Friction HC Verma Exercise Questions Solutions Ch-6

= 2.69 = 2.7 m/s²

^{(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.}

Question-19 :-

Answer-19 :-

^{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),
T − M₂ 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)a − g(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}

Question-20 :-

Answer-20 :-

^{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:}

^{Dividing numerator and denominator by cos θ, we get}

Question-21 :-

The friction coefficient between the board and the floor shown in the following figure is μ. Find the maximum force that the man can exert on the rope so that the board does not slip on the floor.

Answer-21 :-

^{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}

Question-22 :-          (Friction HC Verma Exercise)

Answer-22 :-

^{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
and⇒ 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
therefore⇒ a = 2 m/s2

Question-23 :-

Find the accelerations a₁, a₂, a₃ of the three blocks shown in the following figure if a horizontal force of 10 N is applied on (a) 2 kg block, (b) 3 kg block, (c) 7 kg block. Take g = 10 m/s².

Answer-23 :-

^{Given:
μ₁ = 0.2,  μ₂ = 0.3 and  μ₃ = 0.4
from 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.

Question-24 :-  (Friction HC Verma Exercise)

The friction coefficient between the two blocks shown in the following figure is μ but the floor is smooth. (a) What maximum horizontal force F can be applied without disturbing the equilibrium of the system? (b) Suppose the horizontal force applied is double of that found in part (a). Find the accelerations of the two masses.

Answer-24 :-  

^{From the free body diagrams of the two blocks, we have
R₁ = mg    …(i)
F = μR₁+T    …(ii)
T − μR₁ = 0    ..(iii)
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)

Question-25 :-

Answer-25 :-

^{From the free body diagram, we have:
R₁ + ma − mg = 0
⇒ R₁ = m (g − a)
= mg − ma                    …(i)
Now,
F − T − μR₁ = 0 and
T − μR₁ = 0
⇒ F − [μ (mg − ma)] − μ(mg − ma) = 0
and⇒ F − μ mg − μma − μmg + μma = 0
⇒ F = 2 μmg − 2 μma
hence= 2 μm (g − a)
(b) Let the acceleration of the blocks be a₁}

R₁ = mg − ma               ….(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(g − a)] − (μmg − μma + Ma₁) − μmg + μma = ma₁
⇒ 4μmg − 4μma − 2μmg + 2μma= ma₁ + Ma₁

Question-26 :-

Consider the situation shown in the following figure. Suppose a small electric field E exists in the space in the vertically charge Q on its top surface. The friction coefficient between the two blocks is μ but the floor is smooth. What maximum horizontal force F can be applied without disturbing the equilibrium?
[Hint: The force on a charge Q bye the electric field E is F = QE in the direction of E.]

Answer-26 :-

^{From the diagram:}

^{R₁ + QE − mg = 0
where
R₁ is the normal reaction force
Q is the charge
E is the small electric field
⇒ R₁ = mg − QE                         (1)}

F − T − μR₁ = 0
F − T = μR₁
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
and ⇒ F = 2μmg − 2μQE
⇒ F = 2μ (mg − QE)

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

Question-27 :-

A block of mass m slips on a rough horizontal table under the action of a horizontal force applied to it. The coefficient of friction between the block and the table is μ. The table does not move on the floor. Find the total frictional force applied by the floor on the legs of the table. Do you need the friction coefficient between the table and the floor or the mass of the table?

Answer-27 :-

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.}

Question-28 :-  (Friction HC Verma Exercise)

Find the acceleration of the block of mass M in the situation of figure in the following. The coefficient of friction between the two blocks is μ₁ and that between the bigger block and the ground is μ₂.

Answer-28 :-

Let the acceleration of block of mass M is ‘a’ towards right. So, the block ‘m’ must go down with an acceleration ‘2a’.

As the block ‘m’ is in contact with the block ‘M’, it will also have acceleration ‘a’ towards right. So, it will experience two inertial forces as shown in figure (b). From free body diagram

R1 – ma = 0
⇒ R1 = ma (i)

Also,
2ma + T – Mg + μ1R1 = 0
⇒ T = Mg – (2 + μ1) ma (ii)

Using the free body diagram (c), we can write,

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 (c),
T + T – R – Ma – μ2R2 = 0
⇒ 2T – Ma – ma – 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

Question-29 :-

Answer-29 :-

^{Net force on the block =√20²+152 – (0.5) × 40
= 25 – 20 = 5 N}

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

Question-30 :-

A person (40 kg) is managing to be at rest between two vertical walls by pressing one wall A by his hands and feet and the other wall B by his back (in the following figure). Assume that the friction coefficient between his body and the walls is 0.8 and that limiting friction acts at all the contacts. (a) Show that the person pushes the two wall with equal force. (b) Find the normal force exerted by either wall on the person. Take g = 10 m/s².

Answer-30 :-

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

Question-31 :- (Friction HC Verma Exercise)

In the following figure shows a small block of mass m kept at the left end of a larger block of mass M and length l. The system can slide on a horizontal road. The system is started towards right with an initial velocity v. The friction coefficient between the road and the bigger block is μ and that between the block is μ/2. Find the time elapsed before the smaller blocks separates from the bigger block.

Answer-31 :-

^{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}

^{From the free body diagram, we have:}

^{Substituting the values of a₁ and a₂ in equation (i), we get:} 

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