HC Verma Questions and Solutions: Chapter 6- Friction- 1 | HC Verma Solutions - JEE PDF Download

According to the first law of limiting friction,
f = μN
where f  is the frictional force
N is the normal reaction force
μ is the coefficient of static friction
and
N = mg － Fcosθ
where m is the mass of the body
F is the contact force acting on the body

According to the first law of the limiting friction,
f = μN
where f is the frictional force
μ is the coefficient of friction
N is the normal reaction force
When we take smaller steps on ice, the normal reaction force exerted by the ice is small. Therefore, the smaller steps ensure smaller friction.

Let T be the force applied on an object of mass M.
If T = 0, F_min = Mg.
If T is acting in the horizontal direction, then the body is not moving.
∴ T = μ(mg)

Thus, we have:
Mg ≤ F ≤ Mga 

During the time interval ∆t₂, the scooter is moving with a constant velocity, which implies that the force exerted by the seat on the man is 500 N (for balancing the weight of the man).
During the time interval ∆t₁ and ∆t₃, the scooter is moving with constant acceleration and deceleration, which implies that a frictional force is also applied. Therefore, the net force exerted by the seat on the man should be >500 N.

Since the wall is smooth and the surface of A and B in contact are rough, the net vertical force on the system is in the downward direction. Hence, the system cannot remain in equilibrium.

The normal reaction force on the system (comprising of wall and contact surface of Aand B) is provided by F. As can be seen from the figure, the weight of A and B is in the downward direction. Therefore, the frictional force f_A and f_BA (friction on B due to A) is in upward direction.

Given: both the cars have same initial speed.
Let the masses of the two cars be m1 and m₂.
Frictional force on car with mass m₁ = μm₁ g
So, the deceleration due to frictional force = 
Frictional force on car with mass m₂ = μm₂ g
So, the deceleration due to frictional force = 
As both the acceleration are same, from the second equation of motion 
Thus, we can say that both the cars have same minimum stopping distance.

When we apply hard brakes just enough to prevent slipping on wheels, it provides optimum normal reaction force, which gives the maximum friction force between tyres of the car and the road.

We know that
N = mg cos θ˚
f_max = μN = μmg cos θ
where N  =  normal reaction force
f_max = frictional force
θ =  angle of inclination
μ = coefficient of friction
When the block just begins to slide, it means
mg sin θ = f_max
mg sin θ = μmg cos θ
μ = tan θ
and the coefficient of friction depends on the angle of inclination (θ) and does not depend on mass.
Now consider the block sliding condition:
mg sin θ − f_max = ma
mg sin θ − μmg cos θ = ma
∴ a = g(sin θ − μ cos θ)
From the above equation it is clear that acceleration does not depend on the mass but depends on θ and μ.

Let T be the force applied by the boy on the block. 

Free body diagram for the box:

The condition for preventing the slide is
f_max > T
μ'M'g > T         (i)
Now see the free body diagram of a boy of mass M:

f_max = μmg
The condition for preventing the slide is
f_max > T
μmg > T 
The condition for sliding the entire system (block and boy) is
f' > f (block is not slide)
μ'M'g > μmg
μ'M' > μm
μ < μ'
m < M'

The system is in equilibrium condition when F = f.
Hence, the net horizontal force is zero.
f = μF_N
F > F_N
f = F_N and 0 ≤ μ ≤ 1
Therefore, we can say that F > f. So the net horizontal force is nonzero.
F > f, and so the net horizontal force is zero.
F_N > f ⇒ F_N > μF_N ⇒ μ < 1
Here, the given relation between F and f i.e 
F > f and f = μF_N  will not be satisfied So it cannot be said that the net horizontal force is zero or nonzero.
F_N − f < F < F_N + f
∵ f = μF_N

For the above relation, we can say that F ≠ f and so the net horizontal force is nonzero.

The contact force exerted by a body A on another body B is equal to the normal force between the bodies. Therefore, we can conclude that the force of friction between the bodies is zero or the bodies may be rough but they don't slip on each other.

All the above statements are correct. The static friction is sometimes less than the kinetic friction.

When force F is applied on the block, the force of friction f comes into play. As we increase the  applied F,  the static friction force adjusts itself to become  (equal) to the applied force F  and goes upto  its maximum value equal to limiting friction force.After this ,it is treated as a constant force (i.e . now its value does not change until and unless the body starts moving). If the applied force F is greater than the limiting friction force, then the kinetic friction force comes into play at that time. The kinetic friction force is always less than the limiting friction force.

When the vehicle is accelerating, the force is applied (by the tyre on the road) in west direction .That causes a net resultant  frictional force acting in east direction. Due to this  force of friction only ,the  car is moving in east direction.
When the vehicle is moving with a uniform velocity, the force of friction on the wheels of the vehicle by the road is zero.