Module 3: Friction
Types of friction:
There are two types of friction,
1) Static friction
2) Dynamic friction
1) Static friction:
If a body is kept on a ground (or on the surface of other body) and is in equilibrium, then the friction experienced by a body is called as static friction.
2) Dynamic friction:
If a body is moving on the surface of ground (or on the surface of other body), then the friction experienced by a body is called as dynamic friction.
There are two types of dynamic friction:
i) Sliding friction
ii) Rolling friction
i) Sliding friction:
If a body slide on the surface of ground (or on the surface of other body), then the friction experienced by a body is called as static friction.
ii) Rolling friction:
If a body rolls on the surface of ground (or on the
surface of other body), then the friction experienced by a body is called as rolling
friction.
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When the body is in limiting equilibrium then from the above figure,
∑Fx = 0; P – F = 0 or P = F
∑Fy = 0; R – W = 0 or R = W
If P = F
The body is in limiting equilibrium
If P < F
The body remains at rest
If P ≥ F The
body continues in motion
Coefficient of friction (μ) :
Coefficient of friction is defined as the ratio of limiting friction to the normal reaction at the surface of contact.
Experimentally it is proved that the limiting
friction F is directly proportional to the normal reaction R.
∴ F = μ R (where μ is constant)
This constant is known as coefficient of friction between the surface of contact.
For smooth surface the coefficient of friction is
zero.
Angle of Friction or limiting angle (Φ):
Angle of Friction or limiting angle (Φ) is defined as the angle made by resultant reaction S with the normal reaction R, when the body is in limiting equilibrium.
Consider a body of weight
W resting on a rough horizontal surface shown in figure below.
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From figure the forces acting on the body are,
i) Self weight (W) of the body in downward
direction
ii) Normal rection R
iii) applied pull force P
iv) Limiting friction F
If we considered only two forces R and F and apply
the parallelogram law of forces, then R and F represents the two adjacent sides
of parallelogram in magnitude and direction and S represents the resultant in
magnitude and direction. Therefore, S is called as resultant reaction (or total
reaction). The S makes an angle Φ with the normal reaction R.
The angle Φ is maximum when the body is in limiting
equilibrium. This maximum angle Φ is known as angle of friction (or limiting
angle).
From the above figure,
tanΦ = F / R
tanΦ = μR / R
tanΦ = μ
Hence, the coefficient
of friction (μ) is equal to the tangent of angle of friction (Φ).
Magnitude of resultant reaction (S):
S2 = F2
+ R2
S = √ (F2 + R2)
S = √ (μR2 + R2)
Angle of repose (α):
Angle of
repose (α) is defined as the angle made by the inclined plane with the
horizontal plane at which the body placed on an inclined plane is just on the
point of moving down the plane under the action of its own weight.
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Consider a body of weight
W is resting on an inclined. The plane is inclined at an angle α with the horizontal as shown in figure above. The
weight W of the body acting vertically downward. This weight W is resolved along
the plane and perpendicular to the plane. The component of weight W along the
plane is W sin α and perpendicular to
the plane is W cos α. As the motion of the body is down
of the plane, the force friction F act up the plane. As the body is placed on
the inclined plane, the normal reaction R is perpendicular to the inclined
plane.
For limiting equilibrium condition,
∴ ∑Fx = 0;
F - W sin α = 0
F = W sin α
μ R = W sin α ------------------------(I)
∴ ∑Fy = 0
R - W cos α = 0
R = W cos α -------------------------(II)
On dividing equation (I) by equation (II) we get,
μ R / R = W sin α / W cos α
∴ μ = tan α
But
we know that, μ = tanΦ,
∴ μ = tan α
= tanΦ
Coefficient of friction (μ) = tangent of angle of repose (tan
α) = tangent of angle of friction (tan Φ)
Or α = Φ
Hence from the above relation, the angle of repose
(α) is also called as angle of friction (Φ).
