Laws of Motion

5

Laws of Motion Introductory Exercise 5.1

1. N = Normal force on cylinder by plank R

θ

w

N ′ sin θ + w A = N

Force acting on cylinder f

N

R = Normal force on cylinder by ground, f = Force of friction by ground by cylinder, w = Weight of cylinder. N cos θ = f w + N sin θ = R, N ( R) = Reaction to N,

N (R) B R'

θ

Force of sphere B

wB

wall,

O (Force acting on plank)

R' w

θ P

f'

Ground

i.e.,normal force on plank by cylinder R′ = Normal force on plank by ground, w = Weight of plank, f ′ = frictional force on plank by ground. Resultant of f ′ and R′, N ( R) and w pass through point O.

N ( R) = Reaction to N i.e. normal force on sphere B by sphere A, wB = Weight of sphere B, R′, N ( R) and wB pass through point O, the centre sphere B. 3. N = Normal force on sphere by wall, C θ

N

2.

B A

A θ

O

R Force on sphere A w

N'

R

O

R′ = Normal force on sphere B by right

N(R)

Q

R = Normal force on sphere A by left wall, N = Normal force on sphere A by ground, N ′ = Normal force on sphere A by sphere B, w A = Weight of sphere A. N ′ cos θ = R

w

N

Laws of Motion w = Weight of sphere, T = Tension in string. 4. Component

7. R cos 30° + 3 = f cos 60° R

f

→ of F1

60°

30° 3N

along x-axis : 4 cos 30° = 2 3 N along y-axis : 4 sin 30° = 2 N Component

i.e.,

along y-axis : 4 sin 120° = 2 3 N

or and

→ of F3

along x-axis : 6 cos 270° = 0 N

i.e.,

along y-axis : 6 sin 270° = − 6 N

or

Component

10 N

60°

→ of F2

along x-axis : 4 cos 120° = − 2 N

Component

→ of F4

along x-axis : 4 cos 0° = 4 N along y-axis : 4 sin 0° = 0 N 5. Taking moment about point A T

O A

T sin 30°

30° AB = l

B

w

R 3 f +3= 2 2 R 3 +6= f R sin 30° + f sin 60° = 10 1 3 R +f = 10 2 2 R + f 3 = 20

( T sin 30° ) l = w

…(i)

…(ii)

Substituting the value of f from Eq. (i) in Eq. (ii) R + ( R 3 + 6) 3 = 20 4 R + 6 3 = 20 20 − 6 3 ⇒ R= = 2.4 N 4 ∴ f = (2.4) 3 + 6 = 10.16 N 8. At

point B (instantaneous acceleration only)

vertical

N

AB = l

A

l 2

T T

T=w

⇒

75

B

45°

6. See figure (answer to question no. 3)

OA OB + BC a 1 = = a+a 2

sin θ =

or or

T cos 30° = w 3 T =w 2 2 T= w 3

mg

∴

mg − T sin 45° = ma

…(i)

At point A (instantaneous horizontal acceleration only) …(ii) ∴ T cos 45° = ma Combining Eqs. (i) and (ii) mg − ma = ma g ⇒ a= 2

76 | Mechanics-1 Introductory Exercise 5.2 1. Acceleration of system

a=

( + 120) + ( − 50) 1+ 4 +2

n co v(

sta

nt)

= 10 m/s2 Let normal force between 1 kg block and 4 kg block = F1 ∴ Net force on 1 kg block = 120 − N 120 − F1 ∴ a= 1 or 10 = 120 − F1 i.e., F1 = 110 N Net force on 2 kg block = 2 × a = 2 × 10 = 20 N 2. As, 4 g sin 30° > 2 g sin 30°

The normal force between the two blocks will be zero. mg 3. N ( R) = 4

T θ mg

θ =30°

Tension in spring = mg = 1 × 10 = 10 N 5. Pseudo force ( = ma) on plumb-bob will be

as shown in figure a T φ ma θ °– 90 mg

N a

θ = 30°

ma A mg N (R) = N

∴

N=

mg 4

As lift is moving downward with acceleration a, the pseudo force on A will be ma acting in the upward direction. For the block to be at rest w.r.t. lift. N + ma = mg mg or + ma = mg 4 3 ⇒ a= g 4 4. Angle made by the string with the normal

to the ceiling = θ = 30° As the train is moving with constant velocity no pseudo force will act on the plumb-bob.

T cos φ = mg + ma cos (90° − θ) i.e., …(i) T cos φ = mg + ma sin θ and T sin φ = ma cos θ Squaring and adding Eqs. (i) and (ii), T2 = m2 g2 + m2 a2 sin2 θ + 2m2 ag sin θ …(iii) + m2 a2 cos2 θ 2 2 2 (Qθ = 30°) T = m g + m2 a2 + m2 ag g g2 g = m2 g2 + m2 + m2 ⋅ g (Q a = ) 4 2 2 7 m2 g2 = 4 mg 7 or T= 2 =5 7 N Dividing Eq. (i) by Eq. (ii), ma cos θ tan φ = mg + ma sin θ

Laws of Motion a cos θ = g + a sin θ cos θ = 2 + sin θ cos 30° = 2 + sin 30°

6.

2kg

1kg

F=8N

8 = 2 m/s 2 2+1+1 B

T1 T1

1kg T2 T2

A

Net force on 1 kg mass = 8 − T2 ∴ 8 − T2 = 1 × 2 ⇒ T2 = 6 N Net force on 1 kg block = T1 ∴ T1 = 2a = 2 × 2 = 4 N

 3 φ = tan −1    5 

i.e.,

B

a=

3 5

=

A

2kg

Introductory Exercise 5.3 1. F = 2 g sin 30° = g T1

T2

T1 T1

T2 F

T1

T1

T1

mg

T1

T2 T2

4g

T2

2g

1g

For the system to remain at rest

or

3g

T2

T2 = 2 g

…(i)

T2 + F = T1

…(ii)

T2 + g = T1

…[ii (a)]

T1 = mg

…(iii)

Substituting the values of T1 and T2 from Eqs. (iii) and (i) in Eq. [ii(a)]

∴ and

T1 = 4 g T2 = 1 g

∴

T1 =4 T2

3. 2 g − T = 2a

2g + g = mg i.e.,

77

m = 3 kg

2. As net downward force on the system is

T

zero, the system will be in equilibrium

T

T

T

a

a 1g 2g

8N

78 | Mechanics-1 T − 1 g = 1a Adding above two equations 1g = 3a g ∴ a= 3

=

u2 ( g / 3)2 g = = 2a 2g 18

Downward displacement of 2 kg block 2 1 1 g g = at2 = g ⋅   = 3 2 2 18

Velocity of 1kg block 1 section after the system is set in motion v = 0 + at g = ⋅1 3 g (upward) = 3

As the two are just equal, the string will 1 again become taut after time s. 3 …(i)

F + 1 g − T = 1a

4.

On stopping 2 kg, the block of 1kg will go upwards with retardation g. Time ( t′ ) taken by the 1 kg block to attain zero velocity will be given by the equation. g 0 =   + ( − g) t ′ 3 1 ⇒ t′ = s 3

T

T T

a

F

1g

T

a

2g

If the 2 kg block is stopped just for a moment (time being much-much less than 1 s), it will also start falling down when 3 the stopping time ends. 1 In t′  = s time upward displacement of  3  1 kg block

and

…(ii)

T − 2 g = 2a

Adding Eqs. (i) and (ii), F − 1g = 3a 20 − 10 10 ms −2 ∴ a= = 3 3

Introductory Exercise 5.4 …(i)

2T = 2 × a

1. a 2 kg

T

T

T

T

∴ Acceleration of 1 kg block 2 g 20 2a = = ms −2 3 3 Tension in the string g 10 N T= = 3 3

2T 2T T T 2a 1g

and

1 g − T = 2a

Solving Eqs. (i) and (ii), g a= 3

…(ii)

…(i)

Mg − T = Ma

2.

