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EXERCISES (continued)

28. Let m1 = 1 kg, m2 = 2 kg and m3 = 3 kg in figure (5-E12). Find the accelerations of m1, m2 and m3. The string from the upper pulley to m1 is 20 cm when the system is released from rest. How long will it take before m1 strikes the pulley ?

Figure for problem no. - 28

Answer: Let acceleration of m1 and moving pulley be 'a' and tension in the string attached to m1 and moving pulley be T. Taking Upward direction positive, for m1 we get,

T -m1g = m1a

→T= m1g + m1a = m1 (g + a) =1 x (g+a) N = g+a N

Let the tension in the string joining m2 and m3 be F. Considering the forces on the moving pulley, we get,

2F = T → F =T/2.

Let the acceleration of blocks m2 and m3 be a' with respect to the moving pulley. Acceleration of the block m2 with respect to the upper fixed pulley = a-a'. Similarly acceleration of the block m3 with respect to upper fixed pulley = a+a'.

Consider the forces on block m2,we get,

m2 g - T/2 = m2(a-a') , → a-a' = g - T/2m2 ------------------(A)

Consider the forces on block m3,we get,

m3 g - T/2 = m3(a+a') ,→ a+a' = g - T/2m3 ------------------(B)

Add (A) and (B),

2a = 2g-½T(1/m2+ 1/m3) = 2g -½T(½+1/3) = 2g -5T/12, ...(C)

(Multiply both side by 12)

→24a =24g -5T =24g-5(g+a) =24g-5g-5a = 19g - 5a,

→24a+5a = 19g,

→29a = 19g, → a=19g/29 m/s² . (Acceleration of m1)

From (B), a' = g-a -T/2m3 =g-a -(g+a)/6 =(5g-7a)/6

=(5g-133g/29)/6 =(145g-133g)/174 = 12g/174 m/s².

Acceleration of m2 = a-a' = 19g/29 - 12g/174 = (114g- 12g)/174

=102g/174 =17g/29 m/s².

Acceleration of m3 = a+a' = 19g/29 + 12g/174

= (114g + 12g)/174 = 126g/174 = 21g/29 m/s².

Second Part :--

Length of string of m1 (distance to be travelled) = s = 20 cm 0.20 m

Initial velocity u = 0, acceleration =a= 19g/29 m/s²,

To find time t =?,

From, s=ut+½at², → 0.20 = 0 +½ x(19g/29).t²,

19gt² = 0.2 x2 x29 = 0.4 x29,

t² = 11.6/19x9.8 = 0.623 → t ≈ 0.25 s

So m1 will hit the pulley in 0.25 s.

29. In the previous problem, suppose m2 = 2.0 kg and m3 = 3.0 kg. What should be the mass m so that it remains at rest ?

Answer: From (C),

2a = 2g -5T/12 and T =m1 (g+a), For m1 to be in rest a=0. It gives,

T=m1g. Putting these in (C) → 0=2g-5m1g/12 →5m1/12=2

m1 =24/5 = 4.8 kg

30. Calculate the tension in the string shown in figure (5-E13). The pulley and the string are light and all the surfaces are friction-less. Take g= 10 m/s².

Figure for problem no.- 30

Answer: Let the accelerations of blocks be 'a' m/s² and tension in the string be 'T' N. See the Free Body Diagram for both the blocks below (Forces are taken in the direction of acceleration only):-

FBD of blocks

For the block on the surface, we get, T=ma,

For the hanging block mg-T = ma

→ mg - T = ma (Replace ma =T)

→mg-T=T →2T=mg →T =mg/2 (given m=1 kg and g=10 m/s²)

→T =10/2 = 5 N

31. Consider the situation shown in figure (5-E14). Both the pulleys and string are light and all the surfaces are friction-less. (a) Find the acceleration of the mass M. (b) Find the tension in the string. (c) Calculate the force exerted by the clamp on the pulley A in the figure.

