CONTENTS
5.1
Introduction
5.2
Types of friction
5.3
Graph of friction
5.4
Friction is a cause of motion
5.5
Advantages and disadvantages of friction
5.6
Methods of changing friction
5.7
Angle of friction
5.8
Resultant force exerted by surface on block
5.9
Angle of repose
5.10
Calculation of necessary force in different conditions
5.11
Acceleration of a block against friction
5.12
Work done against friction
5.13
Motion of two bodies one resting on the other
5.14
Motion of an insect in the rough bowl
5.15
Minimum mass hung from the string to just start the motion
5.16
Maximum length of a hung chain
5.17
Coefficient of friction between body and wedge
5.18
Stopping of block due to friction
5.19
Stopping of two blocks due to friction
5.20
Velocity at the bottom of rough wedge
5.21
Sticking of a block with accelerated cart
5.22
Sticking of a person from the wall of rotor
Sample Problems
Practice Problems (Basic and Advance Level)
Answer Sheet of Practice Problems
Chapter
5
In Kargil war Peak - 5179 was conquered
by our army-men. They climbed up vertical
peak by the help of friction.
These pictures are taken from latest movie
'Lakshya' based on this story, directed by
Farhan Akhtar.
5.1 Introduction.
If we slide or try to slide a body over a surface the motion is resisted by a bonding between
the body and the surface. This resistance is represented by a single force and is called friction.
The force of friction is parallel to the surface and opposite to the direction of intended
motion.
5.2 Types of Friction.
(1) Static friction : The opposing force that comes into play when one body tends to move
over the surface of another, but the actual motion has yet not started is
called static friction.
(i) If applied force is P and the body remains at rest then static
friction F = P.
(ii) If a body is at rest and no pulling force is acting on it, force of
friction on it is zero.
(iii) Static friction is a self-adjusting force because it changes itself in accordance with the
applied force.
(2) Limiting friction : If the applied force is increased the force of static friction also
increases. If the applied force exceeds a certain (maximum) value, the body starts moving. This
maximum value of static friction upto which body does not move is called limiting friction.
(i) The magnitude of limiting friction between any two bodies in contact is directly
proportional to the normal reaction between them.
RFl
or
RF sl
(ii) Direction of the force of limiting friction is always opposite to the direction in which
one body is at the verge of moving over the other
(iii) Coefficient of static friction : (a)
s
is called coefficient of static friction and defined
as the ratio of force of limiting friction and normal reaction
R
F
s
(b) Dimension :
][ 000 TLM
(c) Unit : It has no unit.
P
R
F
mg
Friction 3
(d) Value of
s
lies in between 0 and 1
(e) Value of
depends on material and nature of surfaces in contact that means whether
dry or wet ; rough or smooth polished or non-polished.
(f) Value of
does not depend upon apparent area of contact.
(3) Kinetic or dynamic friction : If the applied force is increased further and sets the body
in motion, the friction opposing the motion is called kinetic friction.
(i) Kinetic friction depends upon the normal reaction.
RFk
or
RF kk
where
k
is called the coefficient of kinetic friction
(ii) Value of
k
depends upon the nature of surface in contact.
(iii) Kinetic friction is always lesser than limiting friction
lk FF
sk
i.e. coefficient of kinetic friction is always less than coefficient of static friction. Thus we
require more force to start a motion than to maintain it against friction. This is because once
the motion starts actually ; inertia of rest has been overcome. Also when motion has actually
started, irregularities of one surface have little time to get locked again into the irregularities
of the other surface.
(iv) Types of kinetic friction
(a) Sliding friction : The opposing force that comes into play when one body is actually
sliding over the surface of the other body is called sliding friction. e.g. A flat block is moving
over a horizontal table.
(b) Rolling friction : When objects such as a wheel (disc or ring), sphere or a cylinder rolls
over a surface, the force of friction comes into play is called rolling friction.
Rolling friction is directly proportional to the normal reaction (R) and inversely
proportional to the radius (r) of the rolling cylinder or wheel.
r
R
Frrolling
r
is called coefficient of rolling friction. It would have the dimensions of length and would be
measured in metre.
Rolling friction is often quite small as compared to the sliding friction. That is why heavy
loads are transported by placing them on carts with wheels.
In rolling the surfaces at contact do not rub each other.
The velocity of point of contact with respect to the surface remains zero all the times
although the centre of the wheel moves forward.
5.3 Graph Between Applied Force and Force of Friction.
(1) Part OA of the curve represents static friction
)( s
F
. Its value increases linearly with the
applied force
(2) At point A the static friction is maximum. This
represent limiting friction
)( l
F
.
(3) Beyond A, the force of friction is seen to decrease
slightly. The portion BC of the curve therefore represents the
kinetic friction
)( k
F
.
(4) As the portion BC of the curve is parallel to x-axis
therefore kinetic friction does not change with the applied force, it remains constant, whatever
be the applied force.
