HC Verma Concepts Of Physics Exercise Solutions Of Chapter 6 (Friction)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 6 - Friction:

EXERCISE

1. A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s². What is the coefficient of kinetic friction between the block and the plane?

Sol:

Given: acceleration, a = – 0.4 m/s².

Let mass of the body be ‘m’ and F_f → frictional force.

From the free body diagram,

Vertical motion: R – mg = 0 → R = mg ……… (1)

Horizontal motion: ma = F_f = μR ……. (2)

From equation (1) and (2)

We get, μ = a/g = 4/10 = 0.4

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

2. A block is projected along a rough horizontal road with a speed of 10 m/s. If the coefficient of kinetic friction is 0.10, how far will it travel before coming to rest?

Sol:

Let mass of the body be ‘m’ and deceleration be a.

F_f → frictional force; μ = 0.10.

From the free body diagram,

Vertical motion: R – mg = 0 → R = mg ……… (1)

Horizontal motion: ma = F_f = μR ……. (2)

From equation (1) and (2)

We get, a = μg = 0.10 * 10 = 1 m/s²

Initial velocity u = 10 m/s; Final velocity v = 0 m/s; a = – 1 m/s² (deceleration).

We have, 2as = v² – u²

Or, s = (v² – u²)/2a = (0² – 10²)/2(– 1) = 50 m

So, it will travel 50 m before coming to rest. 

3. A block of mass m is kept on a horizontal table. If the static friction coefficient is μ, find the frictional force acting on the block. 

Sol:

Body is kept on the horizontal table.

If no force is applied, no frictional force will be there

F_s → frictional force

F_a → Applied force

From graph it can be seen that when applied force is zero, frictional force is zero.

4. A block slides down an inclined surface of inclination 30° with the horizontal. Starting from rest it covers 8 m in the first two seconds. Find the coefficient of kinetic friction between the two.

Sol:

Given: distance covered, s = 8 m; time taken, t = 2 s; initial velocity, u = 0.

We know, s = ut + ½ at²

Or, 8 = 0 * 2 + ½ * a * 2² → a = 4 m/s².

Frictional force, F_f = μR

Let, the mass of body be ‘m’.

The equation of motion of the body along perpendicular to the plane: acceleration, a = 0.

→ R – mg cos 30⁰ = 0 → R = mg cos 30⁰ ------- (1)

The equation of motion of the body along the plane: acceleration, a = 4 m/s².

→ mg sin 30⁰ – F_f = ma

Or, mg sin 30⁰ – μR = 4m ------- (2)

Solving equation (1) and (2)

We get, mg sin 30⁰ – μmg cos 30⁰ = 4m

Or, μ = (g sin 30⁰ – 4)/g cos 30⁰ = 0.11

Therefore, the coefficient of kinetic friction between the two is 0.11.

5. Suppose the block of the previous problem is pushed down the incline with a force of 4 N. How far will the block move in the first two seconds after starting from rest? The mass of the block is 4 kg.

Sol:

Given: mass of block, m = 5 kg; Force applied, F = 4 N; μ = 0.11.

The equation of motion of the body along perpendicular to the plane: acceleration, a = 0.

→ R – mg cos 30⁰ = 0 → R = 4g cos 30⁰ ------- (1)

The equation of motion of the body along the plane: acceleration, a.

→ mg sin 30⁰ – F_f + F = ma

Or, 4g * ½ – μR + 4 = 4a ------- (2)

Solving equation (1) and (2)

We get, 4g * ½ – 4* 0.11* g cos 30⁰ + 4 = 4a

Or, a = 5.04 ~ 5 m/s²  

Now, we have u = 0; t = 2 s; a = 5 m/s²

We know, s = ut + ½ at² = 0*2 + ½ * 5 * 2²

Or, s = 10 m

Therefore, the block move in the first two seconds after starting from rest is 10 m.

6. A body of mass 2 kg is lying on a rough inclined plane of inclination 30°. Find the magnitude of the force parallel to the incline needed to make the block move (a) up the incline (b) down the incline. Coefficient of static friction = 0.2.

Sol:

Given: mass of block, m = 2 kg; μ = 0.2; θ = 30⁰.

