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# HC Verma Class XII Physics 6- Friction

Exercise : Solution of Questions on page Number : 95

It is not necessary that the friction coefficient is always less than 1. When the friction is stronger than the normal reaction force, the coefficient of friction is greater than 1. For example, silicon rubber has the coefficient of friction greater than 1. It is easier to push a heavy block from behind than from the top because when we try to push a heavy block from the top, we increase the normal reaction force, which, in turn, increases the friction between the object and the ground (see the figure).

The person started with zero initial velocity, covered a 1 km distance and ended with zero velocity, so the acceleration is zero. Hence, the average friction force is zero.

It is difficult to walk on solid ice because the coefficient of friction between our foot and ice is very less; hence, a person trying to walk on solid ice may slip.

No, we cannot accelerate or stop a car on a frictionless horizontal road. The car will not move on a frictionless surface because rolling is not possible without friction.

In the first case, the normal reaction force is equal to the weight of the stone, hence the stone slides easily because the friction force is very less. However, in the second case, a small piece of wood is sandwiched, which increases the normal reaction force on the wood due to the weight of the door. Hence, greater the normal reaction force on the wood, the greater will be the frictional force between wood and the floor.

Exercise : Solution of Questions on page Number : 96

(a) The coefficient of friction between the card and the coin should be small.
(b) The coin should be heavy.
(c) If the card is pushed gently, the experiment fails because the frictional force gets more to time to act and it may gain some velocity and move with the card.

No, a tug of war cannot be won on a frictionless surface because the tension in the rope on both the sides of both the teams will be same. So, to win, one of the teams must apply some greater force, which is the force of friction.

The normal reaction force on a level road is mg, whereas on an inclined plane it is mg cos θ, which means that on an incline road the friction force between the tyre and the road is less. Hence, tyres have less grip on an incline plane and better grip on a level road.

By throwing the bag in one direction, we gain some velocity in the opposite direction as per the law of conservation of linear momentum. In this way we can come out of the ice easily.

The coefficient of friction increases between two highly smooth surfaces because the atoms of both the materials come very closer to each and the number of bonds between them increase.

Objective I

(b) decrease 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
If we decrease the angle between this contact force and the vertical, then Fcosθ increases and the normal reaction force (N) as well as the frictional force (f) decrease.

(b) smaller friction
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, Fmin = Mg.
If T is acting in the horizontal direction, then the body is not moving.
∴ T=μ(mg) d) >500 N for time ∆t1 and ∆t3 and 500 N for ∆t2.
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.

(d) the system cannot remain in equilibrium. 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.

(a) is upward The normal reaction force on the system (comprising of wall and contact surface of A and 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 fA and fBA (friction on B due to A) is in upward direction.

(c) is same for both cars
Given: both the cars have same initial speed. Thus, we can say that both the cars have same minimum stopping distance.

(b) apply the brakes hard enough to just prevent slipping
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.

(d) all of three are possible. We know that
N = mg cos θ˚
fmax = μN = μmg cos θ
where N = normal reaction force
fmax = frictional force
θ = angle of inclination
μ = coefficient of friction
When the block just begins to slide, it means
mg sin θ = fmax
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 − fmax = 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.

(a) μ < μ’, M < M’
Let T be the force applied by the boy on the block.  Exercise : Solution of Questions on page Number : 97

Objective II (b) the force of friction between the bodies is zero
(d) the bodies may be rough but they don’t slip on each other
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.

(b) Coefficient of static friction is always greater than the coefficient of kinetic friction.
(c) Limiting friction is always greater than the kinetic friction.
(d) Limiting friction is never less than the static friction.
All the above statements are correct. The static friction is sometimes less than the kinetic friction.

(c) The graph is a straight line of slope 45° for small F and a straight line parallel to the F-axis for large F.
(d) There is a small kink on the graph. 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.

(a) is towards east if the vehicle is accelerating
(b) is zero if the vehicle is moving with a uniform velocity
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.

Exercise

Let m be the mass of the body. From the free body diagram,
R − mg = 0
(where R is the normal reaction force and g is the acceleration due to gravity)
⇒ R = mg (1)
Again ma − μkR = 0 Hence, the coefficient of the kinetic friction between the block and the plane is 0.4.

Friction force acting on the block will decelerate it.
Let the deceleration be ‘a’.
Using free body diagram (where R is the normal reaction force)
⇒ R = mg (1) = 1 m/s2 A block of mass m is kept on a horizontal table. If force is applied on the block, a friction force will be there: frictional force and applied force So, friction force is equal to the applied force. One of the case is that the friction force is equals to zero when the applied force is equal to zero.

Free body diagram for the block is as follows: Therefore, the coefficient of kinetic friction between the block and the surface is 0.11.

Free body diagram of the block for this case is as follows: Therefore, the block will move 10 m.

(a) To make the block move up the incline, the applied force should be equal and opposite to the net force acting down the incline.
Applied force = μR + 2g sin 30° (1)
(where μ is the coefficient of static friction)
R = mg cos 30°
Substituting the respective values in Equation (1), we get With this minimum force, the body moves up the incline with a constant velocity as the net force on it is zero.
(b) Net force acting down the incline is given by Because F = 6.41 N, the body will move down the incline with acceleration, hence the force required is zero.

