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#### Page No 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.

#### Page No 95:

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).

#### Page No 95:

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.

#### Page No 95:

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.

#### Page No 95:

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.

#### Page No 95:

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.

#### Page No 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.

#### Page No 96:

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.

#### Page No 96:

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.

#### Page No 96:

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.

#### Page No 96:

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.

#### Page No 96:

(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.

#### Page No 96:

(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.

#### Page No 96:

(c) Mg FMg $\sqrt{1+{\mathrm{\mu }}^{2}}$

Let T be the force applied on an object of mass M.

If = 0, Fmin = Mg.

If T is acting in the horizontal direction, then the body is not moving.
∴ $T=\mathrm{\mu }\left(mg\right)$

Thus, we have:
$Mg\le F\le Mg\sqrt{1+\left(\mathrm{\mu }{\right)}^{2}}$

#### Page No 96:

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.

#### Page No 96:

(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.

#### Page No 96:

(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.

#### Page No 96:

(c) is same for both cars

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 = $\frac{\mathrm{\mu }{m}_{1}\mathrm{g}}{{m}_{1}}=\mathrm{\mu g}$

Frictional force on car with mass m2 = μm2 g
So, the deceleration due to frictional force = $\frac{\mathrm{\mu }{m}_{2}\mathrm{g}}{{m}_{2}}=\mathrm{\mu g}$
As both the acceleration are same, from the second equation of motion
$s=ut+\frac{1}{2}a{t}^{2}$

Thus, we can say that both the cars have same minimum stopping distance.

#### Page No 96:

(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.

#### Page No 96:

(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.

#### Page No 96:

(a) μ < μ', M < M'

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
fmax > T

μ'M'g >        (i)

Now see the free body diagram of a boy of mass M:

fmax = μmg

The condition for preventing the slide is
fmax > 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'

#### Page No 97:

(a) F > FN
(b) F > f
(d) FNf < F < FN + f

The system is in equilibrium condition when F = f.
Hence, the net horizontal force is zero.

f = μFN
F > FN
f = FN 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.

FN > f$⇒$FN > μFN$⇒$μ < 1

Here, the given relation between F and f i.e

F > f and f = μFN  will not be satisfied So it cannot be said that the net horizontal force is zero or nonzero.

FNf < F < FN + f
f = μFN
$\frac{f}{\mu }-f
For the above relation, we can say that F ≠ f and so the net horizontal force is nonzero.

#### Page No 97:

(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.

#### Page No 97:

(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.

#### Page No 97:

(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 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.

#### Page No 97:

(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.

#### Page No 97:

Let m be the mass of the body.

From the free body diagram,
Rmg = 0
(where R is the normal reaction force and g is the acceleration due to gravity)
R = mg                 (1)
Again maμkR = 0
(where μk is the coefficient of kinetic friction and a is deceleration)
or ma = μkR
From Equation (1),
maμkmg
a = μkg
⇒ 4 = μkg
$⇒{\mu }_{\mathrm{k}}=\frac{4}{g}=\frac{4}{10}=0.4$

Hence, the coefficient of the kinetic friction between the block and the plane is 0.4.

#### Page No 97:

Friction force acting on the block will decelerate it.
Let the deceleration be 'a'.

Using free body diagram

Rmg = 0
(where R is the normal reaction force)
R = mg                 (1)
Again, ma − μkR = 0
(where μk is the coefficient of kinetic friction)
From Equation (1),
ma = μkmg
a = μkg = 0.1 × 10
= 1 m/s2
Given:
initial velocity, u = 10 m/s
final velocity, v = 0 m/s (block comes to rest)
a = −1 m/s2 (deceleration)
Using equation of motion v2 $-$ u2 = 2as
(where s is the distance travelled before coming to rest)
$s=\frac{{v}^{2}-{u}^{2}}{2a}$
On substituting the respective values, we get

Therefore, the block will travel 50 m before coming to rest.

#### Page No 97:

A block of mass m is kept on a horizontal table. If force is applied on the block, a friction force will be there: p → frictional force and F → 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.

