# NEWTON’S LAWS

When a body is at rest or moving in a straight line or rotating about an axis, the body
obeys certain laws of motion. These laws are called Newton’s laws of motion. There are three laws of motion. These laws for linear motions and angular motions are given in the next articles.

5.8.1. Newton’s Laws for Linear Motion

First law. It states that a body continues in its state of rest or of uniform motion in a
straight line unless it is compelled by an external force to change that state.

Second law. It states that the rate of change of momentum ofa body is proportional to the external force applied on the body and takes place in the direction of the force.

Third law. It states that to every action, there is always an equal and opposite reaction.

Before discussing Newton’s laws of motion, let us define certain terms like mass, weight and momentum.

1. Mass. The quantity of matter contained in a body is known as mass of the body. Mass is a scalar quantity. In C.G.S. units, the mass is expressed in gram me (gm) whereas in S.I. units the mass is expressed in kilogramme (kg).

2. Weight. Weight of a body is defined as the force, by which the body is attracted
towards the centre of the earth. Mathematically weight of a body is given by

Weight = Mass × Acceleration due to gravity = Mass × g                       … (5.19)

If mass is taken in kilogram (kg) and acceleration due to gravity in metre per
second square (m/s2), then weight is expressed in newton (N). But if mass is taken in gramme (gm) and acceleration due to gravity in centimetre per second square (cm/s2), then weight is expressed in dyne. The relation between newton (N) and dyne is given as

One Newton = 105 dyne.

3. Momentum. The product of the mass of a body and its velocity is known as momentum of the body. Momentum is a vector quantity. Mathematically, momentum is given by

Momentum = Mass × Velocity.

5.8.2. Newton’s First Law of Motion. It consists of two parts. First part states that a
body continues in its state of rest unless it is compelled by an external force to change that state. A book lying on a table remains at rest, unless it is lifted by some external force.

Second part states that a body continues in its state of uniform motion in a straight line unless it is compelled by an external force to change that state. In actual practice, we see that when a body is moving with a uniform velocity in a straight line, the body does not continue in its state of uniform motion but comes to rest after some time. This is due to frictional force acting on the body. For an ideal case i.e., when there is no frictional force acting on the body), the body will continue to move with uniform velocity in a straight line, unless compelled by an external force to change that state.

5.8.3. Newton’s Second Law of Motion. This law enables us to measure a force. Let
a body of mass ‘m’ is moving with a velocity ‘u’ along a straight line. It is acted upon by a force F and the velocity of the body becomes v in time t. Then we have

u = Initial velocity of the body,

v = Final velocity of the body,

m = Mass of the body,

a = Uniform linear acceleration,

F: Force acting on the body, which changes the velocity u to v in time t,

t: Time in second to change the velocity from u to v. Initial momentum of the body

= Mass × Initial velocity = m × u

Final momentum of the body = m × v.

Change of momentum

= Final momentum – Initial momentum = mv – mu = m(v – u)

where k is a constant of proportionality.

In equation (ii), and m (mass of a body) are constants for a given body and hence force acting on a body is proportional to the acceleration produced by the force. This means that for a given body, greater force products greater acceleration while a smaller force produces smaller acceleration. The acceleration produced will be zero if no force is applied on the body.

Two important conclusions are drawn from the first two Newton’s laws of motion:

(i)                           There will be no acceleration, if no external force is applied on the body. This means the body will continue in its state of existing uniform motion in a straight line.

(ii)                        Force applied on the body is proportional to the product of mass of the body and the acceleration produced by the force.

5.8.4. Unit of Force. Let us first define a ‘unit force’. A unit force can be suitably
defined so as to make the value of kin equation (ii) equal to one, A unit force (i.e., Force = 1.0) is that which produces unit acceleration on an unit mass. Then by substituting F = 1.0, m = 1.0 and a 0: 1.0 in equation (ii) (i.e., F = k × m × a), we get

1 = h × 1 × 1 or k = 1.