5.4 Friction is a Cause of Motion.
It is a general misconception that friction always opposes the motion. No doubt friction
opposes the motion of a moving body but in many cases it is also the cause of motion. For
example :
(1) In moving, a person or vehicle pushes the ground backwards (action) and the rough
surface of ground reacts and exerts a forward force due to friction which causes the motion. If
there had been no friction there will be slipping and no motion.
(2) In cycling, the rear wheel moves by the force communicated to it by pedalling while
front wheel moves by itself. So, when pedalling a bicycle, the force exerted by rear wheel on
ground makes force of friction act on it in the forward direction (like walking). Front wheel
moving by itself experience force of friction in backward direction (like rolling of a ball).
[However, if pedalling is stopped both wheels move by themselves and so experience force of
friction in backward direction.]
(3) If a body is placed in a vehicle which is accelerating, the force of friction is the cause of
motion of the body along with the vehicle (i.e., the body will remain at rest in the accelerating
While
pedalling
Pedalling is
stoped
ma
smg
a
Action
Friction
A
C
B
Fl
Fk
Force of
friction
Applied force
O
Fs
Friction 5
vehicle until
).mgma s
If there had been no friction between body and vehicle the body will
not move along with the vehicle.
From these examples it is clear that without friction motion cannot be started, stopped or
transferred from one body to the other.
Sample problems based on fundamentals of friction
Problem 1. If a ladder weighing 250N is placed against a smooth vertical wall having coefficient of
friction between it and floor is 0.3, then what is the maximum force of friction available at
the point of contact between the ladder and the floor [AIIMS 2002]
(a) 75 N (b) 50 N (c) 35 N (d) 25 N
Solution : (a) Maximum force of friction
NRF sl 752503.0
Problem 2. On the horizontal surface of a truck (
= 0.6), a block of mass 1 kg is placed. If the truck is
accelerating at the rate of 5m/sec2 then frictional force on the block will be [CBSE PMT 2001]
(a) 5 N (b) 6 N (c) 5.88 N (d) 8 N
Solution : (a) Limiting friction
NmgRss 88.58.916.0
When truck accelerates in forward direction at the rate of
2
/5 sm
a pseudo force
)(ma
of 5N
works on block in back ward direction. Here the magnitude of pseudo force is less than
limiting friction So, static friction works in between the block and the surface of the truck
and as we know, static friction = Applied force = 5N.
Problem 3. A block of mass 2 kg is kept on the floor. The coefficient of static friction is 0.4. If a force F
of 2.5 N is applied on the block as shown in the figure, the frictional force between the
block and the floor will be [MP PET 2000]
(a) 2.5 N
(b) 5 N
(c) 7.84 N
(d) 10 N
Solution : (a) Applied force = 2.5 N and limiting friction =
mg = 0.4 × 2 × 9.8 = 7.84 N
As applied force is less than limiting friction. So, for the given condition static friction will
work.
Static friction on a body = Applied force = 2.5 N.
F
Problem 4. A block A with mass 100 kg is resting on another block B of mass 200 kg. As shown in
figure a horizontal rope tied to a wall holds it. The coefficient of friction between A and B
is 0.2 while coefficient of friction between B and the ground is 0.3. The minimum required
force F to start moving B will be [RPET 1999]
(a) 900 N
(b) 100 N
(c) 1100 N
(d) 1200 N
Solution : (c) Two frictional force will work on block B.
BGAB ffF
gmmgm BABGaAB )(
= 0.2 × 100 × 10 + 0.3 (300) × 10
= 200 + 900 = 1100N. (This is the required minimum force)
Problem 5. A 20 kg block is initially at rest on a rough horizontal surface. A horizontal force of 75 N is
required to set the block in motion. After it is in motion, a horizontal force of 60 N is
required to keep the block moving with constant speed. The coefficient of static friction is [AMU 1999]
(a) 0.38 (b) 0.44 (c) 0.52 (d) 0.60
Solution : (a) Coefficient of static friction
38.0
8.920
75
R
Fl
S
.
Problem 6. A block of mass M is placed on a rough floor of a lift. The coefficient of friction between the
block and the floor is
. When the lift falls freely, the block is pulled horizontally on the
floor. What will be the force of friction
(a)
Mg (b)
Mg/2 (c) 2
Mg (d) None of these
Solution : (d) When the lift moves down ward with acceleration 'a' then effective acceleration due to
gravity
g' = g – a
0' ggg
[As the lift falls freely, so a = g]
So force of friction
0' mg
5.5 Advantages and Disadvantages of Friction.
(1) Advantages of friction
(i) Walking is possible due to friction.
(ii) Two body sticks together due to friction.
(iii) Brake works on the basis of friction.
(iv) Writing is not possible without friction.
A
B
F
fAB
A
B
F
fBG
Groun
d
Friction 7
(v) The transfer of motion from one part of a machine to other part through belts is
possible by friction.
(2) Disadvantages of friction
(i) Friction always opposes the relative motion between any two bodies in contact.
