Friction
The force of friction is a common but complex force. The exact method
by which friction works is still a topic of great scientific interest but we
can make some general statements about it. We do know that it arises
from the electromagnetic forces between atoms and molecules at the
surfaces of objects.
We can build a simple model of the friction force that is useful in many
situations. The model friction force has the following properties:
There are two types of frictional force. The force of static friction and
the force of kinetic friction:
 The direction of the static frictional force is along the contact surface
and opposite in direction of any applied force.
 The magnitude of the static friction force is given by
Fs  sRn
Where s is the coefficient of static friction and Rn is the normal reaction.
 The direction of the kinetic frictional force is opposite the direction of
motion of the object it acts on.
 The magnitude of the kinetic friction force is given by
Fk  k Rn
Where k is the coefficient of static friction and Rn is the normal reaction.
 The coefficients of friction depend on the nature of the surface.
 The frictional force is nearly independent of the contact area between
the objects.
 The kinetic friction force is usually less than the maximum static
friction force.
Mathematics made simple
M1 Friction
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Friction
 The plot below of the frictional force vs. the applied force illustrates
some of the features of the frictional force.
Note that the frictional force equals the applied force (in magnitude)
until it reaches the maximum possible value msN. Then the object begins
to move as the applied force exceeds the maximum frictional force. When
the object is moving the frictional force is kinetic and roughly constant
at the value mkN which is below the maximum static friction force.
The table below summarizes the main characteristics of the frictional
force.
Static Friction
Kinetic Friction
Symbol
fs
fk
Direction
opposite direction
of applied force
opposite direction of
object's motion
Magnitude
<msN
mkN
Note: In the M1 Module it will be assumed that s = k = 
Mathematics made simple
M1 Friction
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Friction
Limiting equilibrium
If the object is at rest and the forces are in equilibrium with the limiting
friction, the object is said to be in limiting equilibrium. At this point, the
friction force is at its maximum value, called the limiting friction.
Coefficient of friction ()
The magnitude of the maximum frictional force is a fraction of the
normal reaction (Rn). This fraction is called the coefficient of friction 
for the two surfaces in contact.
Fmax = Rn
For a perfectly smooth surface,  = 0.
Note:
The maximum force will only act if:
 there is a state of limiting equilibrium or
 motion is taking place
In other cases the frictional force is equal and opposite the applied force
P (i.e. F = P where P < Rn)
Example
A block of mass 10 kg rests on a horizontal plane, the coefficient of the
friction between the block and the plane being 0.6. Calculate the friction
force acting on the block when a horizontal force P is applied to the block
and the magnitude of P is:
(i) 20 N
(ii) 40 N
(iii) 58.8 N
Solution
Rn N
F
10 kg
P
Resolve upwards  : Rn  10g = 0
 Rn = 98
Maximum friction force Fmax = Rn
Fmax = 0.6  98= 58.8 N
(i) P < 58.8 so F =20 N There is no motion.
10g N
(ii) P < 58.8 so F = 40 N There is no motion.
(iii) P = 58.8 F = Fmax = 58.8 The block is
on the point of moving (equilibrium).
Mathematics made simple
M1 Friction
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Friction
Example
Calculate the maximum frictional force which can act when a block of
mass 5 kg rests on a rough horizontal surface, the coefficient of friction
between the surfaces being
(i) 0.6
(ii) 0.3
Solution
(i) Resolve upwards  : Rn  5g = 0
 Rn = 49
Rn N
Fmax = Rn
Maximum friction force Fmax = Rn
Fmax = 0.6  49 = 29.4 N
(ii) Resolve (as before):
Rn = 49
Fmax = Rn = 0.3  49 = 14.7 N
5 kg
5g N
Example
A block of mass 2 kg rests on a rough horizontal surface. A horizontal
force of 14N is applied to the block. If the block is on the point of
moving, find the coefficient of friction between the block and the
surface.
Solution
Resolve upwards  : Rn  2g = 0
 Rn = 19.6
Maximum friction force Fmax = Rn
Rn N
2 kg
F
14

F
14

 0.714
R 19.6
2g N
Example
A block of mass 5 kg rests on a rough horizontal surface. A horizontal
force of 24.5 N is applied to the block. If the block is on the point of
moving, find the coefficient of friction between the block and the
surface.
Solution
Rn = 5g = 5  9.8 = 40