Substituting the value of k = 1, in equation (ii), we get

F = m × a                                                            … (5.20)

(i) If mass (m) = 1 kg and acceleration produced (a) = 1 m/s2 the unit of force is known as newton (which is written as N). Thus newton is defined as that force which acts on a body of mass one kg and produces an acceleration of 1 m/s2 in the direction of force. Newton is the unit of force in S.L system,

1 N = 1 kg × 1 m/s2 = 1 kg-m/s2,

(ii) If mass (m) 0: 1 gm and acceleration (a) 0: 1 cm/s2, then unit of force is known as ‘dyne’. Thus a dyne may be defined as that force which acts on a body of mass one gm and produces an acceleration of 1 cm/s2. Dyne is the unit of force in C.G.S. system.

1 dyne = 1 gm × 1 cm/s2 = gm cm/s2

By definition,

1 N = 1 (kg) × 1 (m/s2) = 1 x 1000 (gm) × 1 × 100 (cm/s2) = 105 (gm cm/s2)

= 105 dyne                                 (1 dyne = 1 gm × 1 cm/s2 = gm cm/s2)

Note.

(i)                           The body will have acceleration if the external force is acting on the body in the direction of motion of the body.

(ii)                        The body will have retardation if the external force is acting opposite to the direction of motion of the body.

Problem 5.36. A force of unknown magnitude acts on a body of mass 150 kg and produces an acceleration of 3 m / s2 in the direction of force. Find the force.

Sol. Given:

Mass of the body, m = 150 kg

Acceleration, f = 3 m/s2

The force is given by equation (5.20),

Hence F = m × a = 150 (kg) × 3 (m/s2)

= 450 (kg-m/s2) = 450 N. Ans.                                         (kg-m/s2= N)

Problem 5.37. A force of 100 N acts on a body having a mass of 4 kg for 10 seconds. If the initial velocity of the body is 5 m/s, determine:

(i)                           acceleration produced in the direction of force, and

(ii)                        distance moved by the body ill 10 seconds,

Sol. Given:

Force, F = 100 N ;

Mass, m = 4 kg

Time, t = 10 second;

Initial velocity, u = 5 m/s

Let a = Acceleration produced in the direction of force

s = Distance travelled by body in 10 seconds.

(i) Using equation (5.20), we get

F=m × a    or     100 = 4 × a

Problem 5.38. The weight of a body on earth is 980 N. If the acceleration due to gravity on earth = 9.80 m/s2, what will be weight of the body on:

(i)                           the moon, where gravitational acceleration is 1.6 m/s2, and

(ii)                        the sun, where gravitational acceleration is 270 m/s2.

Sol. Given:

Weight of body on earth,   W = 980 N

Acceleration due to gravity on earth, g = 9.80 m/s2.

First calculate the mass of the body on earth. Using equation (5.19), we have

Weight = Mass × g      or       980 = Mass × 9.80

The mass of the body will remain the same on moon as well as on sun.

(i) Weight of the body on moon, where g = 1.6 m/s2

Using equation (5.19),

Weight = Mass × g =100 × 1.6 = 160 N. Ans.

(ii) Weight of the body on sun, where g = 270 m/s2

Weight = Mass × g = 100 x 270 = 27000 N. Ans.

Problem 5.39. A force of 200 N acts on a body having mass of 300 kg for 90 seconds. If the initial velocity of the body is 20 m/s, determine the final velocity of the body:

(i)                           when the force acts in the direction of motion, and

(ii)                        when the force acts in the opposite direction of the motion.

Sol. Given:

Force, F = 200 N

Mass, m = 300 kg

Time, t = 90 seconds

Initial velocity, u = 20 m/s.

Using equation (5.20), we have

F = m × a      or       200 = 300 × a

(- ve sign shows that the body will be moving in the opposite direction).