Figure for problem no. - 31

Answer: From the arrangement it is clear that pulley B moves half the distance the block M moves. So acceleration of B is half than M. Let acceleration of block M be 'a' and that of pulley B be a/2. If tension in the string be T, foe block M we have,

Mg-T = Ma → T = Mg - Ma --------- (A)

and for block 2M,

2T = 2M.a/2 =Ma ----------- (B)

→ 2Mg - 2Ma = Ma (Replacing T)

→ 2g-2a = a → 3a = 2g → a = 2g/3 m/s²

(a) So acceleration of block M is 2g/3 m/s²

(b) From (B), Tension in the string 2T = Ma = M . 2g/3 → T=Mg/3

(c) First calculate the force on the pulley by the strings. As shown in the figure below forces on pulley are T in horizontal and T downwards by strings.

Forces on the pulley and the clamp

Resultant = √(T²+T²) = √(2T²) = √2T = √2Mg/3. The strings exert this force on the pulley and the pulley on the clamp. As per third law of motion equal and opposite force = √2Mg/3 is applied by the clamp on the pulley. The direction of this force will be tanθ =T/T =1 → θ = 45° (From the horizontal downward by the pulley on the clamp) and 45° from the horizontal upward by the clamp on the pulley.

32. Find the acceleration of the block of mass M in the situation shown in figure (5-E15). All the surfaces are friction-less and the pulleys and string are light.

Figure for problem no. - 32

Answer: Let us first find out the tension in the string. Let the tension be T. Consider the mass 2M, it is hanging from a moving pulley and the tension in string on both side is T. So the upward force on this block is 2T and the downward force is its weight 2Mg. It is clear from the arrangement that the displacement of block 2M will always be half the displacement of block M. So acceleration of block 2M will also be half the acceleration of block M. Let the acceleration of block M be 'a' (up the plane) and that of 2M be 'a/2' (Downward). See the 'Free body diagram' of these two blocks in the figure below (Normal force on the block not shown because its component along the slope is zero):-

F.B.D showing Forces and accelerations

Net force in the direction of acceleration for the block 2M is 2Mg-2T and the acceleration is 'a/2'. It gives,

2Mg-2T = 2M. a/2 =Ma

→ 2T = 2Mg - Ma

→ T = Mg - Ma/2 ........................... (A)

Similarly net force in the direction of acceleration for block M is T-Mg.cos60° = T-Mg/2 and the acceleration is 'a'. It gives,

T-Mg/2 = Ma

→ T = Mg/2 + Ma ..............................(B)

Equating (A) and (B) we get,

Mg/2 + Ma = Mg - Ma/2

→ g/2+a = g - a/2

→ a+a/2 = g-g/2

→ 3a/2 = g/2

→ 3a = g

→ a = g/3. (Up along the plane as assumed initially)

33. Find the mass M of the hanging block in figure (5-E16) which will prevent the smaller block from slipping over the triangular block. All the surfaces are friction-less and the pulleys and string are light.

Figure for problem no.-33

Answer: Smaller block tends to slip down under the force that is the component of its weight along the plane = mg.sinθ. In order to prevent the block from slipping, this block along with the triangular block needs to be accelerated by the system so that the component of pseudo force on the block balances the component of weight which is trying to move the block. Let the required acceleration of the triangular block be 'a'. In order to apply Newton's Laws of Motion with respect to the non-inertial frame of the triangular block we add a pseudo force 'ma' on the smaller block opposite to the direction of 'a'. Component of this force along the plane is 'ma.cosθ'. See the diagram below. (Normal force on the block has zero component along the plane hence not shown in the diagram) :-

Accelerations and forces (along the plane) related to different blocks

So in order to prevent the smaller block from slipping,

ma.cosθ = mg.sinθ

→a = g.tanθ ............................... (A)

Let the tension in the string be T. Downward force on the hanging block M is 'Mg-T' and its acceleration is 'a'. It gives,

Mg-T=Ma → T = Mg-Ma. .......... (B)

Consider the forces on the triangular block along the acceleration 'a'. Only horizontal force is T which is causing the movement of both blocks having mass M' and m. It gives,

T = (M'+m)a ................................(C)

Equating (B) and (C).