T T

T

T

M a

T T a mg

Laws of Motion …(ii)

T = Ma Solving Eqs. (i) and (ii) g a= 2 and T = Mg

T T M

or

Substituting value of above value of T in Eq. (i), 24 M= 5 = 4.8 kg T 4. = m1 ⋅ 2a 2

3. Block of mass M will be at rest if

T

T

T/2

Mg

a

T/2 T

2a

T

T/2 T/2

T/2

F

T/2

i.e.,

T/2 T/2

a

79

g T = 4  g +   5 24 g = 5

…(ii)

F − T = m2 a

a

or

3g

or

2g

T = Mg

…(i)

T = 4 m1a

…(i)

For the motion of block of mass 3 kg T …(ii) 3g − = 3a 2 For the motion of block of mass 2 kg T …(iii) − 2 g = 2a 2 Adding Eqs. (ii) and (iii), g = 5a g i.e., a= 5 Substituting above value of a in Eq. (iii), T = 2 ( g + a) 2

F − 4 m1a = m2 a F a= 4 m1 + m2 =

0.40 ( 4 × 0.3) + 0.2

0.40 1.4 2 = ms −2 7 =

∴

T = 4 m1a = 4 × 0.3 × 2.4 7 12 N = 35

=

2 7

80 | Mechanics-1 Introductory Exercise 5.5 1 2 at 2 Displacement of block at time t relative to car would be

1. Block on triangular block will not slip if m1a cos θ

2. (a) Using s = s0 + ut +

N

m1a N sin θ

m1 g

m1g sin θ cos

m1a sin θ

a = 5i ms–2

a = – 5i ms–2 u = 10i ms–1

θ

v=0

(at t = 0)

θ

O

T N

Car

x0

T

1 ( −5) t2 2 x = x0 + 10 t − 2.5 t2 x = x0 + 10 t +

a

M Mg

m1a cos θ = m1 g sin θ i.e., …(i) a = g tan θ N = m1 g cos θ + m1a sin θ …(ii) For the movement of triangular block …(iii) T − N sin θ = m2 a For the movement of the block of mass M …(iv) Mg − T = Ma Adding Eqs. (iii) and (iv), Mg − N sin θ = ( m2 + M ) a Substituting the value of N from Eq. (ii) in the above equation Mg − ( m1 g cos θ + m1a sin θ) sin θ = ( m2 a + Ma) i.e., M ( g − a) = m1 g cos θ sin θ + ( m2 + m1 sin2 θ) a Substituting value of a from Eq. (i) in the above equation, M (1 − tan θ) = m1 cos θ sin θ + ( m2 + m1 sin2 θ) tan θ m cos θ sin θ + ( m2 + m1 sin2 θ) tan θ ∴M = 1 (1 − tan θ) Substituting θ = 30° , m1 = 1 kg and m2 = 4 kg (cos 30° sin 30°+4 + sin2 30° ) tan 30° M= (1 − tan 30° ) 0.443 + ( 4.25) (0.577) = 0.423 = 6.82 kg

x

or

Velocity of block at time t (relative to car) will be dx v= = 10 − 5t dt (b) Time ( t) for the block to arrive at the original position (i.e., x = x0 ) relative to car x0 = x0 + 10 t − 2.5 t2 ⇒ t=4s 3. (a) In car’s frame position of object at time

t would be given by In car’s frame (at t = 0 s) O x0 z0 a = –5 ms–2

x

a = 5i ms–2 v=0

u = 10k ms–1

z

1 ( − 5) t2 2 i.e., …(i) x = x0 − 2.5 t2 and …(ii) z = z + 10 t Velocity of the object at time t would be dx …(iii) vx = = − 5t dt dz and …(iv) vz = = 10 dt (b) In ground frame the position of the object at time t would be given by x = x0 + 0 × t +

Laws of Motion

∴ Net deceleration = (3 + 5) m/s = 8 m/s 2 ∴ Displacement of object at any time t (relative to car) 1 x = x0 + 10t + ( −8) t2 2 or x = x0 + 10t − 4 t2

In ground frame O x0

At t = 0 a = 5i ms–2 v=0

z0 u = 10k ms–1

Thus, velocity of object at any time t (relative to car) dx vx = = 10 − 8 t dt

x = x0 z = z0 + 10 t

and

Velocity of the object at time t would be dx vx = =0 dt dz and vz = = 10 ms −1 dt

The object will stop moving relative to car when 10 − 8 t = 0 i.e., t = 1.25 s ∴ vx = 10 − 8 t for 0 < t < 1.25 s

4. m = 2 kg O

For block not to slide the frictional force ( f ) would be given by

u = 10i ms–1

x0

x

a = 5i ms–2 v=0

f

a = 3 ms–2

ma mg sin θ θ

θ = 37°

z

Normal force on object = mg Maximum sliding friction = µ s mg = 0.3 × 2 × 10 = 6 N 6 Deceleration due to friction = = 3 m/s2 2 Deceleration due to pseudo force = 5 m/s2

f + ma cos θ = mg sin θ f = mg sin θ − ma cos θ 3 4 = m × 10 × − m × 3 × 5 5 18 m 9 = = mg 5 25

or

AIEEE Corner Subjective Questions (Level 1)

T1 and T2 = 2 T At point Q, F2 = 1 and W = 2

1. FBD is given in the answer.

At point P, F1 =

2. FBD is given in the answer. 3. FBD is given in the answer. 4.

5. T2 F1

P

81

2

NB 45° T1

30° T1 45°

NA F2

Q

W

W

N A sin 30° = N B and N A cos 30° = W

T1 2 T1 2

82 | Mechanics-1 3T =W 2 T H= 2 Net moment about O = zero l 3T ∴. W × = ×l 2 2 100 − 40 10. (a) a = = 3 m/s2 6 + 4 + 10

6. T2

V +

45° T1

W

T2 = W and 2 T2 = T1 2

(c) N − 40 = F10 = 30 ∴ N = 70 N. F 60 11. a = = = 1 m/s2 m1 + m2 + m3 60 (a)

40 N x N 10 cm

N = 40 f = 20 f × 10 = N × x

…(i) …(ii) …(iii)

T1 = m1a = 10 N T2 − T1 = m2 a ∴ T2 − 10 = 20 ∴ T2 = 30 N (b) T1 = 0 . New acceleration 60 a′ = = 1.2 m /s2 50 T2 = m2 a ′ = 24 N 12. (a)

N

8.

T

T1

a

30°

0.1 kg

1.9 kg

2g

f

T1 − 2 g = 2a 30°

(b)

T2

W

9.

T + f cos 30° = N sin 30°

…(i)

N cos 30°+ f sin 30° = W T × R = fR

…(ii) …(iii)

V

O

3T 2 H

T/2 W

…(iii)

F6 = 18 N, F4 = 12 N and F10 = 30 N

∴ f

and

…(ii)

(b) Net force = ma

7.

W = 20 N

…(i)

2.9 kg a

0.2 kg 1.9 kg

T2 − 5 g = 5a

5g

Laws of Motion 200 − 16 g 13. (a) a = 16 (b) T1 − 11 g = 11a (c) T2 − 9 g = 9 a

20.

83

a T2 = T1 1 T1

14. If the monkey exerts a force F on the rope upwards, then same force F transfers to bananas also. If monkey releases her hold on rope both monkey and bananas fall freely under gravity.