Problem 5.40. A body of mass 15 kg falls on the ground from a height of 19.6 m. The
body penetrates into the ground. Find the distance through which the body will penetrate into ·the ground, if the resistance by the ground to penetration is constant and equal to 4900 N. Take
g = 9.8 m/s2.

Sol. Given:

Mass, m = 15 kg

Height of body from ground, h = 19.6 m

Resistance to penetration, F* = 4900 N

Acceleration due to gravity, g = 9.8 m/s2.

Let us first consider the motion of the body from a height of 19.6 m to the ground surface.

Initial velocity of the body,  u = 0

Final velocity of the body, when it reaches the ground = v.

Using the equation,

When the body is penetrating into the ground, the resistance to penetration is acting in the upward direction (Resistance always acts in the opposite direction of motion of the body). But the weight of the body is acting in the downward direction.

Weight of the body is given by equation (5.19),

Weight of the body = Mass × g = 15 × 9.80 N = 147 N

Upward resistance to penetration = 4900 N

Net force acting in the upward direction,

F = 4900 – 147 = 4753 N.

As the net force on the body is acting in the opposite direction to the motion of the body, this force will produce retardation.

Using equation (5.20), we have

Distance through which body will penetrate into the ground.

Consider the motion of the body from the ground to the point of penetration into the ground.

Let the distance of penetration = s

Final velocity, v = 0

Initial velocity, u = Velocity of the body on the ground == 19.6 m/s

Retardation,             a == 316.866 m/s2.

Using the relation,

v2 – u2 = – 20a × s                           ( -ve sign is taken due to retardation)

Problem 5.41. A man weighing 637 N dives into a swimming pool from a tower of
height
19.6 m. He was found to go down in water by 2 m and then started rising. Find the average resistance of water. Neglect the resistance of air.

(AMIE Summer, 1982)

Sol. Given:

Weight of the man = 637 N

Height of tower, h = 19.6 m

Distance travelled by man from the water surface into the water = 2 m.

First consider the motion of the man from the top of the tower to the water surface of the swimming pool.

Initial velocity of man, u. = 0

Acceleration due to gravity, g = 9.8 m/s2.

Let the final velocity of the man, when he reaches the water surface = v.

Now using the relation,

v2 – u2 = 2gh            or                 v2 – 02 = 2 × 9.80 × 19.60

Now consider the motion of the man from the water surface of the swimming pool upto the point, from where the man started rising.

Distance traversed,             s = 2 m

Initial velocity of the man on the water surface

u = 19.6m/s

Final velocity, v = 0.

As the velocity of the man becomes zero after travelling a distance 2 m inside the water, hence a force due to water resistance is acting in the opposite direction to the motion of the man. This force will produce retardation.

Let a = Retardation due to water resistance

Using the relation, v2 – u2 = – 2as               (- ve sign is taken due to retardation)

02 – 19.62 = – 2 × a × 2

Average resistance of water

Let      F* = Average resistance of water acting on the man in the upward direction.

Weight of man = 637 N acting in the downward direction.

Net force acting on the man in the upward direction

= F* – Weight of man = (F* – 637) N.

But the net force acting on the man must be equal to the product of mass of the man and retardation.

(F* – 637) = m × a                                             (F = m × a)              … (i)

But mass of man,

Substituting the values of m and a in equation (i), we get

(F* – 637) = 65 × 96.04

F* = 65 × 96.04 + 637

= 6879.6 N. Ans.

Problem 5.42. A bullet of mass 81 gm and moving with a velocity of 300 m l s is fired
into
a log of wood and it penetrates to a depth of 10 cm. If the bullet moving with the same velocity, were fired into a similar piece of wood 5 cm thick, with what velocity would it emerge? Find also the force of resistance, assuming it to be uniform.

(AMIE Winter, 1981)

Sol. Given:

Initial velocity of bullet, u = 300 m/s

Distance travelled, s = 10 cm = 0.10 m

Final velocity, v = 0.