(a) Let F be the required force to make the block moves up the incline: 

The forces acting on the block are shown in the figure.

There is no acceleration perpendicular to the incline.

Hence, R – mg cos 30⁰ = 0 → R = 2g cos 30⁰

Frictional force, F_r = μmg cos 30⁰

The equation of motion of the body along the plane to make the block moves up the incline

→ F = mg sin 30⁰ + μmg cos 30⁰ = 2g * ½ + 0.2*2g cos 30⁰

Or, F = 9.8 + 3.4 = 13.2 N

Therefore, the magnitude of the force parallel to the incline needed to make the block move up the incline is 13.2 N.

(b) Net force acting down the incline is given by,

F = 2 g sin 30° – μR

= 2 × 9.8 × (1/2) – 3.39 = 6.41 N

Due to F = 6.41 N the body will move down the incline with acceleration.

No external force is required.

Therefore, Force required is zero.

7. Repeat part (a) of problem 6 if the push is applied horizontally and not parallel to the incline.

Sol:

Given: mass of block, m = 2 kg; μ = 0.2; θ = 30⁰.

Let F be the required horizontal force to make the block moves up the incline: 

The forces acting on the block are shown in the figure.

There is no acceleration perpendicular to the incline.

Hence, R – mg cos 30⁰ – F sin 30⁰ = 0 → R = 2g cos 30⁰ + F/2

Frictional force, F_r = μ (mg cos 30⁰ + F/2)

The equation of motion of the body along the plane to make the block moves up the incline

→ F cos 30⁰ = mg sin 30⁰ + μ (mg cos 30⁰ + F/2)

Or, F = 2g tan 30⁰ + 0.2 * (2g + F/2cos 30⁰)

Or, F – 0.115 F = 1.15 g + 0.4 g

Or, F = 1.75g = 17.5 N

Therefore, the magnitude of the force horizontal to the incline needed to make the block move up the incline is 17.5 N. 

8. In a children-park an inclined plane is constructed with-an angle of incline 45° in the middle part (figure 6-E1). Find the acceleration of a boy sliding on it if the friction coefficient between the cloth of the boy and the incline is 0.6 and g = 10 m/s².

Sol:

Given: θ = 45⁰; μ = 0.1; g = 10 m/s².

Let the mass of boy be M and acceleration be ‘a’.

The forces acting on the boy are shown in the figure.

There is no acceleration perpendicular to the incline.

Hence, R = Mg cos 45⁰ = Mg/√2

Frictional force, F_f = μR = 0.6Mg/√2

Taking components parallel to the incline and writing Newton’s second law,

We get, Ma = Mg sin 45⁰ – 0.6Mg/√2

Or, a = g/√2 – 0.6g/√2 = 2√2 m/s² 

Therefore, the acceleration of a boy is 2√2 m/s².

9. A body starts slipping down an incline and moves half meter in half second. How long will it take to move the next half meter?

Sol:

Given: s = 0.5 m; t = 0.5 s; u = 0.

We know, s = ut + ½ at²

Or, 0.5 = 0 * 0.5 + ½ * a * 0.5²

Or, a = 4 m/s²

Velocity after 0.5 s is v = u + at = 2 m/s.

For next half meter

Initial velocity, u = 2 m/s; s = 0.5; a = 4 m/s². 

We know, s = ut + ½ at²

Or, 0.5 = 2 * t + ½ * 4 * t²

Or, 2t² + 2t – 0.5 = 0

Or, t = 0.21 or – 1.21

Time can’t be negative,

Therefore, it will take 0.21 s to move the next half meter. 

10. The angle between the resultant contact force and the normal force exerted by a body on the other is called the angle of friction. Show that, if λ be the angle of friction and μ the coefficient of static friction, λ ≤ tan^-1 μ.

Sol:

Applied force → F

Contact force → F_c

Frictional force → F_f

Normal reaction → R

→ μ = tan λ = F_f/R

When F_f = μ R, F_f is the limiting friction (max friction). When applied force increase, force of friction increase upto limiting friction (μ R).

Before reaching limiting friction

F_f < μR

Therefore, tan λ = F_f/R ≤ μR/R

Or, λ = tan^-1 μ.

