t” is the minimum time in which the body is to Body Placed On A Rotating Disc :
be brought to rest, then impulse Consider a body of mass “m” placed at a dis-
I  P  f kt = mv – 0 tance “x” from the centre of a circular disc of radius
v “R”. The disc is rotating in a horizontal plane with a
 k mgt  mv  t   g constant angular velocity “w” about a vertical axis pass-
k ing through its centre. The two forces acting on the body
Problem 6.14 : are, the centrifugal force acting radially outwards trying
When a car of mass 1000 kg is moving with to push the body outwards and the frictional force be-
a velocity of 20ms–1 on a rough horizontal road, tween the surfaces in contact trying to prevent the body
from sliding. For the body to be under equilibrium,
its engine is switched off. How far does the car
move before it comes to rest if the coefficient of s g
mx 2  s mg   
kinetic friction between the road and tyres of the x
car is 0.75 ? The minimum time period of rotation for which
Sol : Here v = 20ms–1, K = 0.75, g = 10ms–2 the body can be under equilibrium is given by
v2 x
Stoping distance S  =26.67m T  2
2K g sg
Application–6.6 :
Application–6.8 :
Vehicle Taking A Turn On A Horizontal Road:
Body Placed On An Accelerating Truck:
Consider a vehicle of mass “m” moving on a
Consider a block of mass “m” placed on an
horizontal rough road with a velocity “V”. The coeffi-
open ended truck as shown in the figure. Let “” be
cient of friction between the tyres and the road is "  s " . the coefficient of friction between the surfaces in con-
When it takes a turn of radius of curvature “r”, the fric- tact. Let the truck starts from rest and accelerates at
tion between the tyres and the road supplies the neces- the rate “aT”. Now the block in the non – inertial frame
sary centripetal force. experiences a pseudo force “maT” in a direction oppo-
site to the direction of motion of the truck.
acceleration of the system a= µS g two blocks is f = mL a
Thus, the maximum force for which both the iv) If F > Fmax the blocks slide relative to each
bodies move together without slipping is other and they move with different accelerations.
In this case, the acceleration of the lower block
Fmax = µS  mU +mL  g
µK mU g
mU is aL =
mL
F
mL The acceleration of the upper block is
F  µK mU g
aU =
ii) If a < µS g the two blocks move together mU
and the force is F =  mU +mL  a v) If the friction between the blocks is absent,
The frictional force acting between the blocks in the lower block remains at rest relative ground. The
F
this case is f = mU a acceleration of the upper block is aU = m
U
iii) If F > Fmax the blocks slip relative to each III) If 1 is the coefficient of friction between the
other and the they move with different accelerations. two blocks and 2 is the coefficient friction between
Here, the acceleration of the upper block is aU = µK g the lower block and the horizontal surface, the maxi-
mum horizontal force for which the two blocks move
The acceleration of the lower block is
together is given by F = mU +mL  µ1 µ2  g
F  µK g Problem 6.17 :
aL =
mL A 2kg block is placed over a 5kg block and
iv) If there is no friction between the two blocks, both are placed on a smooth horizontal surface.
upper block remains at rest relative to the ground and The coefficient of friction between the blocks is
the lower block moves with an acceleration F/mL. 0.10. Find the acceleration of the two blocks if a
II) If the upper block is pulled as shown in the horizontal force of 14N is applied to the upper
figure, lower block is accelerated by the frictional force block (g = 10ms–2).
between the blocks. Sol. Consider the motion of 2kg block. The forces on
F
mU it are (i) gravitational force, 2g = 2 x 10 = 20N. Vertically
downwards, (ii) normal reaction N by the 5kg block,
mL vertically upwards, (iii) force of friction f =  N to the
left and
i) If the two blocks move together, the maximum (iv) applied force 14N.
acceleration of the system is given by   0.10 2kg 14N
µS mU g 5kg
amax =
mL
In the vertical direction, there is no acceleration.
ii) The maximum force required to make the  N = 20N
blocks move together is given by In the horizontal direction, the acceleration of the
µS mU g 2kg block is a.
Fmax = mL +mU 
mL  14 –  N = 2a (  Resultant force = ma)
iii) If a < amax the force on the upper block is 14 – 0.10 x 20 = 2a (  = 0.10)
–2
14 – 2 = 2a; a = 6 ms .
F =  mU +mL  a and the frictional force between the
The system is started towards right with an initial
velocity v. The coefficient of friction between the Problem 6.20:
road and the bigger block is  and that between The coefficient of friction between the board