As the force of resistance is acting in the opposite direction of motion of bullet, hence force of resistance will produce retardation on the bullet.

Let a = Retardation

Now using the relation,

v2-u2 =-20s                                     (- ve sign is taken due to retardation)

02 – 3002 = – 2 × a × 0.1

Let F = Force of resistance offered by wood to the bullet.

Using equation (5.20), we get

F* = m × a = .081 × 450000

= 36450 N. Ans .

Velocity of the bullet with which the bullet will come out from a piece of wood of 5 cm thick.

Let v = velocity with which the bullet emerges from the piece of wood of 5 cm thick.

Initial velocity, u = 300 m/s.

As the resistance offered by wood is uniform, hence retardation will be same as before.

a = 450000 m/s2.

Distance travelled, r = 5 cm = .05 m

Using the relation,

v2– u2 = – 2as                                     (- ve sign is taken due to retardation)

v2 – 3002 = – 2 × 450000 × .05     or    v2 = 300 × 300 – 45000

= 90000 – 45000 = 45000

Problem 5.43. A car, moving on a straight level road, skidded for a total distance of’
60 meters after the brakes were applied. Determine the speed of the car, just before the brakes were applied, if the co-efficient of friction between the car tyres and the road is 0.4. Take g= .9.80 m/s2.

Sol. Given:

Let u = Velocity of car just before applying the brakes.

Final velocity of car, v = 0

Distance travelled, s = 60 m

Co-efficient of friction between car tyres and road,

µ = 0.4

Let W = Weight of car in Newton

R = Normal reaction

= 0.4 W Newton

F=0.4 W.

As frictional force is acting in the opposite direction of motion , hence the frictional force will produce retardation.

Using equation (5.20),

Force = Mass × a

Problem 5.43. A car, moving on a straight level road, skidded for a total distance of’
60 meters after the brakes were applied. Determine the speed of the car, just before the brakes were applied, if the co-efficient of friction between the car tyres and the road is 0.4. Take g= .9.80 m/s2.

Sol. Given:

Let u = Velocity of car just before applying the brakes.

Final velocity of car, v = 0

Distance travelled, s = 60 m

Co-efficient of friction between car tyres and road,

µ = 0.4

Let W = Weight of car in Newton

R = Normal reaction

Fig. 5.17

1st Case. Fig. 5.17 shows the position of the truck, when it is moving down the plane. The road resistance (F1) is acting in the opposite direction of the motion of the truck. The truck is not having any acceleration and hence it is moving with a constant velocity of 10 m/s. Hence the net force on the truck in the direction of motion should be zero. But net force on the truck in the direction of motion (See Fig. 5.17).

= W sin θ – F1

Distance CD = s = 39.12 m.

(ii) Will the car stop at D or move again down the grade?

After reaching point D, the car will move down the grade if the component of the weight of the car along the grade is more than the force of friction due to track. The force of friction along the grade is acting up the plane whereas the component of the weight of the car along the grade is acting down the plane.

But force of friction = 160 N

and component of the weight of the car along the grade = 1000 N.

As the component of the weight of the car along the grade is more than the force of friction, the car will move down from D. Ans.

(iii) Distance travelled by car from B to E on level track while moving down the grade.

First consider the motion from point D to point B.

Initial velocity at D, u = 0

Final velocity at B = v (assumed)

Distance BD, s = Distance CD + Distance BC

= 39.12 + 100 = 139.12 m

Let a = Acceleration when car is moving from D to B.

When the car is moving down from D to B, the frictional resistance of 160 N is acting
upward while the component of the weight of the car along the grade is acting downward. Hence the net downward force along the grade which produces acceleration a’ is (1000 N – 160 N) or 840 N.

Net force down the grade = 840 N.

Now using the relation, Force = Mass × Acceleration

Consider the motion from B to E

Now, from B to E the track is level. The weight of the car is acting vertically downward. The component of the weight of the car along the horizontal plane is zero. The only force acting on the car, when it is moving from B to E, is frictional force. This frictional force of 160 N will oppose the motion of the car and finally the car will stop to E. The frictional force will produce retardation.