Mg-Ma = (M'+m)a

→M = (M'+m)a/(g-a) {Put the value of 'a' from (A)}

→M = (M'+m)g.tanθ/(g-g.tanθ)

→M = (M'+m)tanθ/(1-tanθ)

(Divide numerator and denominator by tanθ)

→M = (M'+m)/(cotθ-1)

It is the required mass of the hanging block to prevent the smaller block from slipping over the triangular block.

34. Find the acceleration of the A and B in the three situations shown in the figure (5-E17).

Figure for problem no.-34

Answer: In each of the cases block B is hanging from a moving pulley and the systems are such that for each displacement of block A, the displacement of block B is its half. So acceleration of B is half the acceleration of A. Let the acceleration of block A be 'a' and that of block B be 'a/2' and the tension in the string is T.

Case (a)

See the F.B.D. of blocks in the diagram below:-

FBD of blocks A and B

Suppose that the acceleration 'a' of block A is downward and of block B is 'a/2' upward. We get,

For block A,

mg-T =ma →4g-T = 4a → T = 4g-4a

For block B,

2T - 5g = 5a/2

→ 2(4g-4a) -5g = 5a/2

→3g-8a = 5a/2

→6g-16a=5a

→21a=6g

→a = 6g/21 =2g/7

So acceleration of block A is 2g/7 m/s² (Downward as assumed) and that of B is g/7 m/s² (upward as assumed).

Case (b)

Let the acceleration of block A is 'a' m/s² towards right and that of block B is 'a/2' m/s² downward. See the free body diagram of the blocks below:-

FBD of block A and B case (b)

Consider block A. The only force along the direction of acceleration is force of tension T. It gives,

T = ma → T = 2a N.

Consider block B. Upward force is 2T by strings on both side of the pulley and downward force is weight 5g. Net downward force is 5g-2T that accelerates a mass of 5 kg by 'a/2' m/s². It gives,

5g-2T=5.a/2

→5g-2x2a=5a/2 (replacing T)

→5g-4a=5a/2

→10g-8a=5a

→13a=10g

→a=10g/13

So the acceleration of block A is 10g/13 m/s² towards right (as assumed earlier) and that of block B is a/2 =10g/26 =5g/13 m/s² downward as assumed.

Case (c)

Let the acceleration of block A (mass 2 kg) is 'a' m/s² downwards and that of block B (mass 1 kg) is 'a/2' m/s² upwards. See the Free Body Diagrams of both blocks below:-

FBD of bothe blocks

Consider block A. Forces on it are weight 2g downward and tension of one string T upward. Net downward force along acceleration 'a' is 2g-T. It gives,

2g-T=2a

→T=2g-2a

Consider block B. Upward force is by two strings =2T and downward force is weight 1xg = g. Net upward force is 2T-g along the direction of acceleration 'a/2'. It gives,

2T-g=1xa/2 =a/2

→4g-4a-g =a/2 (Replacing T)

→3g-4a=a/2

→6g-8a=a

→9a=6g

→a=6g/9

→a=2g/3.

So acceleration of block A is 2g/3 m/s² downward as assumed and that of block B is a/2=g/3 m/s² upward as assumed earlier.

35. Find the acceleration of the 500 g block in figure (5-E18).

Figure for problem no.-35

Answer: Let tension in the string connected to 500 g block be 'T' Newton and its acceleration is 'a' m/s² downward. Forces acting on it are its weight 0.5g N downward and 'T' N upward. See the Free Body Diagram below. Net downward force along acceleration is '0.5g-T' N. It gives,

0.5g-T = 0.5a

T = 0.5g-0.5a

FBD of three blocks

Consider the 50 g block. Let tension in the string connected to 50 g block be 'F' Newton. Its acceleration will also be 'a' m/s² but upward. Its weight will be 0.05g N downward. Net force upward along 'a' is F-0.05g. It gives,