Using the free body diagram, R − mg cos θ − F sin θ = 0
⇒ R = mg cos θ + F sin θ (1)
and
mg sin θ + μR − F cos θ = 0
⇒ mg sin θ + μ(mg cos θ + F sin θ) − F cos θ = 0
⇒ mg sin θ + μmg cos θ + μF sin θ − F cos θ = 0 Therefore, while pushing the block to move up on the incline, the required force is 17.5 N.

Let m be the mass of the boy. Net force acting on the boy, making him slide down
= mg sin 45° − μR
= mg sin 45° − μmg cos 45° Let a be the acceleration of the body sliding down. From the above diagram:
R − mg cos θ = 0
⇒ R = mg cos θ (1)
and
ma + mg sin θ − μR = 0 For the first half metre, u = 0, s = 0.5 m and t = 0.5 s.
According to the equation of motion, Therefore, the time taken to cover the next half metre is 0.21 s.

Let
f be the applied force,
R be the normal reaction force and
F be the frictional force. The coefficient of static friction is given by (where λ is the angle of friction)
When F = μR, F is the limiting friction (maximum friction). When applied force increases and the body still remains still static then the force of friction increases up to its maximum value equal to limiting friction (μR).  From the above diagrams: From Equations (2) and (3) we have  Therefore,
(a) the acceleration of each 1 kg block is 0.4 m/s2,
(b) the tension in the string connecting the 1 kg blocks is
T1 = 0.2g + a + 0.4 = 2.4 N
​ and
(c) the tension in the string attached to the 0.5 kg block is
T = 0.5g − 0.5a
= 0.5 × 10 − 0.5 × 0.4
= 4.8 N.

Exercise : Solution of Questions on page Number : 98

From the free body diagram:   Consider that a 15 kg object is moving downward with an acceleration a. (where R = μg) Given,
initial velocity of the vehicle, u = 36 km/h = 10 m/s
final velocity of the vehicle, v = 0 Let the maximum angle of incline be θ.
Using the equation of motion From the free body diagram R − mg cos θ = 0
⇒ R = mg cos θ (1)
Again,
ma + mg sin θ − μ R = 0
⇒ ma + mg sin θ − μmg cos θ = 0
⇒ a + g sin θ − μg cos θ = 0 ⇒ 30 + 30 sin θ − 40 cos θ = 0
⇒ 3 + 3 sin θ − 4 cos θ = 0
⇒ 4 cos θ − 3 sin θ = 3 On squaring, we get
16 (1 − sin2 θ) = 9 + 9 sin2 θ + 18 sin θ
25 sin2 θ + 18 sin θ − 7 = 0 Therefore, the maximum incline of the road, θ = 16°.

To reach the 50 m distance in minimum time, the superman has to move with maximum possible acceleration.
Suppose the maximum acceleration required is ‘a’.
∴ ma − μR = 0 ⇒ ma = μ mg
⇒ a = μg = 0.9 × 10 = 9 m/s2
(a) As per the question, the initial velocity,
u = 0, t = ?
a = 9 m/s2, s = 50 m
From the equation of motion, (b) After covering 50 m, the velocity of the athelete is
v = u + at The superman has to stop in minimum time. So, the deceleration, a = − 9 m/s2 (max)
R = mg
ma = μR (maximum frictional force)
ma = μmg
⇒ a = μg
= 9 m/s2 (deceleration)
u1 = 30 m/s, v = 0 When the driver applies hard brakes, it signifies that maximum force of friction is developed between the tyres of the car and the road.
So, maximum frictional force = μR
From the free body diagram,
R − mg cos θ = 0
⇒ R = mg cos θ (1)
and
μR + ma − mg sin θ = 0 (2)
⇒ μ mg cos θ + ma − mg sin θ = 0 s = 12.8 m
u = 6 m/s
∴ Velocity at the end of incline Therefore, the harder the driver applies the brakes, the lower will be the velocity of the car when it reaches the ground, i.e. at 36 km/h.

Let a be the maximum acceleration of the car for crossing the bridge. From the above diagram,
ma = μR
(For more accelerations the tyres will slip)
ma = μmg To cross the bridge in minimum possible time, the car must be at its maximum acceleration.
u = 0, s = 500 m, a = 10 m/s2
From the equation of motion, Substituting respective values Therefore, if the car’s acceleration is less than 10 m/s2, it will take more than 10 s to cross the bridge. So, one cannot drive through the bridge in less than 10 s.

(a) From the free body diagram
\ μ2R + m1a − p − m1g sin θ = 0
μ2R + 4a − p − 4g sin 30° = 0
⇒ 0.3 × (40) cos 30° + 4a − p − 40 sin 30° = 20 (2) From Equation (4), (b) In this case, the 4 kg block will move at a higher acceleration because the coefficient of friction is less than that of the 2 kg block. Therefore, the two blocks will move separately. By drawing the free body diagram of 2 kg mass, it can be shown that a = 2.4 m/s2.