the blocks is . Find the time elapsed before the and the floor shown in figure is  . Find the maxi-
2
smaller block separates from the bigger block. mum force that the man can exert on the rope so
m that the board does not slip on the floor.
M v
N1

ma 0
mg / 2
mg
a0 N1 N2 mg / 2
N 2 Sol : Let T is the force exerted by the man on
Mg the rope.
FBD
Sol :
Suppose the acceleration of the lower block is
Along vertical direction,  Fv  0;
a0 towards left (–a0 toward right). The block m expe- or N + T = (M + m)g or N = ( M + m )g - T
riences a pseudo force of magnitude ma0 relative to the The board will not slip over the floor, if T  f .
block M. Let acceleration of the block m is ‘a’ relative For maximum value of T, we have
to M. The situation is shown in figure. T  f  N
For the block m; N1 = mg...............(i)
mg
  (M  m)g  T 
and ma 0   ma...............(ii)
2  (M  m)g T
For block M; N2 = Mg + N1
= Mg + mg = (M + m)g  (M  m)g 
or T   
mg  1   
and N 2   Ma 0 ............(iii)
2 Problem 6.21 :
mg
or  (M  m)g   Ma 0 A block of mass 4 kg is placed on other block
2
 (2M  m)g of mass 5kg, and the block B rests on a smooth
 a0 
2M horizontal table. For sliding the block A on B, a
Substituting the value of a0 in equation (ii), we get horizontal force 12N is required to be applied on
 (M  m )g it. How mucj maximum horizontal force can
a
2M be applied on B so that both A and B move to-
Now using second equation of motion for the gether? Also find out the acceleration produced
1 2 by this force ?
motion of m w.r.t. M, we have   0  at A M a 1
2 M1 M1
2 B 12 N
 t F M2 F M2
a

2 4M
  Sol. Here M1 = 4 kg and M2 = 5 kg.
 (M  m)g  (M  m)g The limiting friction between the blocks,
 2M 

Application–6.11 :
wall on the body is equal to “F”. Weight “mg” of the
A particle moves on a rough horizontal ground body acts vertically down trying to pull the body down
with some initial velocity say  0. If a frac- where as the frictional force “f” acts vertically up trying
tion ‘x’ of its kinetic energy is lost in friction to prevent the body from sliding.
in time t0, then coefficient of friction between For the body to be under equilibrium
the particle and the ground is fs
fL = mg
Sol : If a fraction ‘x’ of kinetic energy is lost then
s N = mg
th
(1–x) of kinetic energy is left N F
s F = mg (  N = F)
Hence the velocity become V  1  x  V0 F=
mg
mg s
E|  1  x  E  V |  1  x  V NOTE 6.6: If a book is held between two hands
and pressed using a force “F” by each hand, then weight
1  xV0  V0   g t 0
is balanced by the total frictional force
V0  fs
 g t 0  V0 1  1  x   
g t0 
1 1  x  fs
 W = 2fL
Application–6.12 : F F
A block of mass m lying on a rough horizontal mg=2s F
plane is acted upon by a horizontal force P and another
mg
force Q inclined at an angle  to the vertical. If the block F=
remains equilibrium, find the coefficient of friction be- 2s mg

tween the block and the horizontal plane. Application–6.13 : A block of mass ‘m’ is pressed
So1: The force Q can be resolved into two com- against a vertical rough wall of coefficient of friction
ponents as shown in the fig.N ‘  ’ with a horizontal force ‘F’.
Q f