#### Page No 97:

Free body diagram for the block is as follows:

From the above diagram:
Rmg cos θ = 0
R = mg cos θ                     (1)

For the block, u = 0 m/s, s = 8 m and t = 2 s.
According to the equation of motion
$s=ut+\frac{1}{2}a{t}^{2}$
$s=0+\frac{1}{2}a{2}^{2}\phantom{\rule{0ex}{0ex}}8=2a$
a = 4 m/s2

Again,
μkR + mamg sin θ = 0
(where μk is the coefficient of kinetic friction)
From Equation (1):
μkmg cos θ + mamg sin θ = 0
mkg cos θ + ag sin θ) = 0
⇒ μk × 10 × cos 30° = g sin 30° − a
$⇒{\mathrm{\mu }}_{\mathrm{k}}×10\frac{\sqrt{3}}{2}=10×\left(\frac{1}{2}\right)-4\phantom{\rule{0ex}{0ex}}⇒\left(5\sqrt{3}\right){\mathrm{\mu }}_{\mathrm{k}}=1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{\mu }}_{\mathrm{k}}=\left(\frac{1}{5\sqrt{3}}\right)=0.11$

Therefore, the coefficient of kinetic friction between the block and the surface is 0.11.

#### Page No 97:

Free body diagram of the block for this case is as follows:

F ma − μkR + mg sin 30° = 0
4 − 4a − μkR + 4g sin 30° = 0               (1)
R − 4g cos 30° = 0                               (2)
R = 4g cos 30° = 0

Substituting the values of R in Equation (1) we get
4 − 4a − 0.11 × 4g cos 30° + 4g sin 30° = 0
$⇒4-4a-0.11×4×10×\frac{\sqrt{3}}{2}+4×10×\frac{1}{2}=0$
4 − 4a − 3.81 + 20 = 0
4 − 4a − 3.18 + 20 = 0
a ≈ 5 m/s2

For the block, u = 0, t = 2 s and a = 5 m/s2.
According to the equation of motion,
​    ​$s=ut+\frac{1}{2}a{t}^{2}$

Therefore, the block will move 10 m.

#### Page No 97:

(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
$=0.2×\left(9.8\right)\sqrt{3}+2×9.8×\left(\frac{1}{2}\right)$
3.39 + 9.8 $\approx$ 13 N

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
F = 2g sin 30° − μR

Because F = 6.41 N, the body will move down the incline with acceleration, hence the force required is zero.

#### Page No 97:

Using the free body diagram,

g = 10 m/s2m = 2 kg, θ = 30 and μ = 0.2

Rmg cos θ − F sin θ = 0
R = mg cos θ + F sin θ              (1)
and
mg sin θ + μRF 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.

#### Page No 97:

Let m be the mass of the boy.

From the above diagram:
Rmg cos 45° = 0

Net force acting on the boy, making him slide down
= mg sin 45° − μR
= mg sin 45° − μmg cos 45°
$=m×10×\left(\frac{1}{\sqrt{2}}\right)-0.6×m×10×\left(\frac{1}{\sqrt{2}}\right)\phantom{\rule{0ex}{0ex}}=m\left(5\sqrt{2}-3\sqrt{2}\right)=m×2×\sqrt{2}$
The acceleration of the boy $=\frac{\mathrm{Force}}{\mathrm{Mass}}$

#### Page No 97:

Let a be the acceleration of the body sliding down.

From the above diagram:
Rmg 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,
v = u + at
= 0 + (0.5)4 = 2 m/s
$s=ut+\frac{1}{2}a{t}^{2}$

For the next half metre, u = 2 m/s, a = 4 m/s2 and s = 0.5.
$⇒0.5=2t+\left(\frac{1}{2}\right)4{t}^{2}$
⇒ 2t2 + 2t − 0.5 = 0
⇒ 4t2 + 4t − 1 = 0

Therefore, the time taken to cover the next half metre is 0.21 s.

#### Page No 97:

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).

