TOPIC - FRICTION
Friction is the force resisting the
relative motion of solid surfaces,
fluid layers, and material elements
sliding against each other.
The constant ratio which the limiting friction bears to
the normal reaction is called CO-EFFICIENT OF
FRICTION.
It is usually denoted by the letter μ
Thus , F=μR
where F= limiting friction
R=normal reaction
The limiting friction F and the normal reaction R
acting at right angles to each others have a resultant
force say S, which is called the RESULTANT
REACTION .
Thus S =√(R²+F²) = √(R²+μ²R²) = R√(1+μ²)
When a body is in limiting equilibrium on another ,
the angle which the resultant reaction S at the point of
contact makes with the normal reaction is called the
ANGLE OF FRICTION and is generally denoted by the
greek letterλ.
Tanλ=μ
Hence the tangent of the angle of friction is equal to
the co-efficient of friction
R
λ
Tanλ = μ
Where F= limiting friction
R= normal reaction
S
F
Let F be the limiting friction , R be the normal reaction and S
be the resultant reaction
. If λ be the angle of friction which S makes with R , then
S cosλ = R ……………(1)
S sinλ = F ……………(2)
Dividing (2) by (1), we have
tan λ = F / R = µ
Hence the tangent of angle of friction is equal to the co-efficient
of friction.
The cone of friction is the cone
which has the point of contact as
its vertex, the normal as its axis and
λ as its semi vertical angle
R
Axis of friction
θ
A heavy body is placed on a rough inclined
plane of inclination ‘α’ greater than the
angle
of friction, being acted upon by a
force parallel to the plane and alone a line of
greater slope. To find the limits between
which the force must lie.
Case-1:Let the body be on the
point of moving up the plane
Let P1 be the force acting parallel to the plane
keeping the body at rest so that the force of
friction R acts down the plane
Resolving along and perpendicular to the
plane
P1=μR+Wsinα…………………(1)
R=Wcosα…………………………..(2)
Eliminating R from (1) and (2)
Body moving upward
μR
α
O
α
P1=W(sinα+μcosα)
=W(sinα +(sinλ/cosλ)cosα)
on solving
P1=Wsin(α+λ)/cosλ
Which gives the amount of force
.
CASE 2:Let the body be on the
point of moving down the plane
 Let P2 be the force required to keep the body at
rest
since body is on the point of moving down the plane
the force of friction acts up the plane
Resolving along and perpendicular to the plane
P2+μR=Wsinα
R=Wcosα
P2=Wsinα-Wcosα
=Wsin(α-λ)/cosλ
Body moving down the plane
P2
μR
α
α
To find the limits between which a force
must lie in order to keep a body in equ. on
rough inclined plane when the force acts
horizontally:
 Let ‘α’be the inclination of the plane to the
horizontal ‘W’ be the weight of the body and r
the normal reaction. Let’μ’ be the co-efficient
of the friction and ‘λ’be the angle of friction’
CASE 1
: LET THE BODY BE ON THE POINT OF
MOVING UP THE PLANE
 Let P1 be the horizontal force required
to keep the body at rest so that the
force of friction μR acts down the plane
B
o
α
μR
α
α
Resolving the force acting on the body
along and perpendicular to the plane
P1cosα=μR+W sinα………………………..(1)
R=W cos α+P1sinα…………………..(2)
Eliminating R from (1) and (2)
P1cosα =μ(W cos α+P1sinα)+W sinα
P1(cos α-tanλsinα)=W(tanλ cosα+sinα)
On solving
P1=W tan(α+λ)
CASE 2:LET THE BODY BE ON THE
POINT OF MOVING DOWN THE
PLANE
 Let P2 be the force required to keep the
body at rest since the body is on the
point of moving down the plane the
force of friction acts up the plane
μR
α
o
α
α
Resolve the force along and perpendicular
P2 cos α+μR=Wsinα…………………..(3)
R=wcosα+P2sinα……………………….(4)
Eliminating R from (3) and (4)
P2cos α+μ (Wcosα+P2sinα)= Wsinα
P2(cosα+tanλ.sinα)=W(sinα-tanλ.cosα)
On solving
P2=Wtan(α-λ)
EXAMPLE
 TWO EQUAL WEIGHTS ARE ATTACHED TO THE ENDS
OF THE STRING WHICH IS LAID OVER THE TOP OF
TWO EQUALY ROUGH PLANES HAVING THE SAME
ALTITUDES AND PLACED BACK TO BACK . THE
ANGLES OF THE INCLINATION OF THE PLANES TO THE
HORIZON BEING 30 ˚AND 60˚ RESP.SHOW THAT THE
WEIGHTS WILL BE ON THE POINT OF MOTION IF THE
COEFFICIENT OF FRICTION BE 2-√3
SOLUTION
 Let R and R’ be the planes with the inclinations of 60˚ and 30˚
resp. and T be the tension of string
The weights W on the plane of inclination 60˚ is on the point of
moving downwards therefore the friction μR on this plane acts
up the plane and the friction μR’ on the other plane acts down
the plane.