Let a* = Retardation from B to E

Force = Mass × Retardation

a = Uniform acceleration of the lift

T = Tension in cable supporting the lift. This is also called the *reaction of the lift.

The lift may be moving Upwards or moving downwards.

*Reaction of lift is equal to the tension (1′) in cable supporting the lift.

1st Case. Let the lift is moving upwards as shown in Fig. 5.22 (a). The weight carried by lift is acting downwards while the tension in the cable is acting upwards. As the lift is moving up, the net force which is equal to (T – W) is acting upwards.

Fig. 5.22 (a) Lift is moving upwards

Net force in upward direction = T – W.

This net force produces an acceleration ‘a’

Hence using,

Net force = Mass × Acceleration

2nd Case. As the lift is moving downwards as shown in Fig. 5.22 (b), the net force is acting downwards. Hence in this case W is more than T (tension in string).

Net force in downward direction = (W – T).

This net force produces an acceleration ‘a’,

Hence using, Net force = Mass × Acceleration

Fig. 5.22. (b) Lift is moving

Problem 5.51. A lift carries a weight of 100 N and is moving with a uniform acceleration of 2.45 m/s2. Determine the tension in the cables supporting the lift, when:

(i)                           lift is moving upwards, and

(ii)                        lift is moving downwards. Take g = 9.80 m/s2.

Sol. Given:

Weight carried by lift, W = 100 N.

Uniform acceleration, a = 2.45 m/s2.

(i) Lift is moving upwards

Let T = Tension in the cables supporting the lift.

= 100 × .75 = 75 N. Ans.

Problem 5.52. A life has an upward acceleration of 1.225 m / s2. What pressure will a
man weighing 500 N exert on the floor of the lift? What pressure would he exert if the lift had an acceleration of 1.225 m/s2 downwards? What upward acceleration would cause his weight to exert a pressure of 600 N on the floor? Take g = 9.8 m/s2.

Sol. Given:

Upward acceleration, a = l.225 m/s2

Weight of man, W = 500 N.

1st Case. The lift is moving up with an acceleration of l.225 m/s2. The pressure exerted by a man on the floor of the lift is equal to the reaction of the lift and it is the same as the tension in the cables supporting the lift.

Let T = Tension in the cables supporting the lift or the reaction of the lift or the pressure exerted by the man on the floor of the lift.

Using equation (5.23)for the lift moving upwards, we have

2nd Case. The lift is moving downwards with an acceleration of 1.225 m/s2. The pressure exerted by the man on the floor of the lift is equal to the reaction of the lift and it is the same as the tension in the cables supporting the lift.

Using equation (5.24),

3rd Case. The lift is moving upwards with an unknown acceleration.

Let a = Acceleration upwards

T = Pressure exerted by man on floor of lift = 600 N
W = Weight of man = 500 N.

Using equation (5.23), we get

or   a = (1.2 – 1.0) × 9.80 = 0.2 × 9.80 = 1.96 m/s2. Ans.

Problem 5.53. An elevator weighs 2500 N and is moving vertically downwards with a constant acceleration. Write the equation for the elevator cable tension. Starting from rest it travels a distance of 35 metres during an interval of 10 seconds. Find the cable tension during this time. Neglect all other resistances to motion. What are the limits of cable tension?

(AMIE Summer, 1979)

Sol. Given:

Weight of elevator, W = 2500 N

Initial velocity, u = 0

Distance travelled = 35 m

Time, t = 10 sec.

Let T = Tension in the cable supporting an elevator in N.

1st Part. The equation for the elevator cable tension is obtained as given below: (see Fig. 5.23). The elevator is moving down.

Net acceleration force in the downward direction

= (W – T) = (2500 – T) N.

The net accelerating force produces an acceleration ‘a’ in the downward direction.