F-0.05g = 0.05a

→F = 0.05g + 0.05a

Consider 100 g block. It will also have acceleration 'a' along the slope upward. Tension in upper string is T and lower string is F. Component of weight (0.10g N) along the slope will be 0.10gxsin30° =0.05g N downward. Normal force on the block by the plane will have zero component along the slope (Hence not shown). So net force along the direction of acceleration is 'T-F-0.05g' N. It gives,

T-F-0.05g= 0.10a

(In these three equations we have three unknowns T, F and a. We eliminate T and F in last equation by replacing them to find the value of 'a'.)

→0.5g-0.5a-.05g-0.05a-0.05g=0.1a

→0.65a=0.4g

→a=0.4g/0.65 =8g/13 m/s²

So acceleration of 500 g block will be 8g/13 m/s² downward as assumed earlier.

36. A monkey of mass 15 kg is climbing on a rope with one end fixed to the ceiling. If it wishes to go up with an acceleration of 1 m/s², how much force should it apply to the rope ? If the rope is 5 m long and the monkey starts from rest, how much time will it take to reach the ceiling ?

Answer: Let the force applied by the monkey be T. It will also be the tension in the rope. Mass of monkey = m = 15 kg and acceleration = a = 1 m/s². Weight of monkey =mg = 15g. Net upward force = 'T-15g' N. It gives,

T-15g= 15 kg x 1 m/s² =15

T=15g+15 =15 x 9.8 + 15 = 147 + 15 = 162 N

Let the time taken to reach the ceiling be 't' s. Distance s= 5 m, Initial velocity u = 0, acceleration a = 1 m/s². From the relation,

s = ut + ½at²

→5 =0 + ½.1.t²

→t² = 10

→t = √10 s .

37. A monkey is climbing on a rope that goes over a smooth light pulley and supports a block of equal mass at the other end (figure 5-E19). Show that whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equal acceleration. If initially both were at rest , their separation will not change as time passes.

Figure for problem no. - 37

Answer: The weight of the monkey is balanced by the weight of equal mass on the other side of the pulley.Whatever force the monkey exerts on the rope equal and opposite force is exerted by the rope on the monkey and this force is the tension in the rope. Tension in the rope is equal on both side of the smooth pulley. Let this tension be T and masses of monkey and the counterweight be m. If the monkey goes up that means net force T-mg is upward ie T>mg. It is also true for the counterweight for which net force T-mg is upward. So its acceleration will also be upward. For both monkey and counterweight acceleration = Force/mass,

= T-mg/m.

Same is the case if the monkey goes downward. In that case net force downward is mg-T for both, and acceleration = mg-T/m. So whatever force the monkey exerts on the rope the monkey and the block move in the same direction with the equal acceleration.

So if the monkey starts from the rest both will travel in same direction equal to a distance s = ut + ½at² =0 +½at² = ½at². So their separation will not change as the time passes.

38. The monkey B shown in figure (5-E20) is holding on to the tail of the monkey A which is climbing up a rope. The masses of the monkeys A and B are 5 kg and 2 kg respectively. If A can tolerate a tension of 30 N in the tail, what force should it apply on the rope in order to carry the monkey B with it ? Take g = 10 m/s².

Figure for problem no.-38

Answer: The force applied by the monkey A on the rope is equal to the tension in the rope. If the monkey A just holds the rope or goes up with a uniform velocity carrying monkey B the tension in the rope is equal to the total weight of both monkeys.

T = 5g+2g = 5 x 10 + 2 x 10 = 50 + 20 = 70 N

If it goes up with some acceleration the force applied by A on the rope (equal to tension T in the rope) will increase proportionately. Let us calculate maximum T. Maximum force allowed in the tail of A = 30 N. Let the acceleration of A for this tension be 'a'. Same will be the acceleration of B as he is holding the tail of A. Mass of B = 2 kg. Weight of B = 2g = 2 x 10 = 20 N. Net upward force on B =30-20 =10 N.

