FRICTION

FRICTION  NEET JEE 2020  PHYSICS NEWTON LAWS OF MOTION 

 

Friction plays dual role in our life. It impedes the motion of object, causes abrasion and wear, and

converts other forms of energy into heat. On the other hand, without it we could not walk, drive cars,

climb roper, or use nails. Friction is a contact force that opposes the relative motion or tendency of

relative motion of two bodies.

Consider a block on a horizontal table as shown in the figure. If we apply a force, acting to the right, the block remains stationary if F is not too large. The force that counteracts F and keeps the block stationary is called frictional force. If we keep on increasing the force, the block will remain at rest and for a particular value of applied force, the body comes to state of about to move. Now if we slightly increase the force from this value, block starts its motion with a jerk and we observe that to keep the block moving we need less effort (force) than to start its motion.

Fig (1)

So from this observation, we see that we have three states of block, first block does not move, second

block is about to move and third block starts moving. The friction force acting in three states are called

static frictional force, limiting frictional force and kinetic frictional force respectively. If we draw the

graph between applied force and frictional force for this observation its nature is as shown in figure.

2.1    STATIC FRICTIONAL FORCE When there is no relative motion between the contat surfaces, frictional force is called static frictional force. It is a self-adjusting force, it adjusts its value according to requirement (of no relative motion). In the taken example static frictional force is equal to applied force. Hence one can say that the portion of graph ab will have a slope of 45°.

Fig (2)

2.2    LIMITING FRICTIONAL FORCE

This frictional force acts when body is about to move. This is the maximum frictional force that can

exist at the contact surface. We calculate its value using laws of friction.

Laws of friction

(i) The magnitude of limiting frictional force is proportional to the normal force at the contact surface.

flim ∝ N ⇒ flim  = μsN     …(3)

Here μs is a constant the value of which depends on nature of surfaces in contact and is called as

‘coefficient of static friction’. Typical values of μ ranges from 0.05 to 1.5.

(ii) The magnitude of limiting frictional force is independent of area of contact between the surfaces.

2.3    KINETIC FRICTIONAL FORCE

Once relative motion starts between the surfaces in contact, the frictional force is called as kinetic

frictional force. The magnitude of kinetic frictional force is also proportional to normal force.

fk = μkN     …(4)

From the previous observation we can say that μk < μs

Although the coefficient of kinetic friction varies with speed, we shall neglect any variation i.e., once

relative motion starts a constant frictional force starts opposing its motion.

Question:    A block of mass 5 kg is resting on a rough surface as shown in the figure. It is acted upon by a force of F towards right. Find frictional force acting on block when (a) F = 5N (b) 25 N (c) 50 N (μs = 0.6, μk = 0.5)   [g = 10 ms−2]    [g = 10 ms-2]

Solution:    Maximum value of frictional force that the surface can offer is fmax = flim = μsN = 0.6 × 5 × 10 = 30 newton Therefore, if F ≤ fmax body will be at rest and f = F of F > fmax body will more and f = fk

(a)    F = 5N < Fmax

So body will not move hence static frictional force will act and,

fs = F = 5N

(b)    F = 25N < Fmax

∴    fs = 25 N

(c)    F = 50 N > Fmax

So body will move and kinetic frictional force will act, it value will be

fk = μk N

= 0.5 × 5 × 10 = 25 newton

Questions:    A block B slides with a constant speed on a rough horizontal floor acted upon by a force which is  times the weight of the block. The line of action F makes 45° with the ground. The coefficient of friction between the block and the ground is μ = n/10. Find the value of n.

Solution:     Let m be the mass of the block. The weight of the block, then is mg. It is given that

F = mg. F can be resolved into two components F cos 30° parallel to the horizontal floor and F sin 30°

perpendicular to it.

Normal reaction  N = mg + F sin θ Hence the friction force f = μN = μ(mg + F sinθ) The body moves with constant speed. This means the force F cosθ is just able to overcome the frictional force f i.e.,  f = μN = F cosθ

μ (mg + F sin θ ) = F cosθ

μ = =

∴    n = 5

2.4    ANGLE OF FRICTION

The resultant of normal reaction and the frictional for is which makes an angle λ with . Now, tan        …(5) The angle λ is called the angle of friction.

Fig. (3)

2.5    ANGLE OF REPOSE

This is concerned with an inclined plane on which a body rests exerting its weight on the plane. The angle of repose of an inclined plane with respect to a body in contact with it is the angle of inclination of the plane with horizontal when the block just starts sliding down the plane under its own weight.

Fig. (4)

The limiting equilibrium of a body resting on the inclined plane is shown in figure.

The forces acting are (i) Its weight mg downward, (ii) Normal reaction, (iii) The force of limiting friction

. Taking α as the angle of repose and resolving the forces along the plane and perpendicular to the

plane, we get for equilibrium

mg cos α =  N    …(i)

mg sin α = f = μN    …(ii)

Dividing equation (ii) by (i),

μ = tan α

∴    angle of repose = α = tan−1 (μ)    …(6)

2.6    MOTION ON A ROUGH INCLINED PLANE

Suppose a motion up the plane takes place under the action of pull P acting parallel to the plane. N = mg cos α Frictional force acting down the plane, F = μN = μ mg cos α Applying Newton’s second law for motion up the plane.

Fig. (5)

P − (mg sin α + f) = ma

P − mg sin α − μ mg cos α = ma

If P = 0 the block may slide downwards with an acceleration a. The frictional force would then act up the plane.

mg sin α − f  = ma

or,    mg sin α − μ mg cos α = ma

Questions:    A 20 kg box is gently placed on a rough inclined plane of inclination 30° with horizontal. The coefficient of sliding friction between the box and the plane is μ =. Find the acceleration of the box down the incline.  (take g = 1000 cm s–2)

Solution:    In solving inclined plane problems, the X and Y directions along which the forces are to be considered, may be taken as shown. The components of weight of the box are (i)      mg sin α acting down the plane and (ii)     mg cos α acting perpendicular to the plane. R = mg cos α

mg sin α − μN = ma

mg sin α − μmg cos α = ma

a = g sin α − μg cos α

= g (sin α − μ cos α) = 250 cm/s2

The box accelerates down the plane at 250 cm/s2.