11. Consider the situation shown in figure (6-E2). Calculate (a) the acceleration of the 1.0 kg blocks, (b) the tension in the string connecting the 1.0 kg blocks and (c) the tension in the string attached to 0.50 kg.

Sol:

Let the acceleration of the blocks be ‘a’.

Assumption: smooth and massless pulley.

From F.B.D. of block A:

There is no acceleration perpendicular to the plane of motion.

Hence, R₁ = Mg = 1g

Frictional force, F_f = μR₁ = 0.2 * 1g = 0.2g

Taking forces along the plane of motion and writing Newton’s second law,

We get, T₁ – 0.2g = 1a ------------------- (1)

From F.B.D. of block B:

There is no acceleration perpendicular to the plane of motion.

Hence, R₂ = Mg = 1g

Frictional force, F_f = μR₂ = 0.2 * 1g = 0.2g

Taking forces along the plane of motion and writing Newton’s second law,

We get, T₂ – T₁ – 0.2g = 1a ------------------- (2)

From F.B.D. of block C:

0.5g – T₂ = 0.5a ---------------- (3)

Solving equation 1, 2 and 3

We get, a = 0.4 m/s²; T₁ = 2.4 N; T₂ = 4.8 N

Therefore, (a) the acceleration of the 1.0 kg blocks is 0.4 m/s²; (b) the tension in the string connecting the 1.0 kg blocks is 2.4 N and (c) the tension in the string attached to 0.50 kg is 4.8 N.

12. If the tension in the string in figure (6-E3) is 16 N and the acceleration of each block is 0.5 m/s², find the friction coefficients at the two contacts with the blocks. 

Sol:                                       

Assumption: all the pulleys are smooth and massless and string is light. So, tension in the string is same at every points.

Given: Tension in the string, T = 16 N; acceleration of each block, a = 0.5 m/s²; g = 10 m/s².

From F.B.D. of block A:

There is no acceleration perpendicular to the plane of motion.

Hence, R₁ = M_Ag = 2g

Frictional force, F_f = μ₁R₁ = μ₁ * 2g = 2μ₁g 

Taking forces along the plane of motion and writing Newton’s second law,

We get, T – μ₁R₁ = M_Aa

Or, 16 - 2μ₁g = 2 * 0.5

Or, μ₁ = 0.75      

From F.B.D. of block B:

There is no acceleration perpendicular to the plane of motion.

Hence, R₂ = M_Bg cos 30⁰ = 2√3g

Frictional force, F_f = μ₂R₂ = μ₂ * 2√3g = 2√3μ₂g 

Taking forces along the plane of motion and writing Newton’s second law,

We get, M_Bg sin 30⁰ – T – μ₂R₂ = M_Ba

Or, 4*g* ½ – 16 – 2√3μ₂g = 4 * 0.5

Or, μ₂ = 0.057 ~ 0.06

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

13. The friction coefficient between the table and the block shown in figure (6-E4) is 0.2. Find the tensions in the two strings.

Sol:

Given: μ = 0.2; M_A = 15 kg; M_B = 5 kg; M_C = 5 kg.

The forces acting on the blocks are shown in the figure and let the acceleration of each block be ‘a’.

From F.B.D. of block A: acceleration = a

The equation of motion of the block A is

                     M_Ag – T₁ = M_Aa

Or, 15g – T₁ = 15a -------------------- (1)

From F.B.D. of block B:

There is no acceleration perpendicular to the plane of motion.

Hence, R = M_Bg = 5g                       

Frictional force, F_f = μR = 0.2 * 5g = g 

Taking forces along the plane of motion and writing Newton’s second law,

We get, T₁ – T₂ – μR = M_Ba

Or, T₁ – T₂ – g = 5a ------------------ (2)

From F.B.D. of block C: acceleration = a

The equation of motion of the block C is

                    T₂ – M_Cg = M_Ca

 Or, T₂ – 5g = 5a --------------------- (3)

Solving equation 1, 2 and 3

We get, a = 9g/25, T₁ = 96 N, and T₂ = 68 N

Therefore, the tensions in the two strings are T₁ = 96 N and T₂ = 68 N.

14. The friction coefficient between a road and the tyre of a vehicle is 4/3. Find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36 km/hr is stopped within 5 m.