P
N F
fL
Fmin
mg
mg
From fig. N = mg - Q cos 
fL = s N = s (mg - Q cos  )
a) The minimum force to be applied parallel to the
wall so as to moved the body upwards is given by
P + Q sin  = s (mg - Q cos  )
Fmin   mg  f    mg  N    mg  F 
P + Q sin 
Coefficient of friction s = .
mg - Q cos  b) Minimum work done to move up the body
6.12 Bodies In Contact with Vertical Surface : through a distance ‘S’ is given by
Consider a body of mass “m” in contact with a W  Fmin S   mg  f  S
vertical wall. Let “ µ ” be the coefficient of friction be-
tween the surfaces in contact. A horizontal force “F” is Application–6.14 :
applied as shown in the figure to hold the body under A block of mass m kg is pushed up against a
equilibrium. The normal reaction “N” exerted by the wall by a force P that makes an angle ‘  ’ with
 6.13.1. Body Sliding Down on a Smooth Inclined
When 'm' is in equilibrium mg =  N
Plane :
MF mg  M  m  Let us consider a smooth inclined plane AB at
mg =   M  m  F 
M an angle  to the horizontal. Let a body of mass m be
Problem 6.23 : kept on the plane as shown in figure. The weight of the
A man of mass 40 kg is at rest between the walls body ‘mg’ acts vertically downwards.
as shown in the figure. If ‘  ’ between the man and the If mg is resolved into two components, com-
walls is 0.8, find the normal reactions exerted by the ponent mg cos  perpendicular to the plane is balanced
walls on the man. by the normal reaction ‘N’. The second unbalanced
Sol : component mg sin  down the plane causes the body
 N1  N2
Since man is at rest, to slide down. Therefore, if ‘a’ is the downward ac-
celeration of the body then
N1–N2 = 0
FR = mg sin   f FR = mg sin   f  0
 N1  N 2  N (say)
N2 FR = mg sin 
 2 N  mg u0
N1 0 A
= 20.8 N  400 mgSin
 mgCos

mg
 N  50 N h
Problem 6.24 : A 2 kg block is in contact with a ver-
v 
tical wall having coefficient of friction 0.5 between the B
surfaces. A horizontal force of 40N is applied on the
block at right angles to the wall. A force of 15N is ap- mg sin 
FR
plied, in the plane of the wall and at right angles to 40N, a  a  g sin 
m m
force. Find the acceleration of the block.
Sol : If ' l ' is the length of the plane, and if the body
starts from rest a A andattains the velocity ‘v’ when it
F=15N f=20N reaches the foot (B) of the plane,

15N
using the kinematical equation v 2  u 2  2as ,
F=40N N=40N
as u=0, s  l and a  g sin , we get
20N 25N=FR
v 2  2( g sin  )l or v  2 gl sin   2 gh .
W=20N
If ‘t’ time taken to travel the distance ' l " with
Resultant of W=20N and 15N = 20 15
2 2
initial velocity u  0 , at the top of the plane, from
= 25N v  u  at , we get
frictional force f   N = 0.5 x 40 = 20N
v 2l sin  2l
This acts in a direction, opposite to 25N force. t (or )t  (or ) t 
 Net force acting on the block, Fnet = 25–20 g sin  g sin  g sin 
5
 acceleration of the block a =2.5ms–2 But Sin 
h
or  
h
2  sin 
6.13 Motion of a body on an inclined plane :
creased, at a particular value of  say  , the body will F mg (sin   k cos )
be just ready to slide down the plane. This angle of a R 
m m
inclination  is called the angle of repose.
a=g sin    k c os  
N fL
 If ‘v’ is the velocity of the body at point ‘B’
and the body starts from rest at A then by the

m gcos 
m g sin  equation v 2  u 2  2as ,

 mg we get v 2  02  2 g (sin   k cos  ) s

1. If    ; the body is at rest. Where ‘s’ is the distance travelled along the in-
clined plane
Threfore frictional force (f) = mg sin 
2. If angle of inclination θ  α ; the body is ready Then v  2 gs (sin   k cos  )
to slide. The body is in limiting equilibrium, the  The time taken by the body to slide down is given
net force on it should be equal to zero. Threfore by s  ut  1/ 2at 2 Here, u  0
mg sin  becomes equal to the limiting friction
1
 f L  . f L = mg Sin , N = mg Cos  s  0(t )  g (sin    k cos  )t 2
2
fL
Dividing them, we get =Tan 2l
N  t
g (sin   k cos  )
 fL 
 S =Tan  = S  NOTE 6.7: Let ‘F’ is the force to be applied parallel
 N  to the inclined plane to prevent the body from sliding
3. If angle of inclination θ  α ; Body slides down down. FR  0 F  mg sin   s cos   0
on a rough inclined plane :
If the angle of inclination is increased beyond  F  mg sin   s cos 
the value of angle of repose (   ) , the body
N fs
starts sliding down the plane, under kinetic friction. F
N 0
u0 fk
A
0
m g S in  
m g C o s
mgSin  h
mgCos mg
mg h
B