T1

a 2

2T1

15. Tension on B = T Tension on A = 3 T 1 Now in these situations a ∝ T

T1 ar

T1 2

16. x A + xC + 2xB = constant.

3

Differentiating twice w.r.t. time we get the acceleration relation.

θ

aA

T

2T a 2

M

aA = sin θ aB

g M

a A = aB sin θ

∴

a

21.

aB θ

…(i) …(ii) …(iii)

T1 = 1a T1 − 20 = 2 ( ar − a / 2) 30 − T1 = 3( ar + a / 2)

17.

18. x + y = 6

30

2M

°

2 Mg

…(i)

T − Mg sin 30° = Ma a 2Mg − 2T = 2M ⋅ 2

…(i)

y−x=4 Solving, we get x = 1 m/s2

n si

…(ii)

…(ii)

22. T = 1a 10 − T = 1a

x

23.

ar

…(i) …(ii)

2T

2T

B

A

50

40

1 x 2a

a y 2 y 3

19.

7g − 3g = 4 m/s2 10 40 − T1 = 4 a 30 + T1 − T2 = 3 a T3 − 10 = 1a

a=

50 − 2T = 5a T − 40 = 4(2a) 24. (a) N = 40 N, µ s N = 24 N F < µsN ∴ f = 20 N and a = 0 (b) N = 20 N, µ s N = 12 N and µ k N = 8 N

…(i) …(ii)

84 | Mechanics-1 F > µsN f = µ kN = 8 N 20 − 8 a= = 6 m/s 2 2

∴ and

27. f = (06 . )(2)(10) = 12 N a2 2 kg

1 kg

µ s N = 8 N and Since, F cos 45° > µ s N ∴ and

f = µ kN = 4 N 20 − 4 a= 6 8 = m/s2 3

25. a = µ g = 3 m/s2 (a) v = at ∴ 6 = 3 t or t = 2 s 1 (b) s = at2 2 1 ∴ s = × 3 × 4 = 6 m. 2 26. f = 0.4 × 1 × 10 = 4 N a1 1 kg f a2 2 kg f

a1 =

f = 4 m/s2 1

a2 =

f = 2 m/s2 2

(a) Relative motion will stop when v1 = v2 or 2 + 4 t = 8 − 2t ∴ t = 1s (b) v1 = v2 = 2 + 4 × 1 = 6 m/s 1 (c) s1 = u1t + a1t2 2 1 s2 = u2 t − a2 t2 2

+ve

a1

(c) N = 60 − 20 = 40 N µ kN = 4 N

–ve

f f

12 = 6 m/s2 2 12 and a1 = = 12 m/s2 1 (a) Relative motion will stop when v1 = v2 or u1 + a1t = u2 + a2 t or 3 − 6t = −18 + 12t 7 ∴ t= s 6 (b) Common velocity at this instant is v1 or v2 . 1 (c) s1 = u1t + a1t2 and 2 1 s2 = u2 t + a2 t2 2 a2 =

28. N = 20 N µ s N = 16 N and µ k N = 12 N Since, W = 20 N > µ s N, friction µ k N will act. 20 − 12 ∴ a= = 4 m/s2 2 29. N = 20 N µN = 16 N Block will start moving when F = µN or 2t = 16 or t = 8 s. After 8 s 2t − 16 a= = t−8 2 i.e., a-t graph is a straight line with positive slope and negative intercept. 30. N = 60 N, µ s N = 36 N, µ k N = 24 N Block will start moving when F = µsN

Laws of Motion or ∴

(b)

4 t = 36 t=9s

85

F

v

12 N

After 9 s a=

4 t − 24 2 = t−4 6 3

52 N

F + 12 = 52 F = 40 N

31. N = mg cos θ = 30 N ∴

mg sin 30° = 30 3 N ≈ 52 N. µ s N = 18 N and µ k N = 12 N F (a)

(c)

F

a

52 N

18 N

12 N

F − 52 − 12 = 6 × 4 ∴ F = 88 N

52 N

F = 52 − 18 = 34 N. Objective Questions (Level-1) Single Correct Option

mg − F F =g− m m m A > mB ∴ a A > aB or ball A reaches earlier. 4 g − 2g g 2. a = = 6 3 g Now, 2 g − T = 2 3 4g ∴ T= = 13 N 3

mg − T = ma

1. a =

∴

Tmax m

2 mg g =g−3 = m 3 10 g − 5 g g 5. a = = 15 3 g 10 g − T = 10 × 3 20 g ∴ T= = Reading of spring 3 balance. 2mg sin 30°− mg 6. a = =0 3m ∴ T = mg

T1

3.

amin = g −

30° T2

7. a1 = g sin θ − µ g cos θ = g sin 45°−µ g cos 45° 100 N

3 T1 = 100 2 T1 = T2 2 4.

T a

mg

…(i) …(ii)

a2 = g sin θ = g sin 45° 2s 1 Now t = or t ∝ a a t1 a2 ∴ = t2 a1 g sin 45° g sin 45°− µ g cos 45° 3 Solving, we get µ = 4 or

2=

86 | Mechanics-1 8. F1 = mg sin θ + µmg cos θ

14. f = mg sin θ

F2 = mg sin θ − µmg cos θ Given that F1 = 2F2

(if block is at rest)

15. 2T cos 30° = F 60°

9. For equilibrium of block, net force from plane should be equal and opposite of weight.

T 30°

a

F

10. No solution is required. 11. Angle of repose θ = tan −1(µ ) = 30°

∴

θ h

F 3 T cos 60° a= m F = 2 3m

T=

F

16. ∴

3  h = R − R cos θ = 1 − R 2  

θ

12. Net pulling force = 15 g − 5 g = 10 g = F Net retarding force = (0.2)(5g) = g = f F−f 9 ∴ a= = g 25 25 9 T1 − 5 g = 5a = g 5 34 ∴ T1 = g 5 27 15 g − T2 = 15a = g 5 48 g ∴ T2 = 5 T1 17 ∴ = T2 24 13.

N = mg − F sin θ µN = (tan φ) N = (tan φ)( mg − F sin θ) F cos θ = µN 17. µmg = 0.2 × 4 × 10 = 8 N At t = 2 s, F = 4 N Since F < µmg ∴ Force of friction f = F = 4 N 18. a1 = g sin θ a2 = g sin θ − µg cos θ t=

t1 = t2

a

1 = 2 θ

v

Relative to lift, ar = ( g + a) sin θ along the plane. 2s Now, t = ar 2L = ( g + a) sin θ

25 1 or t ∝ a a a2 a1

g sin θ − µg cos θ g sin θ

19. Net accelaration of man relative to ground = a + a = 2a T − mg = m (2a) ∴ T = m( g + 2a) 20.

a=0 3 F = (50 + 25) g = 75 g ∴ F = 25 g = 250 N

Laws of Motion 21.

A

N1

1 + µ2 θ

W

B

3m

O

1

µN2

or

22.

tan θ = µ

=

1 + µ2 µ

mg

∴ f = 30 N 30 N > µmg ∴ Normal reaction N = mg = 40 N

N

N2 = W = 250 fmax = µN2 = 75 N

1 Mg g 2 = M 2

Mg −

(always)

∴ Net contact force = (30)2 + ( 40)2 = 50 N 26.