R
T
μR
R’
T
μR’
600
W
300
W
Resolving along and perpendicular to the plane
(i) T+μR=Wsin 60˚
R=Wcos60˚
T=Wsin60˚-μR
=Wsin60˚-μWcos60˚
=W(sin60˚-μcos60˚)
(ii) T-μR’=Wsin30˚
R’=Wcos30˚
T=W(sin30˚+μcos30˚)
Equating the two values of T we get
W(sin60˚-cos60˚)=Wsin30˚+cos30˚
μ=(√3-1)/(√3+1)
=2-√3
EXAMPLE
 THE FORCE ACTING PARALLEL TO A ROUGH
INCLINED PLANE OF INCLINATION ‘α’TO THE
HORIZON JUST SUFFICIENT TO DRAW A WEIGHT
UP THE PLANE IS n TIMES THE FORCE WHICH
WILL JUST LET IT BE ON THE POINT OF SLIDING
DOWN THE PLANE . PROVE THAT
tanα=μ(n+1)/(n-1)
SOLUTION:
Body moving upward
μR
α
α
O
Let P1 be the force acting up the plane
parallelto the plane which is just
sufficient to draw a weight W up the
plane
P1=Wsin(α+λ)/cosλ……………………..(1)
Let P2 be the force which is just sufficient
to support the body
P2=Wsin(α-λ)/cosλ…………………(2)
Body moving down the plane
P2
μR
α
α
Now according to the given condition
P1 =n.P2
From (1) and (2)
Wsin(α+λ)/cos λ=nWsin(α-λ)/cosλ
Sin(α+λ)/sin(α-)λ=n/1
Applying c-d rule, we have
(Sin(α+λ)+sin(α-λ))/(sin(α+λ)-sin(αλ))=(n+1)/(n-1)
On solving
Tanα=tanλ.(n+1)/(n-1)
=μ(n+1)/(n-1)
 A uniform rod of length 2l rests in a vertical plane
against a smooth horizontal peg at a height h , the
lower end of the rod being on level ground . Show
that if the rod be on the point of slipping when its
inclination to the horizontal is θ,then the
co=efficient of friction between the rod and the
ground is
 L sinθ sin2θ/(2h-lcosθ sin2θ)
 Solution
 Let AB be the rod resting with the end A on the
level ground and a point of its lenth resting over
the peg P. let the normal reaction at A and P be R
and S resp. and R the force of friction . Resolving
vertically and horizontally
θ
R+Scosθ=W
And
S sinθ=μW
S=μW/(sinθ+μ cosθ)
Taking moments about A
W l cos θ=S h cosecθ
=(W/(sinθ+μ cosθ))*hcosecθ
μ(h-lsinθcos²θ)=lsin²θcosθ
hence=lsinθ cos2θ/(2h-lcosθsinθ)
ASSIGNMENT
1. A weight of 60 kg can just rest on a rough inclined
plane of inclination 30˚ to the horizon ,when the
inclination is increased to 60˚ find the least horizontal
force which will support it ?
2. Find how high can a particle rest inside a hollow
sphere of radius ‘a’ if the co-efficient of friction be
1/√3?
3. A weight can be just supported on a rough inclined
plane by a force P acting along the plane or by a force
Q acting horizontaly show that weight is PQ/√(Q²sec²øP²)
where ‘ø ‘ is the angle of friction?
4.A body of weight 80 kg rest on rough horizontal plane while a
force of 20 kg is acting on it in a direction making an angle of 60˚
with the horizontal .find the force of friction that is called into
play ?
5.A heavy body is placed on a rough inclined plane to find the
force just sufficient to move the body up the plane the force
acting in a vertical plane through the line of greatest slope to
the body.
6.A uniform rod rest with one extremity against a rough vertical
wall the other being supported by a string of equal length
fastened to point in the wall. Prove that the least angle which
the string can make with the wall is tan‫־‬¹ 3/μ?
TEST
NOTE : do any three.
1.To find the limits between which a force must lie in
order to keep a body in equilibrium on a rough inclined
plane when the force acts horizontally?
2. Find how high can a particle rest inside a hollow
sphere of radius ‘a’ if the co-efficient of friction be
1/√3?
3.A uniform rod rest with one extremity against a rough
vertical wall the other being supported by a string of
equal length fastened to point in the wall. Prove that
the least angle which the string can make with the wall
is tan‫ ־‬¹3/μ?
4.A ladder inclined at 60˚ to the horizon rest between a
rough floor and a smooth vertical wall . Show that if the
ladder begins to slide down when a man has ascended so
that his center of gravity is half way up .co-efficient of
friction between the foot of ladder and this floor is √3/6?
5. A uniform ladder rests in limiting equilibrium with one
end on a rough floor whose co-efficient of friction is and
with the other end against a smooth vertical wall . Show
that the inclination to the vertical Is tan ‫־‬¹2μ?
THANKS