So acceleration of B = a = 10 N/2 kg = 5 m/s² = acceleration of A.

Consider the jointly A and B. Total mass m = 5+2 = 7 kg. Net upward force = T-7g N. It gives,

T-7g=7a → T=7 x 10 + 7 x 5 = 70 + 35 = 105 N.

So to carry the monkey B with it while going up, the monkey can apply a force between 70 N to 105 N on the rope.

39. Figure (5-E21) shows a man of mass 60 kg standing on a light weighing machine kept in a box of mass 30 kg. The box is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight shown by the machine ? What force should he exert on the rope to get his correct weight on the machine ?

Figure for problem no.-39

Answer: Let the weight of the man shown on the weighing machine be F Newtons and the tension in the rope be T Newtons throughout. Consider the box. Total downward force on the rope by the box is weight of the box plus weight of the man shown in the weighing machine (F). it gives,

T= 30g+F Newtons.

Consider the man. His actual weight 60g Newtons is reduced by the apparent F because F is also the force applied by the box on the man. So in this case tension in the rope T is balanced by 60g-F, ie.

T=60g-F Newtons.

Equating these two, we get,

30g+F=60g-F

→2F = 30g

→F = 15g N. In terms of kg the weighing machine will show the weight of the man = 15 kg.

For the weighing machine to show actual weight 60 kg of the man, let the force applied by the man on the rope be P Newtons. So new tension in the rope will be P. Due to this force let the acceleration gained by both the man and the box be 'a'. Consider the case of the man. Upward force on him is by the box the weight shown in the weighing machine equal to 60g N and the tension in the rope P while downward force is his weight 60g N. Net upward Force is

= P-60g-60g = P, it gives,

P = ma = 60 a

a = P/60 m/s²

Consider the case of the box now, Net upward force =P-30g-60g N

P-30g-60g = 30a

P = 90g + 30 xP/60 (Replace a)

→60P - 30P = 90 g x 60

→30P = 5400g

→P = 5400g/30 = 540 x 10/3 =5400/3 = 1800 N.

40. A block A can slide on a friction-less incline of angle θ and length l, kept inside an elevator going up with uniform velocity v (figure 5-E22). Find the time taken by the block to slide down the length of the incline if it is released from the top of the incline.

Figure for problem no.- 40

Answer: Let the mass of the block be M kg. Since the elevator is going up with uniform velocity no additional forces will be added to the bock and it will behave as if the elevator is at rest. The acceleration of the block will be the component of the acceleration due to gravity g along the incline = g.sinθ. Let the time taken by the block to slide down the length 'l' of incline be 't' s. From the relation s=ut+½at² . for u=0 we get,

l=0+½ g.sinθ.t²

gt². sinθ = 2l

t² = 2l/g.sinθ

t=√(2l/g.sinθ) seconds.

41. A car is speeding up on a horizontal road with an acceleration 'a'. Consider the following situations in the car. (i) A ball is suspended from the ceiling through a string and is maintaining a constant angle with the vertical. Find this angle. (ii) A block is kept on a smooth incline and does not slip on the incline. Find the angle of the incline with the horizontal.

Answer: Case (i)

Let the constant angle made by the string with the vertical be θ, mass of the ball be 'm' and tension in the string be T. We want to apply Newton's Laws of Motion with respect to the accelerating car which is a non-inertial frame. So additional pseudo force equal to 'ma' will have to be considered acting on the ball opposite to the direction of acceleration. Now forces on the ball are T, 'ma' and weight of the ball 'mg' as shown in the diagram below:-

Forces on the ball hanging inside a speeding car

As the forces on the ball are in equilibrium, equating the component of forces in vertical direction we get,

T.cosθ = mg ----------(A)

Equating the forces in horizontal direction we get,

T.sinθ = ma ----------(B)

Dividing (B) by (A),

tanθ = a/g

→ θ = tan-1(a/g).