2 kg

a

0.2

8 kg

25 N

F f

f

∴

1 + µ2

Since

θ f

T−

µ2

+

25. µmg = 32 N

N2

∴

1

1 + µ2 1 = mg 2

N1

W

µmg

Fmin =

N1 = µ N2 N2 = W Net moment about B should be zero. 3 ∴ W × = N1 × 4 2

23. a =

µ

N2

4m

87

1 M M Mg × ×g= ×a= 2 2 2 4 Mg T= 2 F

24. θ

N = mg − F sin θ F cos θ = µ N = µ ( mg − F sin θ) µ mg F= cos θ + µ sin θ dF For F to be minimum, =0 dθ

0.5

fmax = (0. 5)(8 + 2)(10) = 50 N > 25 N ∴ Blocks will not move and therefore force of friction between two blocks = 0. 3 27. mg sin θ = 10 × 10 × = 60 N 5 This 60 N > 30 N ∴ Force of friction is upwards. Net contact force is resultant of friction and normal reaction. 28. In first figure T = F and in second figure, T = 2F . mg  29. F = 4   → upwards  2  W = 2mg → downwards F =W ∴ a=0

88 | Mechanics-1 30.

t=

∴

m m

2 µm g a

31. a1 = g sin 30° = g/2

F = at

a2 = g sin 60° =

Maximum acceleration of upper block due to friction µmg amax = =µ g m F at = =µ g 2m 2m

3 g 2 →

→

Angle between a 1 and a 2 is 30°. ∴ →

→

ar = |a 1 − a 2|=

( g/2)2 + ( 3 g/2)2 − 2 ( g/2) ( 3 g/2) cos 30°

= g/2 JEE Corner Assertion and Reason

1. Even if net force = 0, rotational motion can take place. 2. No solution is required. 3. a1 = g sin θ + µ g cos θ a2 = g sin θ − µ g cos θ a1 1 + µ as θ = 45° and sin θ = cos θ = a2 1 − µ 1 we have, 3 a1 1 + 1 / 3 2 = = a2 1 − 1 / 3 1

Substituting µ =

∴

T = m1( g − a) 2m1m2 g = ( m1 + m2 )

Mathematically, we can prove that m2 g < T < m1 g Similarly if m2 > m1, then we can prove that, m1 g < T < m2 g

4. There is no force for providing ( − 2 i$) m/s2 to the block. 5. If we increase F1, maximum value of friction will increase. But if we increase F2 friction acting on the block will increase. 6. No solution is required. ( m1 − m2 ) g 7. a = m1 + m2

m1 g − T = m1a

(if m1 > m2 )

8. If accelerations of both the frames are same then one frame as observed from other frame will be inertial. Further, a frame moving with constant velocity is inertial. 9. No solution is required. 10. No solution is required. 11. Force of friction is in the direction of motion.

Laws of Motion

89

Objective Questions (Level 2) Single Correct Option 3. N sin θ will accelerate B towards left.

1. Tmax = Fmax + fmax + 2T ′ max A T

T

B

T'

T'

f

T T

N

2T

N sin θ

F

T

90° – θ ma (pseudo force)

A

T' T' θ

mg co

N

T mg

= µ ( m A + mB ) g + µ m A g + 2µm A g = µ ( 4m A + mB ) g But, Tmax = M max ⋅ g ∴ M max ⋅ g = µ ( 4 m A + mB ) g i.e., M max = µ ( 4 m A + mB ) = 0.3 × [( 4 × 100) + 70] = 81 kg Option (c) is correct.

Option (b) is correct. 4. If a ≤ µg

v = 8 t ^i − 2 t2 ^j

m2

m1

a

→

→

a=

∴

θ

Let a be the acceleration of B. Due to acceleration of B towards left pseudo force equal to ma will act on block, toward right. Thus, N + ma sin θ = mg cos θ ⇒ N = mg cos θ − ma sin θ

→

2.

sθ

B

dv dt

m2a

^

m1a

^

= 8 i − 4t j y

m1

2i 4j → g

x → a

∴ Force (pseudo) on sphere →

Fs = 1 ⋅ ( − 8 ^i + 4 ^j ) = − 8 ^i + 4 ^j Gravitational force on sphere = − mg ^j = − 10 ^j ∴ Net force on sphere →

F = − 8 ^i + 4 ^j − 10 ^j = − 8 ^i − 6 ^j →

|F |= 10 N Option (b) is correct.

N=0 f2

f1

N=0

→ a

m2

When a is f1

N

f2

Zero µg 2

Zero µg m1 2

Zero Zero

Zero µg m2 2

µg

m1 µg

Zero

m2 µg

(Not yet attained limiting value which is 2 µm1 g)

(Attained limiting value)

Now, if a > µg, the block of mass m2 would be just greater like to move towards left than µg as a then (max) m2 a > f2 While the block of mass m1 will remain at rest as then f1 has margin to increase. Thus, m1 will apply normal force ( N ) or m2 and so will do m1 on m2 to stop its motion (towards left). Option (d) is correct.

90 | Mechanics-1 Fig. 3

5. N + ma sin θ = mg cos θ N

T a3

ma sin θ ma (pseudo force)

a

in gs

m1

mg sin θ mg

The block will fall freely if N = 0 i.e., ma sin θ = mg cos θ ⇒ a = g cos θ

and ∴

Option (c) is correct. 6. Speed of A w.r.t. C = v1

A

Speed of C w.r.t. ground = v0 ∴ Speed of A w.r.t. ground = v1 − v0 Now, speed of B w.r.t. ground = speed of A is w.r.t. ground ∴ v2 = v1 − v0 ⇒ v1 − v2 = v0 Option (a) is correct.

C

B

7. Fig. 1 m1

T

T T T

T

T

T

θ θ

a3

m2g

m2 g − T = m2 a3 T − m1 g sin θ = m1a3 m − m1 sin θ a3 = 2 g m1 + m2

Substituting m1 = 4 kg, m2 = 3 kg and θ = 30° 3 a1 = g 4+3 3 = g 7 4−3 a2 = g 4+3 1 = g 7 1 3 − 42 ⋅ 2g a3 = 4+3 1 = g 7 ∴ a1 > a2 = 3 Option (b) is correct. 8. F − fmax − T = ma (For lower block)

m2g

and

T

T

⇒

T

T T

T mg 2 mg 1

F

T

T − fmax = ma (For upper block) ⇒ F − 2 fmax = 2ma or F − 2 µmg = 2ma F or a= −µ g 2m Option (c) is correct.

Fig. 2 and

f

f

m2 g − T = m2 a1 m2 a1 = g m1 + m2 m1 g − T = m2 a2 T − m2 g = m1a2 m − m2 a2 = 1 g m1 + m2

T

a2

a1

9. a2 cos(90° − θ) = a1

i.e., a2 sin θ = a1 [θ < 0°] Option (b) is correct.

θ

a2 θ

Laws of Motion 10. Had θ been 90°.

⇒ ∴

v1 v1 90° – θ θ m

v2

B w

T = m ( g + a) M−m   = m g + g M + m   M − m  = mg 1 + M + m  

ma cos θ

12. At the position of maximum deflection the

net acceleration of bob (towards its mean position will be) g sin θ − a cos θ as explained in figure.