Therefore extra energy has to be spent in over coming friction. This reduces the efficiency of
machine.
(ii) Friction causes wear and tear of the parts of machinery in contact. Thus their lifetime
reduces.
(iii) Frictional force result in the production of heat, which causes damage to the
machinery.
5.6 Methods of Changing Friction.
We can reduce friction
(1) By polishing.
(2) By lubrication.
(3) By proper selection of material.
(4) By streamlining the shape of the body.
(5) By using ball bearing.
Also we can increase friction by throwing some sand on slippery ground. In the
manufacturing of tyres, synthetic rubber is preferred because its coefficient of friction with the
road is larger.
5.7 Angle of Friction.
Angle of friction may be defined as the angle which the resultant of limiting friction and
normal reaction makes with the normal reaction.
By definition angle
is called the angle of friction
R
F
tan
tan
=
[As we know
R
F
]
or
)(tan 1
Hence coefficient of limiting friction is equal to tangent of the angle of friction.
5.8 Resultant Force Exerted by Surface on Block.
In the above figure resultant force
22 RFS
22 )()( mgmgS
P
R
F
mg
S
1
2
mgS
when there is no friction
)0(
S will be minimum i.e., S = mg
Hence the range of S can be given by,
1
2
mgSmg
5.9 Angle of Repose.
Angle of repose is defined as the angle of the inclined plane with horizontal such that a
body placed on it is just begins to slide.
By definition
is called the angle of repose.
In limiting condition
sinmgF
and
cosmgR
So
tan
R
F
tantan
R
F
[As we know
tan
R
F
]
Thus the coefficient of limiting friction is equal to the tangent of angle of repose.
As well as
i.e. angle of repose = angle of friction.
Sample problems based on angle of friction and angle of repose
Problem 7. A body of 5 kg weight kept on a rough inclined plane of angle 30o starts sliding with a
constant velocity. Then the coefficient of friction is (assume g = 10 m/s2) [JIPMER 2002]
(a)
3/1
(b)
3/2
(c)
3
(d)
32
Solution : (a) Here the given angle is called the angle of repose
So,
3
1
30tan o
Problem 8. The upper half of an inclined plane of inclination
is perfectly smooth while the lower half
is rough. A body starting from the rest at top comes back to rest at the bottom if the
coefficient of friction for the lower half is given [Pb PMT 2000]
(a)
= sin
(b)
= cot
(c)
= 2 cos
(d)
= 2 tan
Solution : (d) For upper half by the equation of motion
asuv 2
22
2/)sin(2022 lgv
singl
[As
]sin,2/,0
galsu
For lower half
)cos(sin20 2
gu
l /2 [As
)cos(sin,2/,0
galsv
]
)cos(sinsin0
glgl
[As final velocity of upper half will be equal to the initial velocity
of lower half]
cossin2
tan2
R
mg cos
F
mg sin
mg
Smooth
Rough
Friction 9
5.10 Calculation of Necessary Force in Different Conditions.
If W = weight of the body,
= angle of friction,
tan
coefficient of friction
then we can calculate necessary force for different condition in the following manner :
(1) Minimum pulling force P at an angle
from the horizontal
By resolving P in horizontal and vertical direction (as shown in figure)
For the condition of equilibrium
cosPF
and
sinPWR
By substituting these value in
RF
)sin(cos
PWP
)sin(
cos
sin
cos
PWP
[As
tan
]
)(cos
sin
W
P
(2) Minimum pushing force P at an angle
from the horizontal
By Resolving P in horizontal and vertical direction (as shown in the figure)
For the condition of equilibrium
cosPF
and
sinPWR
By substituting these value in
RF
)sin(cos
PWP
)sin(
cos
sin
cos
PWP
[As
tan
]
)(cos
sin
W
P
(3) Minimum pulling force P to move the body up an inclined plane
By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the
figure)
For the condition of equilibrium
cossin WPR
sincos PWR
P
R
P sin
P cos
F
W
P
R
P sin
P cos
F
W
P
R + P sin
W cos
F + W sin
P cos
and
cossin PWF
sincos WPF
By substituting these values in
RF
and solving we get
)(cos
)(sin
W
P
(4) Minimum force on body in downward direction along the surface of inclined plane to
start its motion
By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the
figure)
For the condition of equilibrium
cossin WPR
sincos PWR
and
sincos WPF
By substituting these values in
RF
and solving we get
)(cos
)sin(
W
P
(5) Minimum force to avoid sliding a body down an inclined plane
By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the
figure)
For the condition of equilibrium
cossin WPR
sincos PWR
and
sincos WFP
cossin PWF
By substituting these values in
RF
and solving we get
)(cos
)(sin
WP
(6) Minimum force for motion and its direction
Let the force P be applied at an angle
with the horizontal.
By resolving P in horizontal and vertical direction (as shown in figure)
F + P cos
R + P sin
W sin
W cos
W
P
P
P
F
R + P sin
P cos
+
W sin
W cos
W
Friction 11
For vertical equilibrium
mgPR
sin
sinPmgR
….(i)
and for horizontal motion
FP
cos
i.e.