⇒ tan λ ≤ μ
⇒ λ ≤ tan−1 μ

#### Page No 97:

From the above diagrams:
T + mamg = 0
T + 0.5a − 0.5 g = 0                       (1)
μR + ma + T1 T = 0
μR + 1a + T1T = 0                      (2)
μR + 1aT1 = 0
μR + a = T1                                    (3)

From Equations (2) and (3) we have
μR + a = TT1
TT1 = T1
T = 2T1

So, Equation (2) becomes
μR + a + T1 − 2T1 = 0
⇒ μR + aT1 = 0
T1 = μR + a
= 0.2g + a                       (4)

and Equation (1) becomes
2T1 + 0.5a − 0.5g = 0

From Equations (4) and (5)
0.2g + a = 0.25g − 0.25a

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.

#### Page No 98:

From the free body diagram:
μ1R + m1aF = 0
μ1R + 1 − 16 = 0 (R = mg cos θ)

⇒ μ1(2g) + (−15) = 0
${\mathrm{\mu }}_{1}=\frac{15}{20}=0.75$
Again,
μ2R1 + ma = Fmg sin θ = 0
μ2R1 + 4 × 0.5 = 16 − 4g sin 30° = 0
R1 mg cos θ      (θ = 30°)

Therefore, the friction coefficients at the two contacts with blocks are
μ1 = 0.75 and μ2 = 0.06.

#### Page No 98:

Consider that a 15 kg object is moving downward with an acceleration a.

From the above diagram,
T + m1a m1g = 0
T + 15a − 15g = 0
T = 15g − 15a                  (1)

Now,
T1m2gm2a = 0
T1 − 5g − 5a = 0
T1 = 5g + 5a                                   (2)

Again,
T − (T1 + 5a + m2R) = 0
T − (5g + 5a + 5a +m2R) = 0        (3)
(where R = μg)

From Equations (1) and (2),
15g − 15a = 5g + 10a + 0.2 (5g)
⇒ 25a = 90          [g = 10 m/s2]
a = 3.6 m/s2

From Equation (3),
T = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10 = 96 N in the left string.

From Equation (2),
T1 = 5g + 5a
= 5 × 10 + 5 × 36
= 50 + 18
= 68 N in the right string.

#### Page No 98:

Given,
initial velocity of the vehicle, u = 36 km/h = 10 m/s
final velocity of the vehicle, v = 0
s = 5 m, $\mathrm{\mu }=\frac{4}{3}$, g = 10 m/s2

Let the maximum angle of incline be $\theta$.

Using the equation of motion

From the free body diagram

Rmg 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°.

#### Page No 98:

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,
$s=ut+\left(\frac{1}{2}\right)a{t}^{2}$

(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

#### Page No 98:

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,

Rmg cos θ = 0
R = mg cos θ                              (1)
and
μR + mamg sin θ = 0                   (2)
⇒ μ mg cos θ + mamg sin θ = 0

where θ = 30˚

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.

#### Page No 98:

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
a = μg = 1 × 10 = 10 m/s2

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,
$s=ut+\frac{1}{2}a{t}^{2}$
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.

#### Page No 98:

(a) From the free body diagram

R = 4g cos 30°

μ2R + m1apm1g sin θ = 0
μ2R + 4ap − 4g sin 30° = 0
⇒ 0.3 $×$ (40) cos 30° + 4ap − 40 sin 30° = 20             (2)

R1 = 2g cos 30° $=10\sqrt{3}$                        (3)
p + 2a − μ1R1 − 2g sin 30° = 0               (4)

From Equation (2),
$6\sqrt{3}+4a-p-20=0$

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.