v 
Force ‘F’ is required only if angle of incination
f k  k N = k mg cos   is greater than angle of repose  .
The resultant force acting on the body down the Problem 6.25
plane is FR  (mg sin   fk ) A body of mass ‘m’ slides down a smooth
inclined plane having an inclination of 45 o with the
 FR  mg sin   k N  mg sin   k mg cos 
horizontal. It takes 2S to reach the bottom. It the
 FR  mg (sin   k cos  ) body is placed on a similar plane having coeffi-
 The acceleration of the body cient friction 0.5 What is the time taken for it to
reach the bottom ?
 1 3 98 at the rate 10 ms -1 . The resistance due to fric-
410  f c  0.31010 = 4 x 2.7
2 2 tion acting on the train is 10 N per ton.
fc = 10.8 + 10.4 – 20  f c  1.2N 1
; mass, m = 400 × 103 kg
So1. Given sin  =
98
Note 6.8 : If 4 =0.2 and 2 =0.4, then a4>a2. Then, frictional force f = 10 × 400 = 4000 N ;
the blocks move with different accelerations and sepa- velocity v = 10 ms-1
rately. a4 = 3.27ms–2 and a2 = 2.4 ms–2  Power P = (mg sin + f) v
6.14.3 Body moving up on a Rough Inclined plane
 1 
Let us consider a rough inclined plane of coef-  P =  400 × 10 3 × 9.8 ×  + 4000  × 10
  8  
ficient of kinetic friction ‘ k ’. Let  be the angle made = 440000 W = 440 kW.
by inclined plane with horizontal. A body of mass ‘m’ is Problem 6.31 :
placed at the bottom of the inclined plane. A block is placed on a rough inclined plane
A force ‘F’ is applied on the body parallel to the in- of inclination   30 0 . If the force to drag it
clined plane to move the body up the plane with uni-
along the plane is to be smaller than to lift it.
form velocity. The forces involved in the system are
The coefficient of friction  should be less than
shown in the figure. For the body to move up the in-
clined plane with uniform velocity, the force or sum of Sol: mg  sin    cos    mg
components of forces acting up the incined plane must
be equal to the sum of forces or sum of components of sin300 +  cos300 < 1
forces acting down the plane. 1 3 3 1
FR = 0  Fmin  mg sin   fk   0  1    1/ 2   
2 2 2 3
Application–6.19 :
N
F
Body projected up a Rough Inclined Plane
h
0
If a body is projected with an initial velocity ‘u’
to slide up the plane, the kinetic frictional force acts
mgSin  
fk mgCos down the plane and the body suffers retardation due to
mg
a resultant force FR  (mg sin   fk ) down the plane
 Fmin  mg sin   fk
as shown in the figure.
 mg sin   k N  The retardation on the body is given by

 mg sin   k mg cos   N  mg cos  a

(mg sin   k N ) mg sin    k mg cos  )

m m
 F  mg sin    k cos  
( N  mg cos  )
In order to move the body up the inclined plane
 a  g (sin    k cos  )
with uniform acceleration ‘a’ the force to be applied
must be greater than the sum of opposing forces by an If the body comes to rest travelling a distance
amount equal to ‘ma’. “s” up the plane,
FR  F  mg sin   k cos   ma
 F  mg sin   k cos   ma
Problem 6.30:
Find the power of an engine which can draw a
train of 400 metric ton up the inclined plane of 1 in
 to keep the block just stationary, i.e. a= 0.
Here, ma = FH cos  – mg sin  - f
minimum force requried is given by
= FH cos  – mg sin  – k (mg cos  +FH sin  )
mg = FH (cos  –  k sin  ) – mg (sin  +  k cos  ) and
(FN)min = sin   k cos  
k the maximum value of force at which the body remains
1 
 FN min  mg  sin   cos  . stationary FH   mg  sin    k cos  
 k  max  
 cos    k sin  
Application–6.22 : a 
N cos
With horizontal force FH
N FH sin FH
fk F H sin 