RP
cos
….(ii)
Substituting value of R from (i) in (ii)
)sin(cos
PmgP
sincos
mg
P
….(iii)
For the force P to be minimum
)sin(cos
must be maximum i.e.
0]sin[cos
d
d
0cossin
tan
or
frictionof angle)(tan 1
i.e. For minimum value of P its angle from the horizontal should be equal to angle of
friction
As
tan
so from the figure
2
1
sin
and
2
1
1
cos
By substituting these value in equation (iii)
2
2
211
1
mg
P
2
1
mg
2
min 1
mg
P
Sample problems based on force against friction
Problem 9. What is the maximum value of the force F such that the block shown in the arrangement,
does not move (
32/1
) [IIT-JEE (Screening) 2003]
(a) 20 N
(b) 10 N
(c) 12 N
(d) 15 N
Solution : (a) Frictional force
Rf
1
2
1
R + P sin
P cos
F
mg
60o
F
m = 3kg
R
F cos
60°
f
W+F sin 60°
)60sin(60cos FWF
60sin3
32
1
60cos FgF
NF 20
.
Problem 10. A block of mass m rests on a rough horizontal surface as shown in the figure. Coefficient of
friction between the block and the surface is
. A force F = mg acting at angle
with the
vertical side of the block pulls it. In which of the following cases the block can be pulled
along the surface
(a)
tan
(b)
cot
(c)
2/tan
(d)
2/cot
Solution : (d) For pulling of block
fP
Rmg
sin
)cos(sin
mgmgmg
)cos1(sin
2
sin2
2
cos
2
sin2 2
2
cot
5.11 Acceleration of a Block Against Friction.
(1) Acceleration of a block on horizontal surface
When body is moving under application of force P, then kinetic friction opposes its motion.
Let a is the net acceleration of the body
From the figure
k
FPma
m
FP
ak
(2) Acceleration of a block down a rough inclined plane
When angle of inclined plane is more than angle of repose, the body placed on the inclined
plane slides down with an acceleration a.
From the figure
Fmgma
sin
Rmgma
sin
cossin mgmgma
Acceleration
]cos[sin
ga
Note : For frictionless inclined plane
0
singa
.
R
P
Fk
mg
ma
F
mg sin
mg
cos
mg
R
ma
m
mg = F
R+mg cos
mg sin
=p
f
mg
Friction 13
(3) Retardation of a block up a rough inclined plane
When angle of inclined plane is less than angle of repose, then for the upward motion
Fmgma
sin
cossin mgmgma
Retardation
]cos[sin
ga
Note : For frictionless inclined plane
0
singa
Sample problems based on acceleration against friction
Problem 11. A body of mass 10 kg is lying on a rough plane inclined at an angle of 30o to the horizontal
and the coefficient of friction is 0.5. The minimum force required to pull the body up the
plane is [JIPMER 2000]
(a) 914 N (b) 91.4 N (c) 9.14 N (d) 0.914 N
Solution : (b)
)cos(sin
mgF
N4.91)30cos5.030(sin8.910
Problem 12. A block of mass 10 kg is placed on a rough horizontal surface having coefficient of friction
= 0.5. If a horizontal force of 100 N is acting on it, then acceleration of the block will be [AIIMS 2002]
(a) 0.5 m/s2 (b) 5 m/s2 (c) 10 m/s2 (d) 15 m/s2
Solution : (b)
mass
frictionkinetic – force Applied
a
10
10105.0100
= 5m/s2.
Problem 13. A body of weight 64 N is pushed with just enough force to start it moving across a horizontal
floor and the same force continues to act afterwards. If the coefficients of static and dynamic
friction are 0.6 and 0.4 respectively, the acceleration of the body will be (Acceleration due to
gravity = g) [EAMCET 2001]
(a)
4.6
g
(b) 0.64 g (c)
32
g
(d) 0.2 g
Solution : (d) Limiting friction =
RF sl
64 = 0.6 m g m =
g6.0
64
.
Acceleration =
bodyof the Mass
friction Kinetic – force Applied
=
m
mg
K
64
=
g6.0
64 6.0
64
4.064
= 0.2g
Problem 14. If a block moving up at
o
30
with a velocity 5 m/s, stops after 0.5 sec, then what is
[CPMT 1995]
(a) 0.5 (b) 1.25 (c) 0.6 (d) None of these
Solution : (c) From
atuv
0
atu
a
u
t
for upward motion on an inclined plane
)cos(sin
ga
)cos(sin
g
u
t
mg sin
+ F
mg
cos
mg
R
ma
Substituting the value of
sec5.0,30 t
o
and
smu /5
, we get
6.0
5.12 Work Done Against Friction.
(1) Work done over a rough inclined surface
If a body of mass m is moved up on a rough inclined plane through distance s, then
Work done = force distance
= ma s
= mg [sin
+
cos
]s
]cos[sin
smg
(2) Work done over a horizontal surface
In the above expression if we put
= 0 then
Work done = force distance
= F s
=
mg s
It is clear that work done depends upon
(i) Weight of the body.