Case (ii)

Let the angle of the incline with the horizontal be θ, mass of the block "m" and the normal force on the block by the incline N. We want to apply Newton's Laws of Motion with respect to the accelerating car which is a non-inertial frame. So additional pseudo force equal to 'ma' will have to be considered acting on the block opposite to the direction of acceleration. Now forces on the block are weight mg downward, pseudo force ma opposite to 'a' and the normal force N perpendicular to incline and upward as shown in the diagram below:-

Forces on the block resting on the smooth incline inside a speeding car

Since the block can move along the incline let us calculate the component of all the forces along the incline. Taking direction down the plane as positive, total force along the incline

= mg.sinθ - ma.cosθ (Component of N is zero as Tcos90°=0)

For the block not to slip on the incline this force should be zero. Thus,

mg.sinθ - ma.cosθ = 0

g.sinθ = a.cosθ,

→tanθ = a/g,

→ θ = tan-1(a/g).

42. A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s². Find the displacement of the block during the first 0.2 s after the start. Take g= 10m/s².

Answer: Since the elevator descends with an acceleration of 12 m/s² which is greater than the acceleration due to gravity 'g', the block kept on the floor of the elevator will separate and fall under the acceleration due to gravity. Let us calculate the displacement of the block during the first 0.2 s. Here,

u= 0 m/s, t = 0.2 s, displacement 'h' = ?

from the relation, h=ut+½gt²,

→h = 0 + ½ x 10 x (0.2)²

→h = 5x0.04 = 0.2 m = 20 cm.

===<<<O>>>===

CHAPTER-7 - Circular Motion

Questions for Short Answers

Objective - I

Objective - II

Exercises - Q 1 to Q 10

Exercises - Q 11 to Q 20

Exercises - Q 21 to Q 30

CHAPTER-8 - Work and Energy

Questions for Short Answers

Objective - I

Objective - II

Exercises - Q 1 to Q 10

Exercises - Q 11 to Q 20

Exercises - Q 21 to Q 30

Exercises - Q 31 to Q 42

Exercises - Q 43 to Q 54

Exercises - Q 55 to Q 64

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HC Verma's Concepts of Physics, Chapter-6, Friction,

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Click here for → Exercises (11-20)

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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion

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Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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HC Verma's Concepts of Physics, Chapter-4, The Forces

The Forces-

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Sunday, December 13, 2015

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HC Verma's Concepts of Physics, Chapter-6, Friction,

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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion

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HC Verma's Concepts of Physics, Chapter-4, The Forces

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HC Verma's Concepts of Physics, Chapter-3, Kinematics-Rest and Motion:---

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Chapter -2, "Vector related Problems"

Sunday, December 6, 2015

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EXERCISES (continued)

T -m1g = m1a

→T= m1g + m1a = m1 (g + a) =1 x (g+a) N = g+a N

Let the tension in the string joining m2 and m3 be F. Considering the forces on the moving pulley, we get,

2F = T → F =T/2.

Let the acceleration of blocks m2 and m3 be a' with respect to the moving pulley. Acceleration of the block m2 with respect to the upper fixed pulley = a-a'. Similarly acceleration of the block m3 with respect to upper fixed pulley = a+a'.

Consider the forces on block m2,we get,

m2 g - T/2 = m2(a-a') , → a-a' = g - T/2m2 ------------------(A)

Consider the forces on block m3,we get,

m3 g - T/2 = m3(a+a') ,→ a+a' = g - T/2m3 ------------------(B)

Add (A) and (B),

2a = 2g-½T(1/m2+ 1/m3) = 2g -½T(½+1/3) = 2g -5T/12, ...(C)

(Multiply both side by 12)

→24a =24g -5T =24g-5(g+a) =24g-5g-5a = 19g - 5a,

→24a+5a = 19g,

→29a = 19g, → a=19g/29 m/s² . (Acceleration of m1)

From (B), a' = g-a -T/2m3 =g-a -(g+a)/6 =(5g-7a)/6

=(5g-133g/29)/6 =(145g-133g)/174 = 12g/174 m/s².