#### Page No 98:

From the free body diagram

R1 = M1g cos θ                                             (1)
R2 = M2g cos θ                                             (2)
T + M1g sin θ − M1a − μR1 = 0                      (3)
T M2g + M2a + μR2 = 0                             (4)

From Equation (3),
T + M1g sin θ − M1 a − μM1g cos θ  = 0         (5)

From Equation (4),
TM2 g sin θ + M2 a + μM2 g cos θ = 0        (6)

From Equations (5) and (6),
g sin θ(M1 + M2) − a(M1 + M2) − μg cos θ(M1 + M2)
a(M1 + M2) = g sin θ(M1 + M2) = μg cos θ(M1 + M2)
a = g(sin θ − μ cos θ)ag(sin θ − μ cos θ)
∴ The acceleration of the block (system) = g(sin θ − μcos θ)

The force exerted by the rod on one of the blocks is tension, T.
T = −M1g sin θ + M1a + μM1g cos θ
T = −M1g sin θ + M1(g sin θ − μg cos θ) + μM1g cos θ = 0

#### Page No 98:

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(mgP 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:

⇒ μ cos θ − sin θ = 0
θ = tan−1 μ
So,
Dividing numerator and denominator by cos θ, we get

(using the property $1+{\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta$)
Therefore, the minimum force required is  at an angle θ = tan−1 μ.

#### Page No 98:

Let T be the maximum force exerted by the man on the rope.

From the free body diagram,
R + T = Mg
R = MgT                     (1)
Again,
R1Rmg = 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

T (1 + μ) = μMg + μmg

Therefore, the maximum force exerted by the man is $\frac{\mathrm{\mu }\left(M+m\right)g}{1+\mathrm{\mu }}$.

#### Page No 98:

Consider the free body diagram.

(a) For the mass of 2 kg, we have:
R1 − 2g = 0
R1 = 2 × 10 = 20
2a + 0.2 R1 − 12 = 0
⇒ 2a + 0.2 (20) = 12
⇒ 2a = 12 − 4
a = 4 m/s2

Now,
4a − μR1 = 0
⇒ 4a = μR1 = 0.2 (20) = 4
a1 = 1 m/s2

The 2 kg block has acceleration 4 m/s2 and the 4 kg block has acceleration 1 m/s2.

(ii) We have:
R1 = 2g = 20
Ma = μR1 = 0
a = 0
And,
Ma + μmg − F = 0
4a + 0.2 × 2 × 10 − 12 = 0
⇒ 4a + 4 = 12
⇒ 4a = 8
a = 2 m/s2

#### Page No 98:

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,
μ1R1 = μ1 × m1g
μ1R1 = μ1 × 2g = (0.2) × 20
= 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:
μ2 = R2 = μ2m2g = (0.3) × 5 kg
= 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.

#### Page No 98:

From the free body diagrams of the two blocks, we have
R1 = mg    ...(i)
F = μR1+T    ...(ii)
TμR1 = 0    ..(iii)
From equations (i) and (ii), we have
Fμmg = 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
2FTμmg = ma       ....(i)
From the free body diagram of lower block, we have
T = Ma + μmg

Putting the value of T in (i), we get
2FMaμmgμmg = ma
Putting F = 2μmg, we get
2(2μmg) − 2μmg = a(M + m)
⇒ 4μmg − 2μmg = a(M + m)
in opposite directions.

#### Page No 99:

From the free body diagram, we have:
R1 + mamg = 0
R1 = m (ga)
= mgma                    ...(i)
Now,
FT − μR1 = 0 and
T − μR1 = 0
F − [μ (mgma)] − μ(mgma) = 0
F − μ mg − μma − μmg + μma = 0
F = 2 μmg − 2 μma
= 2 μm (ga)

(b) Let the acceleration of the blocks be a1.

R1 = mgma               ....(i)
And,
2F T − μR1 = ma1    ...(ii)
Now,
T = μR1 + Ma1
= μmg − μma + Ma1

Substituting the value of F and T in equation (ii), we get:
2[2μm(ga)] − (μmg − μma + Ma1) − μmg + μma = ma1
⇒ 4μmg − 4μma − 2μmg + 2μma= ma1 + Ma1

Thus, both the blocks move with same acceleration a1 but in opposite directions.