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In above figure the resultant reaction S makes an angle Φ with normal reaction R. If X-axis rotated about Y-axis the resultant reaction S will also rotate correspondingly. The line of action of resultant reaction S will always lie on the surface of right circular cone whose vertex angle is 2Φ. This cone made by rotation of resultant reaction S is called as cone of friction.
Laws of static friction:
1. The frictional force
always acts tangential to the plane of contact and in the opposite direction to
the applied force
2. When the body is in limiting
equilibrium then the ratio of limiting friction to the normal reaction is
called as coefficient of friction
3. Coefficient of
friction is depends upon the nature of surface in contact (not depend upon the
area of surface in contact)
4. The static friction is
always greater than dynamic friction
5. Force of friction increases with the increases in applied force till the limiting equilibrium. Therefore the force of friction is self adjusted.
Force of friction depends
upon the following factors:
1. Nature of surface in
contact. It is zero for smooth surface.
2. Magnitude of applied
force. Force of friction increases with the increase in applied force
i.e. F ∝ P
3. Normal reaction
between between the surface of contact
i.e. F ∝ R
Advantages of friction:
Due to friction, we can
do following things,
1. We can walk on a rough
surface
2. Vehicles can run on
the road and can be stopped
3. Can hammer the nail
4. Can hold pen, book,
scale, bag etc.
Disadvantages
of friction:
1.
Causes wear and tear of machine parts
2.
Energy lost due to friction
3.
More effort required to move the load on rough surface
4.
Vehicle consume more fuel etc.
Equilibrium
of a body placed on a rough horizontal plane:
1.
Force applied horizontally:
Consider a body of weight
W placed on a rough horizontal plane as shown in figure above. Let P be the
force applied to move the body. Weight of the body W acts vertically downward.
Normal reaction R acts
vertically upward. As the body move towards right then the frictional
resistance F acts towards left and if the body move towards left then the
frictional resistance F acts towards right.
For limiting equilibrium,
∑Fx = 0;
P – F = 0 or P = F = μ R
∑Fy = 0;
R – W = 0 or R = W
From the above two equations, we can find out the values of unknown.
2.
Force applied at an angle ϴ with the horizontal plane:
Consider a body of weight W placed on a rough
horizontal plane as shown in figure above. Let P be the force applied at an
angle ϴ
with the horizontal to move the body. Weight
of the body W acts vertically downward. Normal reaction R acts vertically upward. As the
body move towards right then the frictional resistance F acts towards left and
if the body move towards left then the frictional resistance F acts towards
right. Force P can be resolved into horizontal and vertical components.
Horizontal component of P = P cosϴ
Vertical
component of P = P sinϴ
For limiting equilibrium,
∑Fx = 0;
P cosϴ – F = 0 or P cosϴ = F = μ R
∑Fy = 0;
R + P sinϴ – W = 0 or R + P sinϴ = W
From the above two equations,
we can find out the values of unknown.
Equilibrium
of a body placed on a rough inclined plane:
i)
Force applied parallel to the plane:
It
includes following two types,
a) Body
just sliding down the plane
b) Body
just sliding up the plane
a) Body just sliding down the plane:
(Angle of inclination is greater than the angle of repose)
Consider a body of weight
W is resting on an inclined plane. The inclined plane is inclined at an angle α with the horizontal as shown in figure above. The
force P is applied parallel to the plane. The inclination of the plane is
greater than the angle of repose, the body will move down the plane. And an upward
force P will be required to prevent it from sliding down. As the body tends to
move down the plane force of friction F = μ R will act up the plane.
The weight W of the body
acting vertically downward. This weight W is resolved into the two components along
the plane and perpendicular to the plane. The component of weight W along the
plane is W sin α and perpendicular to
the plane is W cos α.