Sol:

Given: μ = 4/3; initial velocity, u = 36 km/hr = 10 m/s; final velocity, v = 0; distance travelled, s = 5 m.

We know, 2as = v² – u²

Or, 2 * a * 5 = 0² – 10²

Or, a = – 10 m/s² 

Let, θ be the maximum angle and the forces acting on the block are shown in the figure.

From F.B.D. of block:

There is no acceleration perpendicular to the plane of motion.

Hence, R = Mg cos θ

Frictional force, F_f = μR = 4/3 Mg cos θ

Taking forces along the plane of motion and writing Newton’s second law,

We get, Ma = μR – Mg sin θ

Or, M * 10 = 4/3 M*10* cos θ – M*10* sin θ

Or, 3 = 4 cos θ – 3 sin θ

Or, 4√ (1 – sin² θ) = 3 sin θ + 3

Squaring on both sides

We get, 16 (1 – sin² θ) = 9 sin² θ + 18 sin θ + 9

Or, 25 sin² θ + 18 sin θ – 7 = 0

Solving the equation we get,

Sin θ = (– 18 ± 32)/50 = 14/50 = 0.28 [Taking +ve sign only]

Or, θ = sin^-1 0.28 = 16⁰

Therefore, the maximum incline of the road is 16⁰.

15. The friction coefficient between an athelete's shoes and the ground is 0.90. Suppose a superman wears these shoes and races for 50 m. There is no upper limit on his capacity of running at high speeds, (a) Find the minimum time that he will have to take in completing the 50 m starting from rest, (b) Suppose he takes exactly this minimum time to complete the 50 m, what minimum time will he take to stop?

Sol:  

Given: μ = 0.90; distance travelled, s = 50 m; initial speed, u = 0; g = 10 m/s².

To reach in minimum time, he has to move with maximum possible acceleration.

Let, the maximum acceleration is ‘a’.

The equation of motion of the athelete:

There is no acceleration perpendicular to the plane of motion.

So, R = mg, and

Taking forces along the plane of motion and writing Newton’s second law,

We get, ma = μR = μmg → a = μg = 0.9 * 10 = 9 m/s².

We know, s = ut + ½ at²

Or, 50 = 0 * t + ½ * 9 * t²

Or, t = 10/3 s

(a) The minimum time that he will have to take in completing the 50 m starting from rest is 10/3 s.

After moving 50 m, velocity of the athelete is given by

→ v² = u² + 2as or, v = 30 m/s.

To stop in minimum time, deceleration should be maximum.

So, a = – 9 m/s²

Now, u = 30 m/s; v = 0; a = – 9 m/s².

We know, v – u = at

Or, 0 – 30 = – 9 * t → t = 10/3 s.

(b) The minimum time will he take to stop is 10/3 s.

16. A car is going at a speed of 21.6 km/hr when it encounters a 12.8 m long slope of angle 30° (figure 6-E5). The friction coefficient between the road and the tyre is 1/2√3. Show that no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than 36 km/hr. Take g = 10 m/s².

Sol:

Given: initial velocity, u = 21.6 km/hr = 6 m/s; μ = 1/2√3; g = 10 m/s²; s = 12.8 m.

There is no acceleration perpendicular to the plane of motion.

So, R = mg cos 30⁰, and

Taking forces along the plane of motion and writing Newton’s second law,

We get, ma = mg sin 30⁰ - μR = mg sin 30⁰ - μ mg cos 30⁰

Or, a = 10 * ½ – 1/2√3 * 10 * √3/2 = 2.5 m/s².

We know, v² – u² = 2as

Or, v = √ (6² + 2 * 2.5 * 12.8)

Or, v = 10 m/s = 36 km/hr (proved).

17. A car starts from rest on a half kilometer long bridge. The coefficient of friction between the tyre and the road is 1.0. Show that one cannot drive through the bridge in less than 10 s.

Sol:

Given: initial speed, u = 0; μ = 1.0; distance travelled, s = 500 m.

To reach in minimum time, he has to move with maximum possible acceleration.

Let, the maximum acceleration is ‘a’.

The equation of motion of the car:

There is no acceleration perpendicular to the plane of motion.