θ 10° 20°

F (force of friction) mg sin 10° = 0.174 mg mg sin 20° = 0.342 mg

30° 40°

µmg cos 30° = 0.500 mg µmg cos 40° = 0.442 mg

60° 90°

µmg cos 60° = 0.287 mg Zero

F

θ a cos θ O a 90°– θ

T − mg = ma

T m

T

φ = 30°

O

ma cos θ > mg sin θ i.e., a > g tan θ ∴ amin = g tan θ [cylinder will be at the point of rising up the inclined plane]

13. Mg − T = Ma

T

T

For θ ≥ φ frictional force ( F ) = µ mg cos θ [φ = angle of repose] For example 1 1 Let µ = ∴ tan φ = 3 3

inclined plane if

g

2T 2T

14. For 0° < θ ≤ φ frictional force ( F ) = mg sin θ

11. The cylinder will start rising up the

ma (Pseudo furce)

91

mg = mg [1 + 1] {as m ( f1) max + ( f2 ) max , the system will remain at rest and the values frictional forces on the blocks will be given T = 4 + f1 and T = 15 − f2 4 + f1 = 15 − f2

…(i)

f1 + f2 = 11 N

→

|T1|= |T2|= T → mg mg T1 = (cos 30° ) ^i + (sin 30° ) ^j 3 3 mg 3 ^ mg ^ = i+ j 6 6 → mg ^ |T2|= j 3 Force by clamp on pulley P y → T2

Let direction being + ive for Eq. (i)

→ T1

Option (a) f1 = − 4 N, f2 = = 5 N ⇒ f1 + f2 = 1 N

wrong

60° x

Option (b) f1 = − 2 N, f2 = + 5 N ⇒ f1 + f2 = 3 N

wrong

Option (c) f1 = 0 N,f2 = + 10 N ⇒ f1 + f2 = 10 N

wrong

Option (d) f1 = + 1 N, f2 = + 10 N ⇒ f1 + f2 = 11 N correct. OR As the likely movement would be towards right f2 will be at its maximum. ∴

f2 = 10 N

⇒

f1 = 1 N

Option (d) is correct.

m

T

T = ma 2 mg sin α = 3 ma 2 g sin 30° g a= = 3 3 mg T= 3

⇒

( f2 ) max = 0.5 × 2 × 10

x

i

T T

93

=

→ T1

+

→ T2

mg 3 ^ mg ^ mg ^ i+ j+ j 6 6 3 mg 3 $ 3 mg $ mg = i+ j = ( 3 ^i + 3 ^j) 6 6 6 =

Option (b) is correct. 21. f1(max) = 0.3 × 4 × 10 = 12 N 4 kg

2 kg f2

µ1 = 0.6

f1

µ2 = 0.3

f1 and f2 are frictional forces. f2 (max) = 0.6 × 2 × 10 = 12 N As, f1(max) + f2 (max) < 16 N ( Fext )

16N

94 | Mechanics-1 T

2 kg

T

f2

4 kg

16N

x=

∴

f1

The system will remain at rest. For the equilibrium of 4 kg mass : …(i) ∴ 16 = T + f1 As f1 will be at its maximum value f1 = 12 N ∴ T = 16 − 12 = 4 N [from Eq. (i)] Further, for the equilibrium of 2 kg mass. T = f1 ∴ f1 = 4 N Option (c) is correct. 22. For the rotational equilibrium of rod

Option (a) is correct.

A ma mg

…(i)

25. N = ma sin θ + mg cos θ N ma sin θ ma a mg sin θ

mg cos θ

θ

Now, as the block does not slide ma cos θ = mg sin θ i.e., a = g tan θ Substituting the found value of a in Eq. (i) N = m ( g tan θ) sin θ + mg cos θ sin2 θ  = mg  θ + cos θ = mg sec θ cos  

Taking moment about O. l l R × cos θ = s sin θ 2 2 s (= ma) R (= mg) O

1 m 2

When, the block stops a = 0, the value of normal force will be N ′ = mg cos θ N ′ mg cos θ ∴ = N mg sec θ

a

θ

or mg cos θ = ma sin θ ⇒ a = g cotθ Option (d) is correct.

Option (c) is correct. 26. For the rotational equilibrium of the block N

2

23. v = 2 t

dv d = (2 t2 ) = 4 t dt dt At t = 1 s, a = 4 ms −2 As a = µsg a 4 µs = = = 0.4 g 10 a=

∴

φ

A mg

be zero. P

R2 8m

B

1m 10 g

x 1m

∴ Taking moment about point Q (10 g) ( 4) = (80 g) ( x)

mg cos θ

θ = 45°

24. Just at the position of tipping off, R1 will

A

x

mg sin θ

Option (c) is correct.

R1

f

O

B

Taking moment about O. a Nx = f 2 a or ( mg cos θ) x = ( mg sin θ) 2 a or x = tan θ 2 x or = tan θ a/2 or or

tan φ = tan θ φ=θ

Laws of Motion | Thus, the normal force ( N ) will pass through point A.

9 ms–2

N

Option (a) is correct.

A

[Note : The cube will be just at the point of tilting (about point A). The cube will tilt if θ is made greater than 45°].

N sin θ

Moment of couple ( N , mg) = Moment of couple ( F , f )

mg F (= 3 )

f=F

a

mg

Option (b) is correct.

→

→ aA

aA = | a A |

28. Taking moment about point O. N1

→

l/4

l/6

mB a A cos θ sin θ tan θ mB a A = tan2 θ 10 × 9 = = 160 N 2 3    4

l  6

5 37° 4

31. Net downward force on ring = mg − µ ma

= m ( g − µa) T

T

N = ma

f = µN = µma

a = 4 ms–2 L = 1m

mg

ma

∴

mg

( T / 2) = ma → Box ma mg T+ = → Pendulum with 2 2

∴ ∴

respect to box ma mg 2ma + = 2 2 a= g/3

3

Option (c) is correct.

a T

→ aB

N cos θ =

mg

l l l N1  −  = N2  − 2 4 2 l l N1 = N2 4 3 ∴ N1 : N2 = 4 : 3 Option (c) is correct.

aB = | a B |

θ

N2

29.

…(ii)

a A = aB tan θ aA N sin θ = mB tan θ mB a A N= sin θ tan θ ∴ Force on rod by wedge ⇒

N (= mg)

x

N

θ = 37° B

27. For the rotational equilibrium of the cube

∴ mgx = Fa mg or mgx = ⋅a 3 a i.e., x = 3

…(i)

N sin θ = mB aB

30.

95

g eff = g − µ a 2L t= g eff =

2L g −µa

=

2×1 10 − (0.5 × 4)

=

1 = 0.5 s 2

96 | Mechanics-1 32. The direction of the normal reactions

33. T − Mg = Ma

between any one hemisphere and the sphere will be along the centres of the two. The three centres of the hemisphere and that of sphere will form a tetrehadron of edge equal to 2R. In figure, C1, C2 and C3 are the centres of the hemispheres and C is the centre of the sphere C1C2 = C2C3 = C3C1 = C1C = C2C = C3C = 2R N

φ

θ

C

O

C3

C2

∠COC2 = 90° 2R C2O = 3 C2O cos θ = C2C 2R / 3 = 2R 1 = 3 φ is the angle which any N makes with vertical φ = 90° − θ √3 1 sin φ = cos θ = 1 3 φ 2 ∴ cos φ = 2 3 For the vertical equilibrium of the sphere. 3 N cos φ = mg 2 or 3N × = mg 3 mg or N= 2 3 Option (b) is correct.