(ii) Material and nature of surface in contact.
(iii) Distance moved.
Sample problems based on work done against friction
Problem 15. A body of mass 5kg rests on a rough horizontal surface of coefficient of friction 0.2. The
body is pulled through a distance of 10m by a horizontal force of 25 N. The kinetic energy
acquired by it is (g = 10 ms2)
[EAMCET (Med.) 2000]
(a) 330 J (b) 150 J (c) 100 J (d) 50 J
Solution : (b) Kinetic energy acquired by body = Total work done on the body – Work done against
friction
=
mgSSF
= 25 × 10 – 0.2 × 5 × 10 ×10 = 250 – 100 = 150 J.
Problem 16. 300 Joule of work is done in sliding a 2 kg. block up an inclined plane to a height of 10
meters. Taking value of acceleration due to gravity ‘g’ to be 10 m/s2, work done against
friction is [MP PMT 2002]
(a) 100 J (b) 200 J (c) 300 J (d) Zero
Solution : (a) Work done against gravity = mgh = 2 × 10 × 10 = 200J
Work done against friction = Total work done – Work done against gravity = 300 – 200 =
100J.
mg sin
+ F
mg
cos
mg
R
ma
s
P
F
mg
R
s
Friction 15
Problem 17. A block of mass 1 kg slides down on a rough inclined plane of inclination 60o starting
from its top. If the coefficient of kinetic friction is 0.5 and length of the plane is 1 m,
then work done against friction is (Take g = 9.8 m/s2) [AFMC 2000; KCET (Engg./Med.) 2001]
(a) 9.82 J (b) 4.94 J (c) 2.45J (d) 1.96 J
Solution : (c)
2
1
8.915.0.cos SmgW
= 2.45 J.
Problem 18. A block of mass 50 kg slides over a horizontal distance of 1 m. If the coefficient of friction
between their surfaces is 0.2, then work done against friction is [CBSE PMT 1999, 2000; AIIMS 2000; BHU 2001]
(a) 98 J (b) 72J (c) 56 J (d) 34 J
Solution : (a)
JmgSW 9818.9502.0
.
5.13 Motion of Two Bodies One Resting on the Other.
When a body A of mass m is resting on a body B of mass M then two conditions are possible
(1) A force F is applied to the upper body, (2) A force F is applied to the lower body
We will discuss above two cases one by one in the following manner :
(1) A force F is applied to the upper body, then following four situations are possible
(i) When there is no friction
(a) The body A will move on body B with acceleration (F/m).
mFaA/
(b) The body B will remain at rest
0
B
a
(c) If L is the length of B as shown in figure A will fall from B after time t
F
mL
a
L
t22
F/mata and
2
1
sAs 2
(ii) If friction is present between A and B only and applied force is less than limiting friction
(F < Fl)
(F = Applied force on the upper body, Fl = limiting friction between A and B, Fk = Kinetic friction
between A and B)
(a) The body A will not slide on body B till
l
FF
i.e.
mgFs
(b) Combined system (m + M) will move together with common acceleration
mM
F
aa BA
(iii) If friction is present between A and B only and applied force is greater than limiting
friction (F > Fl)
In this condition the two bodies will move in the same direction (i.e. of applied force) but
with different acceleration. Here force of kinetic friction
mg
k
will oppose the motion of A
while will cause the motion of B.
F
m
A
M
B
L
Ak amFF
i.e.
m
FF
ak
A
m
mgF
ak
A)(
Free body diagram of
A
Bk aMF
i.e.
M
F
ak
B
M
mg
ak
B
Free body diagram of
B
Note : As both the bodies are moving in the same direction.
Acceleration of body A relative to B will be
mM
MmmgMF
aaa k
BA )(
So, A will fall from B after time
)(
2
2
MmmgMF
MLm
a
L
t
k
(iv) If there is friction between B and floor
(where
gmMFl)(
= limiting friction between B and floor, Fk = kinetic friction between
A and B)
B will move only if
lk FF
and then
Blk aMFF
However if B does not move then static friction will work
(not limiting friction) between body B and the floor i.e. friction
force = applied force (= Fk) not
l
F
.
(2) A force F is applied to the lower body, then following four situations are possible
(i) When there is no friction
(a) B will move with acceleration (F/M) while A will remain
at rest (relative to ground) as there is no pulling force on A.
M
F
aB
and
0
A
a
(b) As relative to B, A will move backwards with acceleration (F/M) and so will fall from it
in time t.