So, R = mg, and

Taking forces along the plane of motion and writing Newton’s second law,

We get, ma = μR = μmg 

Or, a = μg = 1.0 * 10 = 10 m/s².

We know, s = ut + ½ at²

Or, 500 = 0 * t + ½ * 10 * t²

Or, t = 10 s

Therefore, the minimum time required to cross the bridge is 10 s. So one cannot drive through the bridge in less than 10 s.

18. Figure (6-E6) shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg and the incline is μ₁, and that between the block of mass 4.0 kg and the incline is μ₂. Calculate the acceleration of the 2.0 kg block if (a) μ₁ = 0.20 and μ₂ = 0.30, (b) μ₁ = 0.30 and μ₂ = 0.20. Take g = 10 m/s². 

Sol:

The equation of motion of the 4 kg block:

There is no acceleration perpendicular to the plane of motion.

So, R₂ = 4g cos 30⁰ = 2√3g, and

Taking forces along the plane of motion and writing Newton’s second law,

We get, 4a = 4g sin 30⁰ + R – μ₂R₂

Or, 4a = 2g + R – 2√3μ₂g ------------------- (1)

The equation of motion of the 2 kg block:

There is no acceleration perpendicular to the plane of motion.

So, R₁ = 2g cos 30⁰ = √3g, and

Taking forces along the plane of motion and writing Newton’s second law,

We get, 2a = 2g sin 30⁰ – R – μ₁R₁

Or, 2a = g – R – √3μ₁g ------------------- (2)

Solving equation (1) and (2)

We get, a = {3g – √3g (2μ₂ + μ₁)}/6

(a) μ₁ = 0.20 and μ₂ = 0.30

Then, the acceleration of the 2.0 kg block is = a = 2.7 m/s².

(b) μ₁ = 0.30 and μ₂ = 0.20

Then, the acceleration of the 2.0 kg block is = a = 2.4 m/s².

19. Two masses M₁ and M₂ are connected by a light rod and the system is slipping down a rough incline of angle θ with the horizontal. The friction coefficient at both the contacts is μ. Find the acceleration of the system and the force by the rod on one of the blocks.

Sol:

The equation of motion of the M₁ kg block:

There is no acceleration perpendicular to the plane of motion.

So, R₁ = M₁g cos θ, and

Taking forces along the plane of motion and writing Newton’s second law,

We get, M₁a = M₁g sin θ – R – μR₁

Or, M₁a = M₁g sin θ – R – μ M₁g cos θ ------- (1)

The equation of motion of the M₂ kg block:

There is no acceleration perpendicular to the plane of motion.

So, R₂ = M₂g cos θ, and

Taking forces along the plane of motion and writing Newton’s second law,

We get, M₂a = M₂g sin θ + R – μR₂

Or, M₂a = M₂g sin θ + R – μ M₂g cos θ ------- (2)

Solving equation (1) and (2)

We get, a = g (sin θ – μ cos θ) and R = 0

So, the acceleration of the system and the force by the rod on one of the blocks g (sin θ – μ cos θ) and 0 respectively. 

20. A block of mass M is kept on a rough horizontal surface. The coefficient of static friction between the block aid the surface is μ. The block is to be pulled by applying a force to it. What minimum force is needed to slide block? In which direction should this force act?

Sol:

Let the minimum applied force be F.

The equation of motion of the M kg block:

There is no acceleration perpendicular to the plane of motion.

So, R = Mg – F sin θ, and

Taking forces along the plane of motion and writing Newton’s second law,

We get, Ma = F cos θ – μR

Or, Ma = F cos θ – μ (Mg – F sin θ)

For minimum force acceleration should be zero.

So, F cos θ – μ (Mg – F sin θ) = 0

Or, F = μ Mg/ (cos θ + μ sin θ)

Applied force F should be minimum, when μ sin θ + cos θ is maximum.

Again, μ sin θ + cos θ is maximum when its derivative is zero. 

Therefore, d/dθ (μ sin θ + cos θ) = 0

Or, θ = tan^-1 μ and

F = μ Mg/√ (1 + μ²)

Therefore, the minimum force is needed to slide block is F = μMg/√ (1 + μ²) at an angle tan^-1 μ with horizontal.

