Chapter 4 : Friction

Question 1: Define Friction. Explain Laws of Solid Friction.

Answer:

The opposing nature of force which reduces actual motion of the body acts on sliding surface , called friction force and the acting forces is called friction. The force of friction exists between the surfaces of contact of two bodies. The friction acts in opposite to direction of motion of the body.

Laws of Friction:
1) The dynamic friction is less than the limiting friction.
2) The force of friction is directly proportional to weight of the body sliding over the surface.
3) The force of friction depends upon nature of the surfaces in contact.
4) The rolling friction is less than the sliding friction for the same weight of the body.

Question 2: Define Impulse and Linear Momentum. Describe their relation.

Answer:

Impulse: It is defined as the product of magnitude of applied force and time taken to reach from one point to another point. It is denoted by 'J'. ( i.e. J = applied force * time taken). It's unit is Newton sec or kgm/sec

Linear Momentum: If a body of mass 'm' is moving with velocity 'v' then linear momentum is defined as product of its mass and obtained velocity due to effect of force is linear momentum.

Linear Momentum (P) = v * m  where 'v' is an velocity and 'm' is an mass of an object. 

The momentum of a body gives measure of motion of the body. The momentum is related to the force applied in the body. If there is change in momentum, there is change in velocity. The change in velocity is acceleration. Hence, acceleration and applied force are related term.

Relation between Impulse and Momentum :

From Newton Second Law of Motion,

f = dp/dt
Or, f*dt=dp
Integrating in both sides,
Or, (t2-t1)∫f*dt=(p2-p1)∫dp
Or,  F(t2-t1)∫dt = (p2-p1)∫dp
Or, F[t](t2-t1)=[p](p2-p1)
Or, F(t2-t1)=(p2-p1) 
Or, J=(p2-p1)
Hence, Impulse on a body is equal to change of momentums.

Question 3: A car of mass 1000 kg moves at a constant speed of 20m/sec along a horizontal road where the friction force is 200N. Calculate the power developed by the engine.

Solution:

Here we have,
Force of Friction (Fc)=200N
Mass of car (m)=1000kg
Velocity of car (v)=20m/sec

Angle made by car and surface (Ɵ)=0°

Then, Total force applied (F)= mgsinƟ + Fc

=1000kg*10sin0° + 200N

=1000*0+200N

=200N

Hence,

Power developed by car (P)=Force applied (F) * Speed (v)

=200N * 20m/sec

=4000watt 

Question 4: Define coefficient Friction. Describe about Angle of Friction and Angle of Repose.

Answer: 

Coefficient of Friction:

The force of friction (Fc) when a body of mass 'm' slides over another body is directly proportional to normal reaction (R).

i.e. Fc=µR where 'µ' is called proportianlity constant and it is also called coefficient of friction.

µ=Fc/R

Hence,It is defined as the ratio of friction force to normal reaction.

Angle of Friction: 

Let, 'm' be the mass of the block (i.e. body), 'R' is normal reaction on the block by the surface. If 'Fc' is force of friction which acts in the direction opposite to the applied force.

Hence, It is defined as the angle made y the resultant of normal reaction and the force of friction with normal reaction. It is denoted by '∝'. Then completed rectangle OCBA. 
In right angled triangle OAB,
tan∝=AB/OA 
Or, tan∝=Fc/R
i.e. tan∝ = Fc/R

Or, ∝=tan^-l  (Fc/R)..................(i)

But, the coefficient of friction is given by:
µ=Fc/R.......................(ii)

Hence, comparing equation (i) and (ii),

tan∝= µ

The coefficient of friction is equal to the tangent of angle of friction.

Angle of Repose: 

It is defined as the angle made by the inclined plane with horizontal surface such that a body placed over this inclined plane just begins to slide.

Let a body of mass 'm' place over the surface of inclined plane OB. Here, 'mg' is the weight of the body acting vertically downward. 'R' is the normal reaction, 'Fc' is the friction force acting upward along inclined plane. When the body just begins to slide down on inclined plane, the component 'mgsinƟ' of an mg balances the friction force 'Fc'.

i.e. mgsinƟ = Fc..................(i)

 and, the component 'mgcosƟ' of mg balances the normal reaction 'R'

i.e. mgcosƟ = R.................(ii)   

Dividing equation (i) by equation (ii) we get,

mgsinƟ/mgcosƟ = Fc/R

Or, tanƟ = Fc/R....................(iii)

i.e. Ɵ=  tan^-l  (Fc/R)

Again, The coefficient of friction (µ) is given by:

µ=Fc/R................(iv)

Then, comparing equation (iii) and equation (iv), we get

tanƟ = µ

Hence, Coefficient of friction (µ) is given by:

µ= tanƟ = tan∝

Or, tanƟ = tan∝

i.e. Ɵ = ∝

Hence, Angle of repose is equal to angle of friction.