( 500 + 80n )g

i.e.,

N N

C1

T

T = M ( g + a)

∴

2 × 104 ≥ M ( g + a)

or

2 × 104 ≥ (500 + 80n)(10 + 2)

or

14.58 ≥ n

or

n = 14

Option (b) is correct. [Note : Tension in lift cable will increase when the lift is accelerated upwards]. 34. Normal reaction between the surface and

the particle will be zero throughout the motion if the path of the particle is that of a projectile motion (particle is free from surface). v sin φ

v φ

u sin θ u

v cos φ

h

+

a=–g

φ – + u cos θ

v2 = u2 + 2as (v sin φ)2 = (u sin θ)2 + 2 ( − g) h v sin φ = u2 sin2 θ − 2 gh = (20)2 (sin2 60° )2 − 2 ⋅ 10 ⋅ 5 3 =  400 ×  − 100  4 = 10 2 v cos φ = u cos θ = 20 cos 60° = 10 v sin φ 10 2 = v cos φ 10

Laws of Motion | tan φ = 2 φ = tan −1 2 Option (c) is correct. 35. Acceleration

of block B will be g throughout its motion while that of block A will increase from 0 to g and as such t A < tB Option (b) is correct.

vertical

Option (c) is correct. 38. String is winding on the motor shaft the

36. f1 (max) = 0.5 × 10 × 10 = 50 N 100 N

A

µ = 0.5f1

Velocity along PQ = 20 m/s. ∴ Velocity along PO = 20sin θ 3 = 20 × = 12 m/s 5 Velocity of sphere (along direction) 12 12 = = = 15 m/s cos θ 4 / 5

97

B

µ = 0.25f2

C

block B will move up. Further, as shaft is also moving down, B will further.

f1

2 m/s f2

Here, f1 and f2 are friction forces. As, f1 (max.) < Fext . (100 N), block A will move. f2 (max.) = 0.25 (10 + 20) 10 = 75 N As, f1 (max.), [driving force for block B] < f2 (max.), the block will not slop over block C. As, there is no friction between block C and surface below it, both the blocks B and block C will move together with acceleration f1 (max.) a= (mass of B + C) 50 = (20 + 30) ∴

= 1 ms −2 a A = 1 ms −2

B

2 m/s

Thus, Velocity of block B = Velocity of lift + Velocity of winding of string on shaft + Velocity of moving down of shaft q= 2 m/s + 2 m/s + 2 m/s = 6 m/s. Option (d) is correct. 39. F ′ ′ = F ′ cos θ

F is resultant of two equal forces F ′ and F ′  2 φ F = 2F ′ cos    2

Option (c) is correct. R R− 5 =4 37. cos θ = R 5

F φ

F' 20 m/s N O

θ

Q

P

R/5

F''

F''

= 2F ′ cos φ = 2F ′ cos (90° − θ) = 2F ′ sin θ  F′ ′  = 2  sin θ  cos θ 

5 4 θ 3

98 | Mechanics-1 = 2F ′ ′ tan θ = 2( ma ′ ′ ) tan θ 5 4 = 2 × 03. × × = 2 N 2 3

For A to remain in contact with B, B must accelerate in the − ive x-direction. → aA (new)

15 j

Option (b) is correct. φ

40. Velocity of block A = 2 v cos 37 °

bi

15 i

Let acceleration of B = − b ^i

2v

Due to this pseudo force with act on A in the + ive direction. 15 tan φ = 15 + b 3 15 or = 4 15 + b

37° A v

B

= 2 × 10 ×

4 5

= 16 m/s Option (d) is correct. 41. As

the mass is applying maximum possible force without moving, the blocks would at the point of slipping, fA T' T'

T T

A

fC

fB

θ = 45°

As,φ < θ the block will leave contact with B. aA θ

B 2F

A N sin θ

N sin θ θ

N

Fnet on block A = Fnet on block B F − N sin θ = N sin θ − 2F 2 N sin θ = 3 F 3F N= 2 sin θ (as θ = 30°)

= 3F Option (d) is correct.

→

a A = |5 ^i + 15 ^j|

15 j

Option (d) is correct.

F

Option (d) is correct. [Note : The value fB will be 300 N and the values of T and f A will be zero] ∴

∴ Acceleration of B = − 5 ^i

N

T ′ = fC (max) = 0.5 × 60 × 10 = 300 N fB (max) = 0.3 × (60 + 60) × 10 = 360 N As, T ′ < fB (max) the value of T will be zero.

42.

45 + 3 b = 60 3 b = − 15 b= −5

43. For block A not to slide on block B

fAT T

B

or or or

44. Equation to circle is Y (x, y) B

y

φ

15 i x

O θ

B φ = 37°

A 3 tan φ = tan 37°= 4

x

Laws of Motion | 2

2

2

x + y = r (where r = OB) dx dy 2 +2 =0 dt dt dx dy =− dt dt

∴ ⇒

…(i)

l1

99

l2 2T 2T

M1

= − ( −u) = + u =u

T

T

T

T

Speed of bead B 2

 dx   dy  =   +   dt   dt  2

M3g

2

M2 g

M1 gl1 = 2Tl2  2M1M 3  M1 gl1 = 2  g l2  M2 + M 3 

2

= (u) + ( − u)

or

=u 2

M2    M2 ⋅  3 3l M1l1 = 4  1  M2 + M2    3 M1 =3 M2

Option (a) is correct. 45. At maximum acceleration value of a, the

or

block would be in a position to move upwards. ma cos θ

⇒

N

47. fmax = µ k N 2

ma ma sin θ

N√2 mg sin θ fm

mg cos θ

ax

a

Trongh

θ = 45°

f = frictional force N = ma sin θ + mg cos θ ma cos θ = fmax + mg sin θ or

ma cos θ = µ N + mg sin θ

or

ma cos θ = µ ( ma sin θ + mg cos θ) + mg sin θ (Qθ = 45° )

or

a = µ ( a + g) + g

or

a (1 − µ ) = (1 + µ ) g 1+µ a= g 1−µ

i.e.,

Block N = mg cos θ

NN

Option (b) is correct. 46. For the beam to have no tendency to

= µ k 2 mg cos θ ∴ ma = mg sin θ − 2 µ k mg cos θ i.e., a = g (sin θ − 2 µ k cos θ) Option (c) is correct. 48. f1 (max) = 0.5 × 3 × 10

= 15 N f2 (max) = 0.3 × (3 + 2) × 10 = 15 N F 3kg f1

f1 2kg

f2

f2 1 kg

B f3

rotate f3 (max) = 0.1 × (3 + 2 + 1) × 10 =6N

100 | Mechanics-1 Value of maximum frictional force is between block 1 kg and the ground. Increasing from zero when F attains 6 N, the block of mass 1 kg will be at the point of slipping over ground below it. Option (c) is correct. 49. f2 (max) = µ ( m1 + m2 ) g m1 µ1

f1 m2 f2

B

A

f1 F (= 30N)

Option (a) is correct. 50. F = µ s mg cos θ − mg sin θ

 4 3 1 = mg  −  2 3 3 2 mg = 6 Option (b) is correct.

51. µ s = 2 µ k F

a

µ2 s mg

= 0.5 (1 + 2) 10 = 15 N aS = Acceleration of both as one a A = Acceleration of A f1 (max) = µ 1m1 g = 0.2 × 1 × 10 = 2N F − f2 (max) 30 − 15 aS = = = 5 m/s2 m1 + m2 3 µ m g a A = 1 1 = µ 1 g = 0.2 × 10 = 2 m/s2 m1 As, F > f2 (max.) both will move. Further, as aS > a A both will accelerate as one unit. F − µ 2 ( m1 + m2 ) g − µ 1m1 g aB = m2

2s 2.1 2 = = s a AB 9/2 3

∴ Required time t =

in

µ SN

θ

θ = 30°

Force required downwards a=

to 0

in gs m 2

just

slide F

θ

µ KN

θ = 30°

not

Acceleration of A w.r.t. B a AB = a A − AB f _ µ 2 ( m1 + m2 ) g − µ 1m1 g = µ1 g − m2 µ 1m2 g − F + µ 2 ( m1 + m2 ) g + µ 1m1 g =− m2 F − (µ 1 + µ 2 ) ( m1 + m2 ) g =− m2 30 − (0.2 + 0.5) (1 + 2) 10 =− 2 9 2 = − m/s 2 Negative sign indicates that the direction of a AB will be opposite to that of a A .