For limiting equilibrium condition,
∴ ∑Fx = 0;
P + F - W sin α = 0
P + F = W sin α
P + μ R = W sin α ------------------------(I)
∴ ∑Fy = 0
R - W cos α = 0
R = W cos α -------------------------(II)
By using above equation (I) and (II) we can find out the values of unknowns.
b) Body just sliding up the plane:
(Angle of inclination is less than the angle of repose)
As the angle of
inclination is less than the angle of repose the body
will be in equilibrium. Consider the force P is required to move the body up
the plane. As the body tends to move up the plane, force of friction will act
down the plane.
For limiting equilibrium condition,
∴ ∑Fx = 0;
P – W sin α – F = 0
P = F + W sin α
P = μ R + W sin α ------------------------(I)
∴ ∑Fy = 0
R - W cos α = 0
R = W cos α -------------------------(II)
By using above equation (I) and (II) we can find out the values of unknowns.
Ladder
Friction:
Consider
a ladder of length L and weight W is resting on the rough ground and leaning
against a rough wall.
The
weight W of the ladder acts vertically downward through its centre.
Let,
μg =
coefficient of friction between the ladder and the ground
Rg = normal
reaction at the ground
Fg
= force of friction between the ladder and the ground
μw =
coefficient of friction between the ladder and the wall
Rw = normal reaction at the wall
Fw = force of friction
between the ladder and the wall
As
the upper end of the ladder tends to move downwards, the force of friction
between the ladder and the ground (Fw) will be produced in the
upward vertical direction.
As
the lower end of the ladder tends to move away from the wall the force of
friction between the ladder and the ground (Fg) will be produced towards
the wall. As the system is in equilibrium all the conditions will be applied.
i.e. ∑Fx = 0; ∑Fy = 0; and
∑M = 0;
1. The force of friction
between the ladder and the wall is given by,
Fw
=
μw
Rw
If
the wall is smooth then,
μw
=
0. and Fw = 0
Hence
the force of friction between the ladder and the wall is zero.
2.
The force of friction between the ladder and the ground is given by,
Fg
=
μg
Rg
If
the ground is smooth then,
μg
=
0; and Fg = 0
Hence
the force of friction between the ladder and the ground is zero.
3.
Rw
and Rg are the normal reactions at the point of contact
between the ladder and the wall and between the ladder and the ground
respectively.
Numerical on ladder friction
(Q-1) A ladder of weight
400 N and length 10 m is supported on smooth wall with its lower end 4 m from
the wall. The coefficient of friction
between the floor and the ladder is 0.3. Show the forces acting on the ladder
and find frictional force at floor.
Solution:
Guven data,
Weight of ladder W = 400 N
Length of ladder L = 10 m
For smooth wall μw =
0
For rough floor μg =
0.3
i) Forces acting on the
ladder:
Let,
μf =
coefficient of friction between the ladder and the floor
Rf = normal
reaction at the floor
Ff
= force of friction between the ladder and the floor
μw =
coefficient of friction between the ladder and the wall
Rw = Normal
reaction at the wall
Fw
= force of friction between the ladder and the wall
The
angle ϴ can be calculated as,
sinϴ
= 4 / 10
ϴ
= sin-1 (4 / 10)
ϴ = 23.580
The Length BD = DC = 4 / 2 = 2 m
By
applying conditions of equilibrium,
∴ ∑Fx = 0;
Rw
- Ff = 0
Rw = Ff ------------------------(I)
∴ ∑Fy = 0
Fw + Rf
– W = 0
0 + Rf – W = 0 ------------------ (For
smooth wall μw
= 0, therefore Fw = 0)
Rf – 400 = 0
Rf = 400 N -------------------------(II)
Taking
moment of all the forces about point A,
∑MA = 0;
(Ff x L cosϴ)
– (Rf x 4) + (400
x 2) + (Rw x 0) = 0
(Ff x 10 cos23.580)
– (400 x 4) + (400 x 2) + 0 = 0
Ff = [(400 x 4) - (400 x 2)] / (10 cos23.580)
Ff = 87.29 N
Answer:
Rw = Ff =
87.29 N,
Rf = 400 N,
Fw = 0.
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