F 24.5

 0.5
R
49
Mathematics made simple
M1 Friction
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Friction
Example
A block of mass 6 kg rests on a rough
horizontal surface. The coefficient of
friction between block and the surface is
0.5. A force P is applied at an angle of 30
to the horizontal. Find the value of P when
the block is about to move.
Solution
Components diagram
Rn P sin30
F
Rn N
P
30
F
6 kg
6g N
Resolving : P cos 30  F = 0
 F = P cos 30
Resolving : R + P sin 30  6g = 0
 R = 6g  P sin30
F = R
[3]
Sub [1] and [2] into [3]
P cos30 = 0.5(58.5  0.5P)
P cos 30
P cos 30 = 29.4  0.25P
P(cos 30 + 0.25) = 29.4
P = 26.3 N
[1]
[2]
6g N
Example
A block of mass 6 kg rests on a rough
horizontal surface. The coefficient
of friction between block and the
surface is 0.5. A force P is applied
at an angle of 30 to the horizontal.
Find the value of P when the block is
about to move.
Solution
Components diagram
P cos 30
6g P sin30
Mathematics made simple
6 kg
F
6g N
Resolving : R  P sin 30  6g = 0
 R = 6g + P sin30
F = R
[3]
Sub [1] and [2] into [3]
P cos30 = 0.5(58.5 + 0.5P)
P cos 30 = 29.4 + 0.25P
P(cos 30  0.25) = 29.4
P = 47.7 N
M1 Friction
P
30
Resolving : P cos 30  F = 0
 F = P cos 30
Rn
F
Rn N
[1]
[2]
-5-
Friction
Example
A block of mass 6 kg rests on a rough
PN
horizontal surface. The coefficient of
30
friction between block and the surface F
is 0.5. The forces P and 2P are applied
at an angle of 30and 60 to the
horizontal. Find the value of P when
the block is about to move.
R
6 kg
2PN
60
Solution
Components diagram
Resolve:  2P cos60 + P cos30  F = 0
 F = 2P cos60 + P cos30
Rn 2Psin60
Resolve:  R + 2P sin60  P sin30  6g = 0
 R = 6g  2P sin60 + P sin30
F = R
[3]
Sub [1] and [2] into [3]
P + 0.8660P = 0.5(58.8 +P(1.732+0.5))
1.8660P = 29.4  0.616P
2.482P = 29.4
P = 11.8 N
2Pcos60
P cos30
F
P sin30 6g
Example
A block of mass 6 kg rests on a rough
horizontal surface. A force of 25 N is
applied at an angle of 30 to the horizontal.
If the block is about to move, find the
coefficient of friction between the block
and the surface.
Solution
Components diagram

25 cos 30
25 N
30
F
6 kg
6g N
Resolving : 25 cos 30  F = 0
 F = 25 cos 30
 F = 21.651
Resolving : R + 25 sin 30  6g = 0
 R = 46.3
Rn 25 sin30
F
Rn N
[1]
[2]
F 21.651

 0.468
R
46.3
6g N
Mathematics made simple
M1 Friction
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Friction
Example
A block of mass m kg rests on a rough plane inclined at 30 to the
horizontal. The coefficient of friction between the block and the plane is
0.5. A force P acts on the block up the plane along the line of greatest
slope. Find the possible values of P if the block remains stationary.
Solution
There are two possibilities: (a) The block is about to move down the slope,
(b) The block is about to move up the slope.
(a)
Force diagram
Parallel and Perpendicular components
Rn
Rn N
F
PN
mg sin30
30
P
30
mN
mg cos30
: P + F  mg sin 30 = 0
 P = 4.9m  F
[1]
Resolving perpendicular to the plane: Rn  mg cos 30 = 0
 Rn = 8.487 m
[2]
Resolving parallel to the plane
F = R
[3]
 P = 4.9m  0.5(8.487m)
 P = 0.657 m
(b)
Parallel and Perpendicular components
Rn
P
F
mg sin30
Perpendicular: Rn – mg cos30 = 0
 Rn = 8.487 m
F = R
 P = 0.5  8.487 m + 4.9 m
 P = 9.14 m
30
mg cos30
So, the range of values of P is:
Mathematics made simple
Parallel: P  F - mg sin30 = 0
 P = F + 4.9m
0.657m  P  9.14 m.
M1 Friction
-7-
Friction
Example
A block of mass m kg rests on a rough plane inclined at 30 to the
horizontal. The coefficient of friction between the block and the plane is
0.5. A horizontal force P acts on the block. Find the possible values of P if
the block remains stationary.
Rn
PN
30
mg N
Solution
There are two possibilities: (a) The block is about to move down the slope,
(b) The block is about to move up the slope.
(a)
Rn
F
Pcos30
mg sin30
Resolving parallel:
P cos30 + F  mg sin30 = 0
[1]
Resolving perpendicular:
Rn - mg cos 30  Psin30 = 0
[2]
F = R
P cos30  mg sin30 +0.5(mg cos30 + Psin30) = 0
mg cos30 Psin30
0.8660P  0.5mg + 0.4330mg + 0.25P = 0
1.116P = 0.067mg  P = 0.06 mg
(b)
Rn
Pcos30
mg sin30
F
mg cos30 Psin30
Resolving parallel:
P cos30  F  mg sin30 = 0
[1]
Resolving perpendicular:
Rn - mg cos 30  Psin30 = 0
[2]
F = R
P cos30  mg sin30  0.5(mg cos30 + Psin30) = 0
0.8660P  0.5mg  0.4330mg  0.25P = 0
0.616P = 0.933mg  P = 1.51 mg
So, the range of values of P is:
Mathematics made simple
0.06 mg  P  1.51 mg.
M1 Friction
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