21. The friction coefficient between the board and the floor shown in figure (6-E7) is μ. Find the maximum force that the man can exert on the rope so that the board does not slip on the floor.

Sol: 

Let, the max force exerted by the man is T and the board does not slip on the floor. So acceleration a = 0.

Writing the equilibrium equation of the broad

We get, R₁ – R₂ – mg = 0 -------------- (1)

And T – μR₁ = 0 ------------------ (2)

Writing the equilibrium equation of the man

We get, R₂ + T – Mg = 0 ---------------- (3)

Solving equation (1), (2), and (3)

We get, T = μg (M + m)/ (1 + μ).  

22. A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. Find the acceleration of the two blocks if a horizontal force of 12 N is applied to (a) the upper block, (b) the lower block. Take g = 10 m/s².

Sol:

Applied force, F = 10 N, g = 10 m/s².

  

(a)  

Maximum frictional force, f_max = μR₁

= 0.2 * 2 * 10 = 4 N

The maximum acceleration of 4 kg block is

= f_max/mass = 4/4 = 1 m/s²

If the acceleration of the 2 kg block greater than or less than 1 m/s² and the applied force is greater than maximum frictional force, then there will be sliding in between two blocks. Otherwise, both will move together. So f = f_max.

From the free body diagram,     

The equation of motion of block of 2 kg mass:

Ma = F – f → 2a = 12 – 4

Or, a_{2 kg} = 4 m/s². And

The equation of motion of block of 4 kg mass:

Ma = f → 4a = 4

Or, a_{4 kg} = 1 m/s².

(b)

Maximum frictional force, f_max = μR₁

= 0.2 * 2 * 10 = 4 N

The maximum acceleration of 2 kg block is

= f_max/mass = 4/2 = 2 m/s²

If the acceleration of the 4 kg block greater than or less than 2 m/s² and the applied force is greater than maximum frictional force, then there will be sliding in between two blocks. Otherwise, both will move together. So f = f_max.

From the free body diagram,

The equation of motion of block of 2 kg mass:

Ma = f → 2a = 4

Or, a_{2 kg} = 2 m/s². And

The equation of motion of block of 4 kg mass:

Ma = 12 – f → 4a = 12 – 4

Or, a_{4 kg} = 2 m/s².

23. Find the accelerations a₁, a₂, a₃ of the three blocks shown in figure (6-E8) 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.

Sol: 

Applied force, F = 10 N, g = 10 m/s². f₁ R₂ a₁

(a) Horizontal force of 10 N is applied on 2 kg block

The equation of motion of 2 kg block:

R₁ = 2g = 20 N

And 10 – f₁ = 2a₁ 

Or, a₁ = (10 - 4)/2 = 3 m/s²   [f₁ = μ₁R₁ = 4]

Maximum frictional force between 2 kg and 3 kg block, f_{1, max} = μ₁R₁ = 4 N, and maximum frictional force between 3 kg and 7 kg block, f_{2, max}

= μ₂R₂ = 15 N.

As f_{1, max} < f_{2, max}, therefore, 3 kg block and 7 kg block move together.

And a₂ = a₃

The equation of motion of 3 kg block and 7 kg block together

→ 10 a₂ = f₁ = 4 or, a₂ = a₃ = 0.4 m/s².

(b) Horizontal force of 10 N is applied on 3 kg block:

As the sum of maximum frictional forces f_{1, max} and f_{2, max} is greater than applied force. Therefore, three blocks are move together (a₁ = a₂ = a₃ = a).

The equation of motion of three blocks together,

→12a = 10 or, a = 5/6 m/s².

(c) Horizontal force of 10 N is applied on 7 kg block:

Similarly, it can be proved that when the 10 N force is applied to the 7 kg block, all the three blocks will move together.

Again a₁ = a₂ = a₃ = (5/6) m/s².

24. The friction coefficient between the two blocks shown in figure (6-E9) 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. 

Sol: 

 (a)

In limiting case, f = f_max = μR₁ = μmg

The equation of motion of the m kg block:

There is no acceleration perpendicular to the plane of motion.