a FH mg sin  mg mg cos 
fk

FH FH sin
mg sin 
  mg mg cos Thus a body will remain stationay when the value
of external horizontal force is in between (FH)min and
When a horizontal force FH acts on the body as (FH)max   FH   FH   FH max
shown in the fig. one of its component FH cos  acts min
 sin   k cos    
along the plane upwards and the other component FH  mg    FH  mg  sin   k cos  
 cos   k sin    cos   k sin  
sin  acts perpendicular to plane downwards then    
normal reaction acting on the body is given by Problem 6.32 :
N = mgcos  + FH sin  . A 30kg block is to be moved up an inclined
plane at an anglue 300 to the horizontal with a
fk = k N = k (mgcos  + FH sin  )
velocity of 5ms–1. If the frictional force retarding
If we consider the block to be still moving the motion is 150N find the horizontal force required
downwards with an acceleration a then FR = ma to move the block up the plane. (g=10ms–2.)
mg sin  –[fk +FHcos  ] = ma Sol. The force required to a body up an inclined plane is
 mg sin  –  k(mg cos  +FHsin  )–FHcos  =ma F = mg sin  + frictional force
 mg(sin  -  k cos  )–FH(  k sin  + cos  ) = ma = 30(10) sin 300+150 = 300N. If P is the horizontal
F force, F = P cos 
(or) a = g (sin  –  kcos  )– H (  k sin  + cos  )
m F 300 300 2
To keep the block stationary by external force, P   = 200 3  346N .
a = 0 and cos  cos  3
 sin  -k cos   Problem 6.33 :
FH min  mg   In the figure shown acceleration of the block is
 cos   k sin  
0
40 2 N )45
Now if the external force is increased from this
value, frictional force starts decreasing. Frictional force 10 Kg
 k  0.4
becomes zero, when, mg sin   FH cos  0
 s  0.5
) 60
or, FH  m g tan  SOLUTION:
If the external force is further increased, the body N = mg cos  – 40 2 sin 450
acquires a tendency to move upward and friction starts
acting downward.
 V2
cot   1  3    cot 1  3 V  2g  sin   cos  AC  CD 
 2g
Problem 6.35: 2 g  sin    cos   AC
A rough inclined plane is inclined at 300 to CD = 2 g
the horizontal as shown in the figure. A uni-
form chain of length L is partly on the inclined  AB  BC 
 CD =   AC
plane and partly hanging from the top of the  AC AC 
incline. If the coefficient of friction between AB
chain and inclined plane is  , the maximum  CD  AB   BC   
BD
length of the hanging part to prevent the Application–6.25 :
chain from falling vertically is A block of mass m slides down a rough in-
clined plane of inclination  with horizontal
Sol: with zero initial velocity. The coefficient of
friction between the block and the plane is
300
 with   tan1    . Rate of work done by
the force of friction at time t is
SOLUTION:
 xg =  (L – x) g sin30 a = g (sin  –  cos  )
    L  x  g cos30 V = at = gt (sin  –  cos  )
1 3 f =  mg cos 
x  L  x   L  x
2 2 Rate of wrok done by friction = fV

x 

1  3 L     mg cos   gt  sin    cos  


3  3    mg2 t cos  sin    cos 
Problem 6.36 : Application–6.26 :
In the above problem, the maximum length of A body takes “n” times as much time to slide
the hanging part to prevent the chain from slid- down a rough inclined plane as it takes to slide down
ing down the incline is an identical but smooth inclined plane. If the angle of
Sol: m1g sin30 = m2g +  m1gcos30 inclination of the inclined plane is “ ”. What is the coef-
1 ficient of friction between the body and the rough plane?
 L  x g  xg    L  x g 3 / 2 Solution :
2
Let l be the length of both the inclined planes
1
 L  x  x    L  x
3
x 
1 3 L   and  be the inclination. Then the times taken by a
2 2
Application–6.24 :
3  3   body sliding the smooth and rough inclined planes are
2l
tsmooth 
A body slides down a rough inclined plane g sin 
and then over a rough plane surface. It starts
at A and stops at D. The coefficient of fric- 2l
and trough  g (sin    k cos  )
tion is :
A
Sol: If trough  ntsmooth