From the free body diagram From Equations (5) and (6), The force exerted by the rod on one of the blocks is tension, T. Let P be the force applied to slide the block at an angle θ. From the free body diagram,
R + P sin θ − mg = 0
⇒ R = −P sin θ + mg (1)
μR = P cos θ (2)
From Equation (1),
μ(mg − P sin θ)−P cos θ = 0
⇒ μmg = μP sin θ + P cos θ The applied force P should be minimum, when μ sin θ + cos θ is maximum.
Again, μ sin θ + cos θ is maximum when its derivative is zero:
ddθμ sin θ+cos θ=0
⇒ μ cos θ − sin θ = 0
θ = tan−1 μ Dividing numerator and denominator by cos θ, we get Let T be the maximum force exerted by the man on the rope. From the free body diagram,
R + T = Mg
⇒ R = Mg − T (1)
Again,
R1-R-mg=0
⇒R1=R+mg      (2)
and
T- μR1=0
From Equation (2),
T − μ(R + mg) = 0
⇒ T − μR − μ mg = 0
⇒ T − μ(Mg − T) − μmg = 0
T − μMg + μt − μmg = 0 Consider the free body diagram. (a) For the mass of 2 kg, we have: ⇒ 2a + 0.2 (20) = 12
⇒ 2a = 12 − 4
⇒ a = 4 m/s2
Now, The 2 kg block has acceleration 4 m/s2 and the 4 kg block has acceleration 1 m/s2.
(ii) We have: a = 0
And,
Ma + μmg − F = 0
4a + 0.2 × 2 × 10 − 12 = 0
⇒ 4a + 4 = 12
⇒ 4a = 8
⇒ a = 2 m/s2

Given:
μ1 = 0.2
μ2 = 0.3
μ3 = 0.4
Using the free body diagram, we have: (a) When the 10 N force is applied to the 2 kg block, it experiences maximum frictional force.
Here, = 4 N (From the 3 kg block)
Net force experienced by the 2 kg block = 10 − 4 = 6 N But for the 3 kg block (Fig. 3), the frictional force from the 2 kg block, i.e, 4 N, becomes the driving force and the maximum frictional force between the 3 kg and 7 kg blocks.
Thus, we have: = 15 N
Therefore, the 3 kg block cannot move relative to the 7 kg block.
The 3 kg block and the 7 kg block have the same acceleration (a2 = a3), which is due to the 4 N force because there is no friction from the floor. (b) When the 10 N force is applied to the 3 kg block, it experiences maximum frictional force of (15 + 4) N, i.e., 19 N, from the 2 kg block and the 7 kg block.
So, it cannot move with respect to them.
As the floor is frictionless, all the three bodies move together. (c) Similarly, it can be proved that when the 10 N force is applied to the 7 kg block, all three blocks move together with the same acceleration.  From the free body diagrams of the two blocks, we have From equations (i) and (ii), we have
F − μmg = T …(ii)
From equations (i) and (iii), we have
T = μmg
Putting T = μmg in equation (ii), we have
F = μmg + μmg = 2μmg (b) From the free body diagram of upper block, we have
2F − T − μmg = ma ….(i)
From the free body diagram of lower block, we have
T = Ma + μmg
Putting the value of T in (i), we get
2F − Ma − μmg − μmg = ma
Putting F = 2μmg, we get
2(2μmg) − 2μmg = a(M + m)
⇒ 4μmg − 2μmg = a(M + m) Exercise : Solution of Questions on page Number : 99 From the free body diagram, we have: (b) Let the acceleration of the blocks be a1. Substituting the value of F and T in equation (ii), we get: Thus, both the blocks move with same acceleration a1 but in opposite directions.

From the free body diagram: where F is the maximum horizontal force required
From Equation (1),
F − T − μ(mg − QE) = 0
F − T = μ(mg − QE)
⇒ F − T − μmg + μQE = 0 (2)
T − μR1 = 0
⇒ T = μR1 = μ (mg − QE) (3)
From Equation (2),
F − μmg + μ QE − μmg + μQE = 0
⇒ F − 2 μmg + 2μQE = 0
⇒ F = 2μmg − 2μQE
⇒ F = 2μ (mg − QE)
Therefore, the maximum horizontal force that can be applied is 2μ (mg − QE).

When a block slips on a rough horizontal table, the maximum frictional force acting on it can be found from the free body diagram (see below).
R = mg From the free body diagram,
F − μR = 0
⇒ F = μR = μmg
But the table is at rest, so the frictional force at the legs of the table is also μR. Let this be f, so from the free body diagram
f − μR = 0
⇒ f = μR = μmg
Therefore, the total frictional force on the table by the floor is μmg.

Let us the acceleration of the block of mass M be a and let it be towards right. Therefore, the block of mass m must go down with acceleration 2a. As both the blocks are in contact, it (block of mass m) will also have acceleration a towards right. Hence, it will experience two inertial forces as shown in the free body diagram given below.    Therefore, the block will move at 53° angle with the 15 N force.   