F
ML
a
L
t22
(ii) If friction is present between A and B only and F < Fl
(where F = Pseudo force on body A and Fl = limiting friction between body A and B)
(a) Both the body will move together with common acceleration
mM
F
a
(b) Pseudo force on the body A,
Mm
mF
maF
and
mgFsl
F
m
A
M
B
L
F
A
Fk
maA
FK
MaB
B
FK
MaB
B
Fl
Friction 17
(c)
l
FF
mg
Mm
mF s
gMmF s)(
So both bodies will move together with acceleration
Mm
F
aa BA
if
gMmF s][
(iii) If friction is present between A and B only and F > Fl
(where Fl =
s (m + M)g = limiting friction between body B and surface)
Both the body will move with different acceleration. Here force of kinetic friction
mg
k
will oppose the motion of B while will cause the motion of A.
mgma kA
i.e.
ga kA
Free body diagram of
A
Bk MaFF
i.e.
M
mgF
ak
B][
Free body diagram of
B
Note : As both the bodies are moving in the same direction
Acceleration of body A relative to B will be
M
MmgF
aaa k
BA )(
Negative sign implies that relative to B, A will move backwards and will fall it after
time
)(
22
MmgF
ML
a
L
t
k
(iv) If there is friction between B and floor : The system will move only if
'
l
FF
then
replacing F by
l
FF
. The entire case (iii) will be valid.
However if
l
FF
the system will not move and friction between B and floor will be F while
between A and B is zero.
Sample problems based on body resting on another
Problem 19. A 4 kg block A is placed on the top of a 8 kg block B which rests on a smooth table. A just slips
on B when a force of 12 N is applied on A. Then the maximum horizontal force on B to make
both A and B move together, is
(a) 12 N (b) 24 N (c) 36 N (d) 48 N
Solution : (c) Maximum friction i.e. limiting friction between A and B,
l
F
= 12 N.
If F is the maximum value of force applied on lower body such that both body move
together
It means Pseudo force on upper body is just equal to limiting friction
A
Fk
maA
FK
MaB
B
F
l
FF '
Mm
F
m
12
84
4
F
.36 NF
Problem 20. A body A of mass 1 kg rests on a smooth surface. Another body B of mass 0.2 kg is placed
over A as shown. The coefficient of static friction between A and B is 0.15. B will begin to
slide on A if A is pulled with a force greater than
(a) 1.764 N
(b) 0.1764 N
(c) 0.3 N
(d) It will not slide for any F
Solution : (a) B will begin to slide on A if Pseudo force is more than limiting friction
l
FF '
R
Mm
F
ms
mg
Mm
F
m15.0
NF 764.1
Problem 21. A block A of mass 2 kg rests on another block B of mass 8 kg which rests on a horizontal
floor. The coefficient of friction between A and B is 0.2, while that between B and floor is
0.5. When a horizontal force of 25 N is applied on the block B, the force of friction between
A and B is [IIT-JEE 1993]
(a) Zero (b) 3.9 N (c) 5.0 N (d) 49 N
Solution : (a) Limiting friction between the block B and the surface
RF BSBS .
gMm 5.0
10)82(5.0
N50
but the applied force is 25 N so the lower block will not move
i.e. there is no pseudo force on upper block A. Hence there will
be no force of friction between A and B.
5.14 Motion of an Insect in the Rough Bowl.
The insect crawl up the bowl up to a certain height h only till the component of its weight
along the bowl is balanced by limiting frictional force.
Let m = mass of the insect, r = radius of the bowl,
= coefficient of friction
for limiting condition at point A
cosmgR
......(i) and
sinmgFl
......(ii)
Dividing (ii) by (i)
R
Fl
tan
RFl
As
y
yr 22
or
2
1
r
y
So
2
1
1
1
ryrh
,
2
1
1
1
rh
Fl
mg sin
mg
mg cos
A
R
r
O
y
h
A
B
8kg
2kg
25
N
A
B
Surface
Friction 19
Problem 22. An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of
friction between the insect and the surface is 1/3. If the line joining the centre of the
hemispherical surface to the insect makes an angle
with the vertical, the maximum
possible value of
is given by [IIT-JEE (Screening) 2001]
(a)
3cot
(b)
3tan
(c)
3sec
(d)
3cosec
Solution : (a) From the above expression, for the equilibrium
sinandcos mgFmgR
.
Substituting these value in
RF
we get
tan
or
3
1
cot
.
5.15 Minimum Mass Hung From the String to Just Start the Motion.
(1) When a mass m1 placed on a rough horizontal plane : Another mass
2
m
hung from the
string connected by pulley, the tension (T) produced in string will try to start the motion of
mass
1
m
.
At limiting condition
l
FT
Rgm
2
gmgm 12
12 mm
this is the minimum value of
2
m
to start the motion.
Note : In the above condition Coefficient of friction
1
2
m
m
(2) When a mass m1 placed on a rough inclined plane : Another mass
2
m
hung from the
string connected by pulley, the tension (T) produced in string will try to start the motion of
mass
1
m
.