B C D
 Sol: F  n 
  tan  
 n  1 

Application–6.32 :
A body is pushed down with velocity ‘u’ from
N1 = mgcos  + Fsin  the top of an inclined plane of length ‘L’ and angle of

N1 F sin  1
inclination ‘  ’. The top of plane of length L  n  1 is
N2 = mgcos  N  1  mg cos  n
2
rough and the remaining part is smooth. If the body
N1 m g tan  sin  reaches the bottom of the plane with a velocity equal to
 1 the initial velocity ‘u’, then find the value of coefficient
N2 m g cos  of friction of rough plane.
N
N1 Solution :
 1 Tan2  sec2  u
0
N2
Application–6.31 : mgSin  mgCos
L mg
A body is released from rest from the top of an
n h
inclined plane of length ‘L’ and angle of inclination ‘  ’.
 n  1
1  n  L
L  n  1 is smooth and u 
The top of plane of length
n
the remaining part is rough. If the body comes to rest 2 2 L
For rough part : v  u  2a1 ......... (1)
on reaching the bottom of the plane then find the value n
of coefficient of friction of rough plane.
2 2  1
Solution : N For smooth part : u  v  2a 2 L  1   .....(2)
u0 n
0 adding equations (1) and (2) we get
mgSin  mgCos L  n  1
L mg a1  a 2 L 
n n  n 
h
 n  1 g  sin    K cos    g sin   n  1
 L
v  n 
solving we get   K  n  tan 
For smooth part : 2 L
V  2a1 QUESTIONS
n
2  n  1 Long Answer Question :
For rough part : 0  V  2a 2  L
n  1. Define coefficients of static and dynamic friction.
Mention the laws of static and dynamic friction.
L  n  1
2a1  2a 2  L Derive an expression for the acceleration of the
n  n 
body sliding on a horizontal plane having
g sin    g  sin    c o s    n  1  coefficient of friction  k
g sin  1  n  1  g cos   n  1 2. Describe an experiment to explain static friction.
A.U : A. The iron rod has a larger cross sectional area.
1. Does the frictional force between two surfaces Hence the same force applied to it produces a
depend on (i) the normal force between them (ii) small pressure than the pressure at the tip of the
the area of contact. pin.
A. Frictional force depends on normal force and is force
independent of area of contact. Pressure 
area
2. To pull a body from one place to another place
11. Is friction a reactio;n of the applied force?
at what angle the body is to be pulled to apply
least pulling force? A. No
A. The angle made by the pulling force with the 12. If a table is to be taken from one corner of a
horizontal must be equal to the angle of friction. room to another corner which is better pulling or
pushing?
3. When a paper boat is kept on a flowing water
surface will there be friction between the paper A. Pulling
boat and the water surface? 13. What are the units and dimensions of coefficient
A. Yes. If there were no friction, it would not have of friction ?
been carried away by the water flow. F

4. If the normal force is doubled, what is the change A. It is a ratio of two forces  µ   and hence it
 R
in the coefficient of friction.
is no units or dimensions.
A. No change
14. Does friction also occur as action - reaction pair?
5. Is it possible to slide a body down a rough plane
A. Yes
without retardation?
15. What is the influence of normal reaction on the
A. Yes. By applying a suitable force up the plane.
area of contact?
6. Is the coefficient of friction ever greater than
A. Area of contact increases with normal reaction.
unity ? Why ?
16. Can the coefficient of friction exceed unity ?
A. No, frictional force is less than one, except in the
case of hyperfine surfaces. A. For normal plane surfaces, the coefficient of
friction is less than unity. But when the surfaces
7. Is coefficient of friction ever equal to zero ?
are so irregular that sharp minute projections and
A. No, as we can never get a pair of surfaces which cavities exist in the surfaces, the coefficient of
are perfectly smooth. friction may be greater than one.
8. If the angle of friction is  what will be the 17. In general, if static friction is the cause of rolling
minimum pulling force required? friction, how is it less than static friction?
A. mg sin  where m is the mass of the body pulled A. Rolling starts before the static friction reaches its
on the rough horizontal surface. maximum value
9. When a running athlete wants to stop immediately 18. How can you move a solid a cylinder such that it
which force does he seek? slides up a rough inclined plane without rolling?
A. Kinetic friction A. By applying a pulling force
10. Why is it easier to push a pin into wood than it is
to push an iron rod ?