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For no slipping

a aA B=

F mg

m

g=

2

i.e., F mg=3

2

=2

3

F

mg

Slipping will obviously be m2there if is

greater than above mentioned value min=

2

3

F

mg

For no slipping.

Option (c) is correct.

16. Fnet (downward) = +mg masin cos

= +m g a( sin cos ) g g aeff= +sin cos Time ( )T required to cover 2L distance

along inclined would be

T L

g=

2

eff

=+

2L

g a( sin cos )

Option (c) is correct.

17. Fnet on block along incline in the upward

direction

= ma mgcos sin = m a g( cos sin )

g a geff= cos sin

Time ( )t to move s distance would begiven by

s g t=1

2

2eff

i.e., t s

g

s

a g= =

2 2

eff ( cos sin )

Substitutings= 1m, = 30 ,

a= 10 3 m/s2and g= 10 m/s2

t=

2 1

10 33

210

1

2_

=1

5s

Option (b) is correct.

18. N w

w w

+ = +2

302

sin

f=frictional force

N w

=5

4

f N w

max= = 5

4

The block will remain stationary ifw w

2 30

5

4cos

or w w

2

3

2

5

4

or 3 5

or3

5

Block will move if +( ) ( )max max1 2 , the system willremain at rest and the values frictional

forces on the blocks will be given

T f= +4 1and T f= 15 24 151 2+ = f f (i)

f f1 2 11+ = N

Let direction being + ive for Eq. (i)Option (a)f1 4= N,f2 5= = N

f f1 2 1+ = N wrong

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

f f1 2 3+ = N wrong

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

f f1 2 10+ = N wrong

Option (d)f1 1= + N,f2 10= + N

f f1 2 11+ = N correct.

ORAs the likely movement would be towardsrightf2 will be at its maximum.

f2 10= N

f1 1= N

Option (d) is correct.

20. 2 2mg T masin =

and T ma= 2 3mg masin =

a g

= 2 303

sin=

g

3

T mg=

3

Force ( )R applied by clamp on pulleywould be

| | | |T T1 2

= =T

T13

303

30

= + mg mg

(cos ) (sin )^ ^i j

= +mg mg3

6 6i j^ ^

| |^

T23

=

mgj

Force by clamp on pulley P

= + T T1 2

= + +mg mg mg3

6 6 3i j j^ ^ ^

= +mg mg3

6

3

6 i j = +

mg

63 3( )

^ ^i j

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

f1andf2 are frictional forces.

f2 2 10 12(max)= =0.6 NAs,f f1 2 16(max) (max)+ < N ( )Fext

Laws of Motion | 93

f1 = 0.61 = 0.52f2

F= 15NF = 2t1T T

15N4N T T

B A

= 30mTT

T

T

PT

T2m

60

xi

y

j

T1

x

y T2

60

f1 = 0.61 = 0.32f2

16N2 kg 4 kg

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The system will remain at rest.

For the equilibrium of 4 kg mass : 16 1= +T f (i)Asf1will be at its maximum value

f1 12= N T= 16 12

= 4N [from Eq. (i)]Further, for the equilibrium of 2 kg mass.

T f= 1 f1 4= NOption (c) is correct.

22. For the rotational equilibrium of rod

Taking moment about O.

R l

sl

=2 2

cos sin

or mg macos sin = a g= cotOption (d) is correct.

23. v t= 2 2

a dv

dt

d

dtt t= = =( )2 42

At t= 1 s, a= 4 2msAs a gs=

sa

g= = =

4

100.4

Option (c) is correct.

24. Just at the position of tipping off, R1will

be zero.

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

x=1

2m

Option (a) is correct.

25. N ma mg= +sin cos (i)

Now, as the block does not slide

ma mgcos sin =i.e., a g= tan

Substituting the found value of ain Eq. (i) N m g mg= +( tan )sin cos

= +

=mg mg

sin

coscos

2 sec

When, the block stops a= 0, the value ofnormal force will be

N mg = cos

N

N

mg

mg

=

cos sec

Option (c) is correct.

26. For the rotational equi librium of the block

Taking moment about O.

Nx fa= 2or ( cos ) ( sin )mg x mg

a =

2

or x a=

2tan

or x

a/tan

2=

or tan tan = or =

94 | Mechanics-1

O

mg

ma

s(= ma)

R(= mg) aA

8 mP

R1 R2

1 m

BA

1 m

x10 g

mgcos

a

ma

masin

N

mgsin

= 45

mg

N

mgsin

B O f

mgcos x

A

f1f2

16NT T

4 kg2 kg

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Thus, the normal force ( )N will pass

through pointA.

Option (a) is correct.

[Note : The cube will be just at the point of

tilting (about point A). The cube will tilt if ismade greater than 45].

27. For the rotational equilibrium of the cube

Moment of couple ( , )N mg

=Moment of couple ( , )F f mgx Fa=

or mgx mg

a= 3

i.e., x a=

3

Option (b) is correct.

28. Taking moment about point O.

N l l

N l l

1 22 4 2 6

=

N

l

N

l1 24 3=

N N1 2 4 3: :=Option (c) is correct.

29.

( / )T ma2 = Box

T ma mg+ =

2 2Pendulum with

respect to

box

22 2

ma ma mg+ =

a g= / 3

30. N m aB Bsin = (i)

a aA B= tan (ii) N m

aB

Asintan

=

N m aB A=sin tan

Force on rod by wedge

N m aB Acos

cos

sin tan

=

=m aB Atan2

=

=10 9

3

4

1602

N

Option (c) is correct.31. Net downward force on ring = mg ma

= m g a( )

g g aeff =

t L

g=2

eff

=2L

g a

=

2 1

10 4( )0.5

= =1

20.5 s

Laws of Motion | 95

l/6l/4

N1 N2

mg

a

TT

ma

mgmg

= 37

A

Nsin

N

NB

29 ms

aB

aA

a

a

A A

B B

==

| |

| |

a

a

L= 1m

2a= 4 ms

f = N

= ma

mg

T

N = ma

a

F( )=mg3

N(= mg)

mg

x

f = F

3

4

37

5

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32. The direction of the normal reactions

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, C C1 2, and C3are the centres of

the hemispheres and Cis the centre of the

sphere

C C C C C C1 2 2 3 3 1= = = = = =C C C C C C R1 2 3 2

= COC2 90

C O R

2

2

3=

cos =C O

C C2

2

= 2 32

RR/

=1

3

is the angle which any N makes withvertical

= 90

sin cos = =1

3

cos =2

3

For the vertical equilibrium of the sphere.

3N mgcos =

or 32

3N mg =

or N mg=

2 3

Option (b) is correct.

33. T Mg Ma =

i.e., T M g a= +( )

2 104 +M g a( )

or 2 104 + +( )( )500 80 10 2n

or 14.58n

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 u as2 2 2= +

( sin ) ( sin ) ( )v u g h = + 2 2 2

v u ghsin sin = 2 2 2

=

( ) (sin )20 60 2 10 52 2 2

=

400

3

4100

= 10 2

v ucos cos cos = = = 20 60 10

v

v

sin

cos

=10 2

10

96 | Mechanics-1

O C3

C1C

NN

N

C2

T

( 500 + 80n)g

+ ucos

+

u

h

vsin

v

vcos

a = gusin

31

2

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Laws of Motion | 97

tan = 2 = tan 1 2

Option (c) is correct.

35. Acceleration of block B will be g

throughout its motion while that of blockAwill increase from 0 togand as such

t tA B