So, R₁ = mg, and

Taking forces along the plane of motion and writing Newton’s second law, in equilibrium a = 0

We get, ma = F – T – f  

Or, ma = F – T – μmg = 0

Or, F = T + μmg --------------- (1)

The equation of motion of the M kg block:

There is no acceleration perpendicular to the plane of motion.

So, R₂ = Mg – R₁ = Mg - μmg, and

Taking forces along the plane of motion and writing Newton’s second law, in equilibrium a = 0

We get, Ma = T – f = 0

Or, T = μmg

Putting the value of T in equation (1)

We get, F = 2μmg.

(b)

In limiting case, f = f_max = μR₁ = μmg and F = 4μmg

The equation of motion of the m kg block:

There is no acceleration perpendicular to the plane of motion.

So, R₁ = mg, and

Taking forces along the plane of motion and writing Newton’s second law,

We get, ma = F – T – f 

Or, ma = 4μmg – T – μmg 

Or, ma = 3μmg – T ---------- (1)

The equation of motion of the M kg block:

There is no acceleration perpendicular to the plane of motion.

So, R₂ = Mg – R₁ = Mg – μmg, and

Taking forces along the plane of motion and writing Newton’s second law,

We get, Ma = T – f 

Or, Ma = T – μmg ------------- (2)

Solving equation (1) and (2)

We get, a = 2μmg/ (M + m)

So, the accelerations of the two masses are a = 2μmg/ (M + m) in opposite direction. 

25. Suppose the entire system of the previous question is kept inside an elevator which is coming down with an acceleration a < g. Repeat parts (a) and (b).

Sol:

(a) From the free body diagram

Vertical motion of ‘m’ kg mass

→mg - R₁ = ma

Or, R₁ = mg – ma = m (g – a)

Horizontal motion of ‘m’ kg mass

→ T + f – F = 0

Or, F = T + μR₁ = T + μm (g - a) ------------ (1)

Vertical motion of ‘M’ kg mass

→ Mg + R₁ – R₂ = Ma

Or, R₂ = M (g – a) + m (g – a) = (g – a) (M + m)

Horizontal motion of ‘M’ kg mass

→ T – f = 0

Or, T = f = μR₁ = μm (g – a)

Putting the value of T in equation (1)

We get, F = μm (g – a) + μm (g – a) = 2 μm (g – a).

(b) F = 4μm (g – a)

From the free body diagram

Vertical motion of ‘m’ kg mass

→mg – R₁ = ma

Or, R₁ = mg – ma = m (g – a)

Horizontal motion of ‘m’ kg mass

→ F – T – f = ma₁

Or, ma₁ = 4μm (g – a) – T – μR₁

Or, ma₁ = 4μm (g – a) – T – μm (g – a)

Or, ma₁ = 3μm (g – a) – T ------------ (1)

Vertical motion of ‘M’ kg mass

→ Mg + R₁ – R₂ = Ma

Or, R₂ = M (g – a) + m (g – a) = (g – a) (M + m)

Horizontal motion of ‘M’ kg mass

→ T – f = Ma₁

Or, T – μm (g – a) = Ma₁ -------------- (2)

Solving equation (1) and (2)

We get, a₁ = 2μm (g – a)/ (M + m).

26. Consider the situation shown in figure (6-E9). Suppose a small electric field E exists in the space in the vertically upward direction and the upper block carries a positive 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 by the electric field E is F = QE in the direction of E.]

Sol:

From the free body diagram

The equation of motion of m kg block, a = 0

Vertical motion

→ R₁ = mg – EQ

Horizontal motion

→ F = T + f = T + μR₁

Or, F = T + μ (mg – EQ) ----------- (1)

The equation of motion of M kg block, a = 0

Vertical motion

→ R₂ = R₁ + Mg = mg – EQ + Mg

Horizontal motion

→ T = f = μR₁ = μ (mg – EQ)

Putting the value of T in equation (1)

We get, F = 2 μ (mg – EQ).

So, maximum horizontal force F can be applied without disturbing the equilibrium is 2μ (mg – EQ).

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?

Sol:

Because the block slips on the table, maximum frictional force acts on it.