F = µ s N − mg sin θ = µ s mg cos θ − mg sin θ F = mg sin θ − µ k N = mg sin θ − µ kmg cos θ Thus, µ s mg cos θ − mg sin θ = mg sin θ − µ kmg cos θ (µ s + µ k) mg cos θ = 2mg sin θ µ + µ s  = 2 tan θ  s   2 3 2 4 ⇒µs = µs = 2 3 3 3 Option (a) is correct. F + mg sin θ − µ k mg cos θ 52. a = m mg mg 2 3 + − ⋅ mg 2 3 3 2 = 6 m g = 3 Option (d) is correct.

block

Laws of Motion | 53. Minimum force required to start the

For retarted motion

motion upward = mg sin θ + µ k mg cos θ 1 4 3 = mg  + ⋅  2 3 3 2  7 = mg 6

02 = v2max − 2a ′ s2 v2 s2 = max 2a ′ v2max  1 1  s1 + s2 = + 2  a a ′  v2  8 6  25 = max  +  2  g g

⇒

54. Minimum force required to move the block

up the incline with constant speed = mg sin θ + µ k mg cos θ 1 2 3 = mg  + ⋅  2 3 3 2   5 = mg 6 (5.22)2 55. S1 = = 14.3 m 9.8  2 ⋅    8 

vmax =

50 × 9.8 = 5.92 ms −1 14

Option (c) is correct. 57. mg sin θ = 170 × 10 ×

8 = 906.67 N 15 µ = 0.4 T

µ = 0.2 mg

Option (c) is correct. 1200 g − 1000 g 56. a′ = 1200 g = 6

101

sin

θ

si mg

T

A

B

f2

nθ

f1

θ

15 = 300 N 17 15 f2 (max) = 0.4 × 170 × 10 × 17 = 600 N f1 (max) = 0.2 × 170 × 10 ×

17 T' = 1000g

θ

– a'

a

Speed = vmax T = 1350g

15 25 m

1200 g

1200 g

1350 g − 1200 g 1200 g = 8

a=

For accelerated motion

⇒

8

Stops

v2max = 02 + 2as1 v2 s1 = max 2a

The whole system will accelerate as mg sin θ is greater than both f1 (max) and f2 (max). Total force of friction = f1 (max) + f2 (max) = 900 N Option (a) is correct. 58. mg sin θ − 300 − T = ma

and mg sin θ + T = ma

…(i) …(ii)

Substituting Eq. (i) by Eq. (ii), 2T + 300 = 0 T = − 150 N = 150 N, compressive. Option (a) is correct.

102 | Mechanics-1 More than One Correct Options 1. (a) Normal force between A and B = m2 g F 2N

A

T

T 2N

B

i.e., T =3N Option (d) is correct. 2. At point A

T T

α

= 1 × 10 = 10 N ∴ Force of limiting friction by B on A (or by A on B) = µ × 10 = 2 N

T1 T1 A T2

β

Total force opposing applied external force F = 2N + T = 2N + 2N = 4N Thus, if F ≤ 4 N The block A will remain stationary and so block B also. The system will be in equilibrium. ∴ Option (a) is correct. (b) If F > 4 N F − T −2=1a and T − 2 = 1a Adding above equation F − 4 = 2a i.e., F = 4 + 2a

…(i)

T2 B F

mg

…(i)

T1 cos α = T2 cos β + mg and T1 sin α = T2 sin β

…(ii)

At point B T2 cos β = mg

…(iii)

T2 sin β = F − mg

…(iv)

Using Eq. (iii) in Eq. (i), T1 cos α = 2T2 cos β

…(ii)

= mg

…(v)

Dividing Eq. (ii) by Eq. (v), …(vi)

2 tan α = tan β Option (a) is correct.

For F > 4 N

Squaring and adding Eqs. (iii) and (v), 2a + 4 > 4

or

a>0

⇒

T − 2> 0

i.e., T > 2N ∴ Option (b) is incorrect. (c) Block A will move over B only when F > 4 N and then the frictional force between the blocks will be 2 N if a is just 0 [as explained in (b)]. Option (c) is correct. (d) If F = 6 N using Eq. (ii) 2a = 6 − 4 ⇒

a = 1 m/s 2

∴ Using Eq. (i), T −2=1

T12 = 4 T22 cos2 β + T22 sin2 β Dividing Eq. (iii) by Eq. (iv)

…(vii)

√2 1 β 1

∴

tan β = 1 1 cos β = = sin β 2

Substituting the values of sin β and cos β in Eq. (vii) 1 1 T12 = 4 T22   + T22    2  2 5 = T22 2

Laws of Motion | ⇒ 2T1 = 5T2 Option (c) is correct. 3. Displacement of block in 4 s v (ms–1)

φ

θ O

4

t (s)

S = Area under curve = 16 m. ∆K =Workdone by frictional force 1 × 1 × 42 = µ × 1 × 10 × 16 2 ⇒ µ = 0.1 Option (a) is correct. Option (b) is incorrect. Acceleration, a = tan φ = tan ( π − θ) = − tan θ = − 1 m/s 2 If half rough retardation = 0.5 m/s2 1 ∴ 16 = 4 t + ( − 0.5) t2 2 2 i.e., t − 16t + 64 = 0 or t=8s Option (d) is correct. Option (c) is incorrect. 4. Let acceleration of wedge ( A) = a ma sin θ N F = mg cos θ

a

ma (Pseudo force)

N sin θ F θ A

M

103

2

mg θ

N + ma sin θ = mg cos θ N = mg cos θ − ma sin θ Acceleration of N sin θ a= M or Ma = ( mg cos θ − ma sin θ ) sin θ

or a ( M + m sin θ) = mg cos θ sin θ mg cos θ sin θ i.e., a = M + m sin2 θ 0.6 × g × cos 45° sin 45° = 1.7 + (0.6 × sin2 45° ) 3g = 17 + 3 3g = 20 Let aB = Acceleration of block B Net force on B (along inclined plane) maB = ma cos θ + mg sin θ ⇒ aB = a cos θ + g sin θ Thus, ( aB ) V = ( a cos θ + g sin θ) cos θ = a cos2 θ + g sin θ cos θ 1 = ( a + g) 2 3g 1  = + g  20 2 23 g = 40 ( aB ) H = ( a cos θ + g sin θ) sin θ − a 23 g 3 g = − 40 20 17 g = 40 5. f1 (max.) = µ 1m A g T(pull) = 125 N f1

A

f1 B

f2

= 0.3 × 60 × 10 = 180 N Fnet on B = f1 (max.) + T = 180 + 125 = 305 N A will remain stationary as T < f1 (max.) ∴ f1 = 125 N Force of friction acting between A and B = 125 N ∴ Options (c) and (d) are incorrect.

104 | Mechanics-1 (a)

f2 (max) = µ 2 ( m A + mB ) g

7.5 N

= 0.2 (60 + 40) 10

17.5 N F

= 200 N

C

B

T 17.5 N

f1 + T = 125 + 125 = 250 N As, f1 + T > f2 (max. block B /along the A as A is stationary) will move towards right with acceleration. Option (a) is correct. ( f + T) − f2 (max.) aB = 1 mB + m A =

250 − 200 40 + 60

Option (b) is correct.

9. N sin θ = ma = 1 × 5 = 5 N cos θ = mg = 1 × 10 = 10

6. (See solution to Question no. 4). A

T = 17.5 + 7.5 = 25 N F = T + 37.5 + 17.5 = 80 N (c) T − 7.5 − 17.5 = 4 a F − T − 37.5 − 17.5 = 8 a F = 200 N Solving these equations we get, a = 10 m/s 2

m a

N = mg cos θ − ma sin θ Option (c) is correct and option (d) is incorrect. As angle between the directions of a and g sin θ will be less than 90°, acceleration of block A will be more than g sin θ. A

θ anet > g sin θ

Option (a) is correct and option (b) is incorrect.