At limiting condition
For
gmTm 22
...... (i)
For
FgmTm
sin
11
RgmT
sin
1
cossin 11 gmgmT
......(ii)
From equation (i) and (ii)
]cos[sin
12
mm
this is the minimum value of
2
m
to start the motion
Note : In the above condition Coefficient of friction
tan
cos
1
2
m
m
Sample problems based on hung mass
m2
m1
T
R
m1g
F1
T
m2
R
T
m1g sin
+ F
m1g
cos
T
m2g
m1
m1g
Problem 23. Two blocks of mass
1
M
and
2
M
are connected with a string passing over a pulley as shown
in the figure. The block
1
M
lies on a horizontal surface. The coefficient of friction between
the block M1 and horizontal surface is
. The system accelerates. What additional mass m
should be placed on the block
1
M
so that the system does not accelerate
(a)
12 MM
(b)
1
2M
M
(c)
1
2M
M
(d)
)( 12 MM
Solution : (b) By comparing the given condition with general expression
1
2
Mm
M
2
1M
Mm
1
2M
M
m
Problem 24. The coefficient of kinetic friction is 0.03 in the diagram where mass
kgm20
2
and
kgm4
1
. The acceleration of the block shall be
)10(2
msg
(a) 1.8
2
ms
(b) 0.8
2
ms
(c) 1.4
2
ms
(d) 0.4
2
ms
Solution : (c) Let the acceleration of the system is a
From the F.B.D. of
2
m
amFT 2
amgmT 22
aT 20102003.0
aT 206
.....(i)
From the FBD of
1
m
amTgm 11
aT 4104
aT 440
....(ii)
Solving (i) and (ii)
./4.1 2
sma
5.16 Maximum Length of Hung Chain .
A uniform chain of length l is placed on the table in such a manner that its
'l
part is hanging
over the edge of table with out sliding. Since the chain have uniform linear density therefore
the ratio of mass or ratio of length for any part of the chain will be equal.
M2
M1
m
m1
m2
20 kg
T
T
4 kg
T
F
m2
m2a
T
F
m1
m1g
m1 a
Friction 21
We know
table theonlying mass
table thefromhanging mass
1
2 m
m
[From article 5.15]
For this expression we can rewrite above expression in the following manner
table theonlying length
table thefromhanging length
[As chain have uniform linear density]
ll
l
by solving
)1(
l
l
Problem 25. A heavy uniform chain lies on a horizontal table top. If the coefficient of friction between
the chain and the table surface is 0.25, then the maximum fraction of the length of the
chain that can hang over one edge of the table is [CBSE PMT 1990]
(a) 20% (b) 25% (c) 35% (d) 15%
Solution : (a) From the expression
ll
1
'
l
125.0
25.0
[As
= 0.25]
ll 25.1
25.0
'
5
l
= 20% of the length of the chain.
5.17 Coefficient of Friction Between Body and Wedge .
A body slides on a smooth wedge of angle
and its time of descent is t.
If the same wedge made rough then time taken by it to come down becomes n times more
(i.e. nt)
The length of path in both the cases are same.
For smooth wedge
2
2
1attuS
2
)sin(
2
1tgS
.....(i)
and0As[u
]sin
ga
For rough wedge
2
2
1attuS
2
)()cos(sin
2
1ntgS
.....(ii)
)]cos(sinand0As[
gau
From equation (i) and (ii)
2
)sin(
2
1tg
=
2
)()cos(sin
2
1ntg
2
)cos(sinsin n
S
Smooth wedge
S
Rough wedge
l
( l – l
)
2
1
1tan n
Problem 26. A body takes just twice the time as long to slide down a plane inclined at 30o to the
horizontal as if the plane were frictionless. The coefficient of friction between the body
and the plane is [JIPMER 1999]
(a)
4
3
(b)
3
(c)
3
4
(d)
4
3
Solution : (a)
22 2
1
130tan
1
1tan n
4
3
.
5.18 Stopping of Block Due to Friction.
(1) On horizontal road
(i) Distance travelled before coming to rest : A block of mass m is moving initially with
velocity u on a rough surface and due to friction it comes to rest after covering a distance S.
Retarding force
RmaF
mgma
ga
.
From
aSuv 2
22
Sgu
20 2
],0[As gav
g
u
S
2
2
or
gm
P
S2
2
2
[As momentum P = mu]
(ii) Time taken to come to rest
From equation
tauv
tgu
0
],0As[gav
g
u
t
(iii) Force of friction acting on the body
We know, F = ma
So,
t
uv
mF )(
t
mu
F
[As v = 0]
mgF
g
u
t
As
v = 0
S
u
Friction 23
(2) On inclined road : When block starts with velocity u its kinetic energy will be
converted into potential energy and some part of it goes against friction and after travelling
distance S it comes to rest i.e. v = 0.
And we know that retardation
]cos[sin
ga
By substituting the value of v and a in the following equation
Sauv 2
22
Sgu ]cos[sin20 2
)cos(sin2
2
g
u
S
Sample problems based on motion of body on rough surface
Problem 27. A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by
friction in 10s. Then the coefficient of friction is [AIEEE 2003]
(a) 0.01 (b) 0.02 (c) 0.03 (d) 0.06
Solution : (d)
atuv
0 tgu
06.0
1010
6
gt
u
.