Acceleration of m2 = a-a' = 19g/29 - 12g/174 = (114g- 12g)/174

=102g/174 =17g/29 m/s².

Acceleration of m3 = a+a' = 19g/29 + 12g/174

= (114g + 12g)/174 = 126g/174 = 21g/29 m/s².

Second Part :--

Length of string of m1 (distance to be travelled) = s = 20 cm 0.20 m

Initial velocity u = 0, acceleration =a= 19g/29 m/s²,

To find time t =?,

From, s=ut+½at², → 0.20 = 0 +½ x(19g/29).t²,

19gt² = 0.2 x2 x29 = 0.4 x29,

t² = 11.6/19x9.8 = 0.623 → t ≈ 0.25 s

So m1 will hit the pulley in 0.25 s.

29. In the previous problem, suppose m2 = 2.0 kg and m3 = 3.0 kg. What should be the mass m so that it remains at rest ?

Answer: From (C),

2a = 2g -5T/12 and T =m1 (g+a), For m1 to be in rest a=0. It gives,

T=m1g. Putting these in (C) → 0=2g-5m1g/12 →5m1/12=2

m1 =24/5 = 4.8 kg

30. Calculate the tension in the string shown in figure (5-E13). The pulley and the string are light and all the surfaces are friction-less. Take g= 10 m/s².

Figure for problem no.- 30

Answer: Let the accelerations of blocks be 'a' m/s² and tension in the string be 'T' N. See the Free Body Diagram for both the blocks below (Forces are taken in the direction of acceleration only):- FBD of blocks

For the block on the surface, we get, T=ma,

For the hanging block mg-T = ma

→ mg - T = ma (Replace ma =T)

→mg-T=T →2T=mg →T =mg/2 (given m=1 kg and g=10 m/s²)

→T =10/2 = 5 N

Answer: From the arrangement it is clear that pulley B moves half the distance the block M moves. So acceleration of B is half than M. Let acceleration of block M be 'a' and that of pulley B be a/2. If tension in the string be T, foe block M we have,

Mg-T = Ma → T = Mg - Ma --------- (A)

and for block 2M,

2T = 2M.a/2 =Ma ----------- (B)

→ 2Mg - 2Ma = Ma (Replacing T)

→ 2g-2a = a → 3a = 2g → a = 2g/3 m/s²

(a) So acceleration of block M is 2g/3 m/s²

(b) From (B), Tension in the string 2T = Ma = M . 2g/3 → T=Mg/3

(c) First calculate the force on the pulley by the strings. As shown in the figure below forces on pulley are T in horizontal and T downwards by strings. Forces on the pulley and the clamp

Net force in the direction of acceleration for the block 2M is 2Mg-2T and the acceleration is 'a/2'. It gives,

2Mg-2T = 2M. a/2 =Ma

→ 2T = 2Mg - Ma

→ T = Mg - Ma/2 ........................... (A)

Similarly net force in the direction of acceleration for block M is T-Mg.cos60° = T-Mg/2 and the acceleration is 'a'. It gives,

T-Mg/2 = Ma

→ T = Mg/2 + Ma ..............................(B)

Equating (A) and (B) we get,

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Answer: Let the accelerations of blocks be 'a' m/s² and tension in the string be 'T' N. See the Free Body Diagram for both the blocks below (Forces are taken in the direction of acceleration only):-

FBD of blocks

(c) First calculate the force on the pulley by the strings. As shown in the figure below forces on pulley are T in horizontal and T downwards by strings.

Forces on the pulley and the clamp

34. Find the acceleration of the A and B in the three situations shown in the figure (5-E17).

Figure for problem no.-34

See the F.B.D. of blocks in the diagram below:-

FBD of blocks A and B

Let the acceleration of block A is 'a' m/s² towards right and that of block B is 'a/2' m/s² downward. See the free body diagram of the blocks below:-

FBD of block A and B case (b)

Consider block A. The only force along the direction of acceleration is force of tension T. It gives,