…(i) …(ii)

Solving these two equations we get, 1 tan θ = and N = 5 5 N. 2 10. Let f1 = friction between 2 kg and 4 kg f2 = friction between 4 kg and ground ( fs1) max = 0.4 × 2 × 10 = 8 N ( f k1) = 0.2 × 2 × 10 = 4 N ( fs2 ) max = 06 . × 6 × 10 = 36 N Fk2 = 0.4 × 6 × 10 = 24 N (b) At t = 1 s, F = 2 N < ( fs2 ) max ∴ Both the blocks are at rest. ∴ f1 = 0

a

7. Maximum value of friction. f1 = between A and B = 0.25 × 3 × 10 = 7.5 N f2 = between B and C = 0.25 × 7 × 10 = 17.5 N f3 = between C and ground = 0.25 × 15 × 10 = 37.5 N

…(i) …(ii) …(iii)

8. Maximum value of friction available to block is less than the maximum value of friction available to man.

= 0.5 m/s 2

ma

T

37.5 N

(c) At t = 4 s, F = 8 N < ( fs2 ) max ∴ Both the blocks are at rest. f2 = F = 8 N, 11. a = 0, T1 = 10 N, T2 = 20 + T1 = 30 N, T3 = 20 N. 12. fmax = 0.3 × 2 × 10 = 6 N (a) At t = 2 s, F = 2 N ∴ f = 2N (b) At t = 8 s, F = 8 N > 6 N ∴ f =6N

Laws of Motion | (c) At t = 10 s, F = 10 N and f = 6 N 10 − 6 ∴ a= = 2 m/s2 2 (d) Block will start at 6 s. After that, net impulse 1 = × 4 × (6 + 10) + 2 × 10 − 6 × 6 2 = 16 N-s = mv 16 ∴ v= = 8 m/s. 2 13. fmax = 0.4 × 2 × 10 = 8 N (b) At t = 3 s, F = 6 N ∴ Common acceleration

6 a = = 1 m/s 2 6 ∴ Pseudo force on 2 kg =2×1=2 N

105

(backward)

14. N = Mg − F sin θ F cos θ = µN = µ ( Mg − F sin θ) µMg ∴ F= cos θ + µ sin θ For F to be minimum, dF =0 dθ

Match the Columns 1. Acceleration after t = 4 s

At t = 4 s, F = 8 N ∴ Fmax = 8 i.e., µ s mg = 8 8 8 ⇒ µs = = = 0.4 mg 2 × 10 ∴ (a) → (r) At t = 4 s,

a = 1 ms −2

t = 4 s, F =8N F − µ k N = ma F − ma F − ma i.e., µ k = = N mg 8 − (2 × 1) = = 0.3 2 × 10 (b) = (q) At t = 01 . s, F = 0 . 2 N ∴ Force of friction (at t = 01 . s) = 0.2 N (c) → (p) ∴ At t = 8 s, F = 16 N F − µ kmg ∴ a= m 16 − (0.3 × 2 × 10) = =5 2 a i.e., = 0.5 10 (d) → (s). ∴ 2. At θ = 0°, dragging force = 0

∴

Force of force = 0

∴ (a) → (s) At θ = 90° Normal force on block by plane will be zero. ∴ Force of friction = 0 ∴ (b) → (s) At θ = 30° Angle of repose = tan −1 µ = tan −1(1) = 45° As θ < angle of repose, the block will not slip and thus, force of friction = mg sin θ = 2 × 10 × sin 30° = 10 N ∴ (c) → (p) At θ = 60° As θ > angle of repose Block will accelerate and thus force of friction = µ N = 1 × 2 × 10 × cos 60° = 10 N ∴ (d) → (p). 3. All contact forces (e.g., force of friction and

normal reaction) are electromagnetic in nature. ∴ (a) → (q), (r) (b) → (q), (r).

106 | Mechanics-1 Nuclear force is the force between nucleons (neutrons and protons). Between two protons field force also acts.

5. (a) As block A is stationary, f = 10 N T

T

T

T

B

4. (a) N R − 10 = 2 ( + 5)

T T=F

C A

5 m/s2

F = 10N

f

+ ive

∴ (a) → (p) 10 N

(b) As block C is stationary force of friction between C and ground will be zero. ∴ (b) → (s) (c) Normal force ( N C ) on C from ground N C = N B + mC g = mB g + mC g = ( mB + mc ) g = (1 + 1) 10 = 20 N

NR

N R = 20 N

⇒ ∴ (a) → (f)

(b) mg ( = 20 N) > F ( = 15 N)

∴ (c) → (q). (d) As block A is stationary T = F (as shown in figure) = 10 N

f NR 20 N

∴ (a) → (p). 6. If friction force ( f ) is less than the applied

F (= 15 N)

force ( F ).

Block would be slipping in the downward direction. Force of friction will be in the upward direction. Frictional force = 20 − 15 =5N ∴ (b) → (p) (c) If F = 0, the block will slip downwards due to mg ( = 20N) Limiting friction = µ s N R = 0.4 × 20 =8N Minimum value of F for stopping the block moving down = 20 − 8 = 12 N ∴ (c) → (s). (d) F = mg + 8 = (2 × 10) + 8 = 28 N ∴ (d) → (s)

mg F

fmax = 8 N

F

Net force on body = F − f The body will be in motion and thus the friction will be kinetic. ∴ (a) → (q) (b) If friction force ( f ) is equal to the force applied, the body will be at rest. If the body is at the point of slipping the force of force will be limiting too. Emphasis is being given to the word “may be” as when a body is moving and the external force is made just equal to the frictional force, the body would still be moving with friction force at its limiting (kinetic) value. ∴ (b) → (p), (r) (c) If object is moving the friction would be kinetic as explained in (a). (c) → (q). ∴ (d) If the object is at rest, then friction may be static and limiting as explained in (b). ∴ (d) → (p), (r)

Laws of Motion | 7. (a) Normal force between A and B = m A g

f1 (max.) = µ 1m A g (towards left) f1

A

µ1

f1( R) B

3 kg

f2 C

(d) Friction force on 3 kg block due to 5 kg block = f2 (max) = 5 N, towards left. ∴ (d) → (q), (s). 8. (a) and (b)

F µ2

5 kg

T

f2(R)

= 0.2 × 2 × 10 =4N Normal force between B and C = ( m A + mB ) g f2 (max) = µ 2 ( m A + mB ) g = 0.1 (2 + 3) × 10 =5N Total friction force on 3 kg block = f1 (max) + f2 (max) = 4 + 5 = 9 N towards left ∴ (a) → (q), (s) (b) Friction force on 5 kg block = f2 ( R) = f2 (max) = 5 N, towards right ∴ (b) → (p), (s) (c) Friction force on 2 kg block due to 3 kg block = f1( R) = f1 (max) = 4 N, towards right ∴ (c) → (p), (s)

107

2 kg s mg

in

° 30

A

T µ=

√3 2

30°

µ=

T T T B 3kg

3 2

3 = 40.89 ° 2 Now as, angle of incline (30° ) < angle repose (= 40.89 °) The block A and so also B will remain stationary. ∴ (a) → (r), (b) → (r) (c) Tension ( T) in the string connecting 2 kg mass = mg sin 30° 1 = 2 × 10 × 2 = 10 N ∴ (c) → ( s) (d) Friction force on 2 kg mass = zero. ∴ (d) → (r). ∴ Angle of repose = tan −1