Problem 28. A 2 kg mass starts from rest on an inclined smooth surface with inclination 30o and length
2 m. How much will it travel before coming to rest on a surface with coefficient of friction
0.25 [UPSEAT 2003]
(a) 4 m (b) 6 m (c) 8 m (d) 2 m
Solution : (a)
aSuv 2
22
230sin20 g
20v
Let it travel distance ‘S’ before coming to rest
.4
1025.02
20
2
2m
g
v
S
5.19 Stopping of Two Blocks Due to Friction.
When two masses compressed towards each other and suddenly released then energy
acquired by each block will be dissipated against friction and finally block comes to rest
i.e., F × S = E [Where F = Friction, S = Distance covered by block, E = Initial
kinetic energy of the block]
m
P
SF 2
2
[Where P = momentum of block]
m
P
Smg 2
2
[As F =
mg]
m1
A
m1
B
m2
m2
S1
S2
u
v = 0
S
gm
P
S2
2
2
In a given condition P and
are same for both the blocks.
So
2
1
m
S
2
1
2
2
1
m
m
S
S
5.20 Velocity at the Bottom of Rough Wedge.
A body of mass m which is placed at the top of the wedge (of height h) starts moving
downward on a rough inclined plane.
Loss of energy due to friction = FL (Work against friction)
PE at point A = mgh
KE at point B =
2
2
1mv
By the law of conservation of energy
i.e.
FLmghmv
2
2
1
)(
2FLmgh
m
v
5.21 Sticking of a Block With Accelerated Cart.
When a cart moves with some acceleration toward right then a pseudo force (ma) acts on
block toward left.
This force (ma) is action force by a block on cart.
Now block will remain static w.r.t. block. If friction force
mgR
mgma
]As[maR
g
a
g
a
min
This is the minimum acceleration of the cart so that block does not fall.
and the minimum force to hold the block together
minmin )( amMF
g
mMF )(
min
M
a
F
CART
m
F
mg
m
R
ma
u = 0
L
m
m
B
A
v
h
Friction 25
5.22 Sticking of a Person With the Wall of Rotor.
A person with a mass m stands in contact against the wall of a cylindrical drum (rotor). The
coefficient of friction between the wall and the clothing is
.
If Rotor starts rotating about its axis, then person thrown away from the centre due to
centrifugal force at a particular speed w, the person stuck to the wall even the floor is removed,
because friction force balances its weight in this condition.
From the figure.
Friction force (F) = weight of person (mg)
R = mg
mgFc
[Here, Fc= centrifugal force]
mgrm
2
min
r
g
min
Sample problems (Miscellaneous)
Problem 29. A motorcycle is travelling on a curved track of radius 500m if the coefficient of friction
between road and tyres is 0.5. The speed avoiding skidding will be [MH CET (Med.) 2001]
(a) 50 m/s (b) 75 m/s (c) 25 m/s (d) 35 m/s
Solution : (a)
./50105005.0 smrgv
Problem 30. A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The
coefficient of friction between the block and the wall is 0.2. The weight of the block is [AIEEE 2003]
(a) 2 N
(b) 20 N
(c) 50 N
(d) 100 N
Solution : (a) For equilibrium
Weight (W) = Force of friction (F)
102.0 RW
= 2N
Problem 31. A body of mass 2 kg is kept by pressing to a vertical wall by a force of 100 N. The friction
between wall and body is 0.3. Then the frictional force is equal to [Orissa JEE 2003]
(a) 6 N (b) 20 N (c) 600 N (d) 700 N
FC
R
F
mg
10 N
10 N
F
R
W
Solution : (b) For the given condition Static friction = Applied force = Weight of body = 2 × 10 = 20 N.
Problem 32. A fireman of mass 60kg slides down a pole. He is pressing the pole with a force of 600
N. The coefficient of friction between the hands and the pole is 0.5, with what
acceleration will the fireman slide down (g = 10 m/s2) [Pb. PMT 2002]
(a) 1 m/s2 (b) 2.5 m/s2 (c) 10 m/s2 (d) 5 m/s2
Solution : (d) Friction =
R = 0.5 × 600 = 300 N, Weight = 600 N
ma = W – F
60
300600
m
FW
a
2
/5 sma
Problem 33. The system shown in the figure is in equilibrium. The maximum value of W, so that the
maximum value of static frictional force on 100 kg. body is 450 N, will be
(a) 100 N
(b) 250 N
(c) 450 N
(d) 1000 N
Solution : (c) For vertical equilibrium
WT o45sin
1
o
W
T45sin
1
For horizontal equilibrium
o
TT 45cos
12
o
o
W45cos
45sin
= W
and for the critical condition T2 = F
W = T2 = F = 450 N
W
45o
100 kg
600 N
F
R
W
W
T1
45o
T1 cos
45o
T2
F
T1 sin
45o