From the free body diagram

Vertical motion of the block

→ R₁ = mg

Horizontal motion of the block

→ F – μR₁ = 0 → F = μR₁ = μmg

But the table is at rest. So, frictional force at the legs of the table is not μ₁R₂. Let the frictional force at the legs of the table be f, so from the free body diagram of the table

We get, f = μmg.

28. Find the acceleration of the block of mass M in the situation of figure (6-E10). The coefficient of friction between the two blocks is μ₁, and that between the bigger block and the ground is μ₂.

Sol: 

String is inextensible, so length of the string is constant.

Therefore, length of the string (L) = AB + BC + CD + ED

Or, L = x₂ + x₁ + CD + x₁ [BC = ED and x₁, x₂ changing with time]

Differentiate above equation w.r.t time twice 

We get, d²(x₂)/dt² + d²(2x₁)/dt² = 0

Or, a₂ = 2a₁

→ a₁ = a and a₂ = 2a

 The equation of motion of the m kg block:

Taking forces along the horizontal direction and writing Newton’s second law,

We get, R₁ = ma₁ = ma --------------- (1)

Taking forces along the vertical direction and writing Newton’s second law,

We get, ma₂ = mg – T – μ₁R₁

Or 2ma = mg – T – μ₁ma --------------- (2)

 The equation of motion of the M kg block:

Taking forces along the vertical direction and writing Newton’s second law,

We get, R₂ = Mg + T + μ₁R₁ = Mg + T + μ₁ma -------- (3)

Taking forces along the horizontal direction and writing Newton’s second law,

We get, ma₁ = 2T – R₁ – μ₂R₂

Or ma = 2T – ma – μ₂ (Mg + T + μ₁ma) --------------- (4)

Solving equation (2) and (4)

We get, Ma = 2(mg – μ₁ma – 2ma) – ma – μ₂ (Mg + mg – μ₁ma – 2ma + μ₁ma)

Or, Ma = 2mg – 2 μ₁ma – 5ma – μ₂Mg – μ₂mg + 2μ₂ma

Or, a = [2m – μ₂ (M + m)] g/ [M + m {5 + 2(μ₁ – μ₂)}].

29. A block of mass 2 kg is pushed against a rough vertical wall with a force of 40 N, coefficient of static friction being 0.5. Another horizontal force of 15 N, is applied on the block in a direction parallel to the wall. Will the block move? If yes, in which direction? If no, find the frictional force exerted by the wall on the block.

Sol: 

Maximum frictional force = 0.5 * 40 = 20 N.

F_R = √ (15² + 20²) = 25 N.

And, tan θ = 20/15 → 53⁰ 

So, it will move at an angle 53⁰ with the 15 N force.

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 (figure 6-E11). 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 walls with equal force, (b) Find the normal force exerted by either wall on the person. Take g = 10 m/s².

Sol:

Given: mass of the person, M = 40 kg; μ = 0.8; g = 10 m/s².

As the limiting friction acts at all the surface, so the person remains in equilibrium.

Hence, ƩF_H = 0 and ƩF_V = 0

We get, R_A = R_B = R ------------- (1)

And, f_A + f_B = Mg --------------- (2)

(a) He gives equal force on both the walls so gets equal reaction R from both the walls. If he applies unequal forces R should be different he can’t rest between the walls. From equation (1).

(b) As the reaction forces and coefficient of friction are same in both sides. So, f_A = f_B = f = μR = 0.8R

From equation (2) we can write

2f = Mg → 2*0.8*R = 40*10 → R = 250 N.

31. Figure (6-E12) shows a small block of mass m kept at the left end of a larger block of mass M and length I. 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 blocks is μ /2. Find the time elapsed before the smaller block separates from the bigger block.

Sol:

R₁ = mg and ma_m = μR₁/2 = μmg/2

→ a_m = μg/2

R₂ = Mg + R₁ = (M + m) g

Ma_M = μR₂ – μR₁/2 = μ (M + m) g – μR₁/2

→ a_M = μ (M + m/2) g/M

S₁ = vt – ½ a_mt² and S₂ = vt – ½ a_Mt²

S₁ = S₂ + l or, ½ a_Mt² – ½ a_mt² = l

Or, ½ [μ (M + m/2) g/M] t² – ½ (μg/2) t² = l
Or, t = √ [4Ml/ (M + m) μg].

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