Friction : Static friction , Limiting friction , Kinetic friction

Concept of Friction :

When a body slides on a surface, the surface is observed to exert a retarding force on the body, preventing relative motion.

When an external horizontal force is applied to a body resting on a rough surface, the body does not immediately start moving : this is due to the fact that the force of friction applied by the surface on this body prevents the onset of relative motion.

Friction is an important force in several aspects of everyday life.

Sometimes, it is important to minimise friction (e.g. in the engine of a car); in other cases it is because of friction that our machines work (from nails to bicycles and cars).

The force of friction acts tangential to the surface of contact between two bodies, and acts in such a manner as to prevent relative motion.

What are the distinct types of Friction ? 

Static friction: Where there is no relative motion between the bodies in contact.

Kinetic friction : Where there is relative motion kinetic friction may be sliding friction or, rolling friction, according as whether the bodies involved slide or, roll on each other.

Static friction :

Suppose that a body is kept on a rough surface and an external force Fext is applied on the body in the horizontal direction.
The body does not accelerate initially (remains at rest) due to the fictional force f opposing.

Relative motion between the two bodies: The force f  equals the external force .

As Fext is increased,’ f ‘ increases , until it reaches a maximum value flim , also known as the Limiting Friction.

If Fext is increased thereafter, the force of friction ‘ f ‘ suddenly drops and motion commences. Friction is kinetic.

The actual force of static friction (f ) is always less then its limiting value flim before relative motion commences:
|f | < flim.
We must note that flim is not the actual force of static friction, it is merely a maximum value. (upper bound)

Kinetic friction :

Once the body is in motion, the frictional force that acts on it is called kinetic (fk).

In general, for sliding, fk ≤ flim (static)

 Laws of friction  ?

Both the Static limiting frictional force flim and sliding Kinetic frictional force (fk) are proportional to the normal contact reaction (N).

The normal reaction N acts on the body perpendicular to the surface. Suppose that the normal contact reaction force on a body from the ground is N.

Then, flim ∝ N

flim = μs N     …….(i)

where μs is a constant of proportionality called co-efficient of static friction.

It depends on the nature of surfaces in contact.

For kinetic sliding friction,

fk ∝  N

fk = μk N    ……..(ii)

Where μk is known as co-efficient of kinetic friction.


fk ≤ flim  , and μk ≤ μs ……… (iii)

The value of N in equation (i) and (ii) depends on the orientation of surfaces in contact.

What is the Angle of friction ?

The angle subtended by the resultant of the limiting force of static friction and the normal reaction, with respect to the normal reaction is known as the angle of friction.

In figure, the block of mass M is resting under a maximum applied horizontal force F on a rough horizontal surface.

The resultant of $ \displaystyle \vec{f} $ and $ \displaystyle \vec{N} $ is $ \displaystyle \vec{R} $

$ \displaystyle \vec{R} = \vec{f} + \vec{N} $

The angle between $\vec{R}$  and $\vec{N}$  is λ , called the Angle of friction.

In fact , f = μs N and $\displaystyle tan\lambda = \frac{f}{N} $

i.e. tanλ = μs    ………(iv)

What is Angle of Repose ?

The maximum angle of inclination of an inclined plane w.r.t. horizontal for which an object will remain at rest when placed on it, is known as the angle of repose.

Suppose a block of mass M is kept on a rough inclined plane. The angle of inclination of the plane w.r.t. horizontal is gradually increased from 0°.  It is found as the angle increases, the tendency of the block slipping increases.

Ultimately just at a particular maximum angle of inclination the block is on the verge of slipping as shown in the figure.

How to Calculate of Angle of Repose ?

Let λ be the angle of friction according to the definition and , the angle of repose. With reference to the chosen X-Y axis , we obtain for equilibrium under static conditions,

Fr = Mg sinθ   …(v)

N = Mg cosθ    …(vi)

and Fr = μs N    ..(vii)

Therefore from (v), (vi) and (vii)

tanθ = μs …(viii)

Again we already know,

tanλ = μs

Therefore, λ is numerically equal to θ

Solved Example : A block starts slipping on an inclined plane. It moves one metre in one second. What is the time taken by block to cover next one metre ?

Solution :

The forces acting on the block at any moment are :

(i) m g , Weight

(ii) N , normal reaction

(iii) f , friction force

Let a = acceleration of the block (down the plane)

$ \displaystyle S = ut +\frac{1}{2}a t^2 $

For path AB,

$ \displaystyle 1 = 0 \times t +\frac{1}{2}a (1)^2 $

for path AC,

$ \displaystyle 2 = 0 \times t +\frac{1}{2}a (t)^2 $

From (i) & (ii), t2 = 2 ;

t = √2  sec

Time taken to cover the distance BC = (t – 1) sec

= (√2 – 1) sec = 0.41 sec.

Solved Example  : A block of mass m = 2 kg is kept on a rough horizontal surface. A horizontal force F = 4.9 N is just able to slide the block. Find the coefficient of static friction. If F = 4 N, then what is the frictional force acting on the block ?

Solution: Maximum frictional force fmax = F = 4.9 N

fmax = μ N

4.9 = μ (m g )

4.9 = μ (2 x 9.8)

μ = 0.25

If the applied horizontal force is F = 4 N

F < fmax

We known that when F < fmax the block does not slide .

The frictional force on the block is f = F = 4 Newton.

Solved Example : Two blocks A & B are connected by a light inextensible string passing over a fixed smooth pulley as shown in the figure. The coefficient of friction between the block A & B the horizontal table is μ = 0.2; If the block A is just to slip, find the ratio of the masses of the blocks.

Solution : From F.B.D. of A, shown in fig

N + T Sinθ  = mAg …(1)

T Cosθ  = fmax = μ N …(2)

From (1), N = mAg – T Sinθ  …(3)

From (2) and (3),

T Cosθ =  mAg – T Sinθ

T (Cosθ + Sinθ ) =  mAg …(4)

From F.B.D. of B,

T = mBg    …(5)

Taking ratio of (4) and (5)

$ \displaystyle \frac{\mu}{cos\theta +\mu sin\theta} = \frac{m_B}{m_A} $

$ \displaystyle \frac{m_A}{m_B}= \frac{cos\theta +\mu sin\theta}{\mu}$

on putting θ = 60° , μ = 0.2

$ \displaystyle \frac{m_A}{m_B}= \frac{67}{20} $

Solved Example: The block A is kept over a plank B. The maximum horizontal acceleration of the system in order to prevent slipping of A over B is a = 2m /sec2. Find the coefficient of friction between A & B.

Solution: For the block A, fmax = m amax

amax of A = maximum acceleration of plank in order to have no slipping between A and B.

fmax = μ m g

μ m g = m amax

μ = amax / g

μ = 2/10 = 0.2

Also Read :

∗Variation of Friction with respect to applied Force
∗ Friction : Block on Block
∗ Circular Turning & Banking of Roads

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