Static and Kinetic Friction :Class 11 CBSE Physics Chapter 5 - Future Study Point

Static and Kinetic Friction :Class 11 CBSE Physics Chapter 5

static and kinetic friction

Static and Kinetic Friction: Class 11 CBSE Physics Chapter 5

Static and Kinetic Friction: Class 11 CBSE Physics Chapter 5: It is well known to everybody that friction is the force that opposes the motion of any object. Friction is of three kinds static friction, kinetic friction, and rolling friction. Friction is the force of reaction against the force which is responsible for the movement of the object. Static friction is the tendency of relative motion of a body in a rest position as example if we apply a force on an object and it is not moving, here the object has a tendency to move relative to a surface but it is not moving, the frictional force which is exerted by the surface to object is known as static friction. Kinetic friction is the frictional force that opposes the motion of a moving object, kinetic friction exists when there is relative motion.

static and kinetic friction

Static and Kinetic Friction: Class 11 CBSE Physics Chapter 5

Static Friction: The points to be noted about static friction?

(i)Static friction exists when there is a tendency of relative motion

(ii)Static friction varies

(iii) The value of static friction exists within 0≤ Fs ≤ FLimiting

FLimiting = μsN

Where μs is the constant of static friction and N is the normal reaction

A body becomes capable to move only when the applied force is greater than the maximum value of the static friction(i.e FLimiting ).

When the body moves then static friction is replaced by the kinetic friction

If the body is in a rest position then

Static friction is equal to the applied force

Example:A body of 4 kg is in the rest position on a plane and 6N force is applied on it but it is not moving, find static friction exerted by the surface.( μs=0.2,g=10/s²)

 

Solution:

static friction

The forces acting on the object are

Normal reaction = N

The weight of the object =mg =4g

N = 4g, both forces are in opposite directions, so they will cancel each other

We have

FLimiting = μsN =0.2×4g =0.2×4×10=8N

FLimiting = 8N

It is the maximum value of the static friction, since its value should be within 0≤ Fs ≤ FLimiting

The value of  Fs  should be 6N because body is in rest

Static and Kinetic Friction : Class 11 CBSE Physics Chapter 5

Kinetic Friction: The points to be noted about kinetic friction?

(i) It exists when relative motion is there

(ii)Kinetic friction is a constant quantity

(iii)Kinetic friction is directly proportional to the normal reaction

Let the kinetic friction of a moving body is Fk  and the normal reaction on the body is N

Fk  ∝ N

Fk  = μkN

Where μk is the constant of kinetic friction and its value depends on the type of surface on which body is moving.

(iv)Kinetic friction is always opposite to the direction of motion

Example: A body of the mass 5 kg is moving on a surface due to a force of 10N in the forward direction find acceleration produced on the body.(μk =0.2 and g=10/s²)

Solution:

kinetic friction

The forces acting on the object are

The normal reaction by the surface to body=N

The weight of the body = mg = 4g

The friction =Fk

Here N = mg(since the body is moving in the forward direction)

Force applied in forward direction =10N

Applying the formula for the frictional force

Fk  = μkN

= 0.2×5g =0.2×4×10 =8N

The net force is Fnet =10 -8 =2N

Let the acceleration produced is a

Applying Newton’s second law

Fnet = ma = 4a

2 = 4a

a =2/4 =0.5 m/s²

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