#### Page No 99:

From the free body diagram:

R1 + QEmg = 0
where
R1 is the normal reaction force
Q is the charge
E is the small electric field
R1 = mgQE                         (1)

F− μR1 = 0
FT = μR1
where F is the maximum horizontal force required

From Equation (1),
F − T − μ(mgQE) = 0
F − T = μ(mgQE)
F − T − μmg + μQE = 0              (2)
T − μR1 = 0
T = μR1 = μ (mgQE)               (3)

From Equation (2),
F − μmg + μ QE − μmg + μQE = 0
F − 2 μmg + 2μQE = 0
F = 2μmg − 2μQE
F = 2μ (mgQE)

Therefore, the maximum horizontal force that can be applied is 2μ (mg QE).

#### Page No 99:

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

(i) (ii)

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.

#### Page No 99:

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.

(Free body diagram 1)
From the free body diagram 1, we have:

(Free body diagram-2)

R1ma = 0
R1 = ma                               ....(i)
Again,
2ma + TMg + μ1R1 = 0
T = Mg − (2 + μ1) ma          ....(ii)

Using the free body diagram 2, we have:

T + μ1R1 + Mg − R2 = 0
Substituting the value of R1 from (i), we get:
R2 = T + μ1 ma + mg

Substituting the value of T from (ii), we get:
R2 = (Mg − 2ma − μ1ma) = μ1 ma + Mg + ma
R2 = Mg + Ma − 2ma               ....(iii)

Again using the free body diagrams −2,
T + TRMa − μ2R2 = 0
⇒ 2T Mama − μ2(Mg + mg − 2ma) = 0

Substituting the values of R1 and R2 from (i) and (iii), we get:
2T = (M + m)a + μ2 (Mg + mg − 2ma)            ....(iv)

From equations (ii) and (iv), we have:
2T = 2mg − 2(2 + μ1) ma
= (M + m) a + μ1(Mg + mg − 2ma)
⇒ 2mg − μ2 (M + m)g = a[M + m − 2μ2m + 4m + 2μ1m]
Therefore, the acceleration of the block of mass M in the given situation is given by
$a=\frac{\left[2m+{m}_{2}\left(M+m\right)\right]g}{M+m\left[5+2\left({\mathrm{\mu }}_{1}-{\mathrm{\mu }}_{2}\right)\right]}$

#### Page No 99:

Net force on the block$=\sqrt{\left({20}^{2}+{15}^{2}\right)}-\left(0.5\right)×40$
= 25 − 20 = 5 N

Therefore, the block will move at 53° angle with the 15 N force.

#### Page No 99:

Thus, we have:
μR + μR = mg
⇒ 2μR = 40 × 10

Normal force = 250 N

#### Page No 99:

Let a1 and a2 be the accelerations of masses m and M, respectively.
Also, a1 > a2 so that mass m moves on mass M.

Let after time t, mass m is separated from mass M.

Using the equation of motion
During this time, mass m covers $vt+\frac{1}{2}{a}_{1}{t}^{2}$ and ${\mathrm{s}}_{m}=vt+\frac{1}{2}{a}_{2}{t}^{2}$.

For mass m to separate from mass M, we have:

From the free body diagram, we have:

$m{a}_{1}+\frac{\mathrm{\mu }}{2}R=0\phantom{\rule{0ex}{0ex}}⇒m{a}_{1}=-\left(\frac{\mathrm{\mu }}{2}\right)mg=\left(\frac{\mathrm{\mu }}{2}\right)m×10\phantom{\rule{0ex}{0ex}}⇒{a}_{1}=-5\mathrm{\mu }$

Again,
$M{a}_{2}+\mathrm{\mu }\left(M+m\right)g-\left(\frac{\mathrm{\mu }}{2}\right)mg=0$
⇒ 2Ma2 + 2μ (M + m)g − μmg = 0
⇒ 2Ma2 = μmg − 2μmg − 2μmg
$⇒{a}_{2}=\frac{-\mathrm{\mu }mg-2\mu Mg}{2M}$

Substituting the values of a1 and a2 in equation (i), we get:
$t=\sqrt{\frac{4Ml}{\left(M+m\right)\mathrm{\mu }g}}$

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