Hence using the relation,

Net force = Mass × Acceleration.

Fig. 5.23

Hence the above equation (i) represents the equation for the elevator cable tension when the elevator is moving downwards.

2nd Part. Limits of cable tension are obtained from equation (i) as given below:

(i) When a = 0, and this value is substituted in equation (r), the value of T is obtained as

Limits of cable tension (T) are:

At    a = 0, T = 2500 N

At    a = 9.81, T = 0. Ans.

3rd Part. s = 35 m,            u = 0 and· t = 10 seconds.

Problem 5.54.A cage, carrying 10 men each weighing 500 N, starts moving downwards from rest in a mine vertical shaft. The cage attains a speed of 12 metres/s in 20 metres. Find the pressure exerted by each man on the floor of the cage. Take g = 9.80 m/s2.

Sol. Given:

Weight of one man, = 500 N

Total weight of 10 men on the cage, W = 500 × 10 = 5000 N

Initial velocity of cage, u = 0

Final velocity of cage, v = 12 m/s

Distance travelled, s = 20 m

Acceleration due to gravity, g = 9.80 m/s2.

The pressure exerted by the man on the floor of the cage will be same as the tension produced in the cables supporting the cage.

Let T = Total tension produced by 10 men in the cables.

a = Acceleration of the cage.

The acceleration will be obtained by using the relation,

v2 – u2=2as    or        122 – 0=2a × 20

But tension produced by each man in the cable is the same as the pressure exerted by each man on the floor of the cage.

Pressure exerted by each man on the floor = 316.326 N. Ans.

Problem 5.55. An elevator weighing 500’0 N is ascending with an acceleration of
3 m/s
2. During this ascent its operator whose weight is 700 N is standing on the scales placed on the floor. What is the scale reading? What will be the total tension in the cables of the elevator during this motion?

(AMIE Summer, 1982)

Sol. Given:

Weight of the elevator, W1 = 5000 N

Acceleration of elevator, a = 3 m/s2

Weight of the operator, W2 = 700 N

When the operator is standing on the scale, placed on the floor of the elevator, the reading of the scale will be equal to the reaction (R) offered by the floor on the
operator.

Hence let R = Reaction offered by floor on operator. This is also equal to the reading of scale.

T = Total tension in the cables of elevator.

Consider the motion of operator. The operator is moving upwards along with the elevator with an acceleration a = 3 m/s2. The net force on the operator is acting upwards.

Fig. 5.24

Net upward force on operator

= Reaction offered by floor on operator – Weight of operator

= (R -700)

Total tension in the cables of elevator.

Let T = Total tension in the cables of elevator

W = Total weight (i.e., weight of elevator + weight of operator)

= 5000 + 700 = 5700 N.

As the elevator with the operator is moving upwards with an acceleration f = 3 m/s2, the net force will be acting on the elevator and operator in the upward direction.

Net upward force on elevator and operator,

= Total tension in the cables – Total weight of elevator and operator

= (T-5700)

Mass of elevator and operator

5.8.8. Analysis of the Motion of Two Bodies Connected by a String. Fig. 5.25 shows a light and inextensible string passing over a smooth and weightless pulley. Two bodies of weights W1 and W2 are attached to the two ends of the string. Let W1 be greater than W2. As W1 > W2, the weight W1 will move downwards, whereas the smaller weight (W2) will move upwards. For an inextensible string, the upward acceleration of the weight W2 will be equal to the downward acceleration of the weight W1

As the string is light and inextensible and passing over a smooth pulley, the tension* of the string will be the same on both sides of the pulley.

Let T = Tension in both strings,

a = Acceleration of the bodies.

Fig.5.25

*Tension of the string on both sides of the pulley will be different if the string is heavy i.e., weight of string is considered), extensible and does not pass over a smooth pulley.

Consider the motion of weight, W1. The weight W1 is moving downwards with an acceleration a. The forces acting on W1 are (i) its weight W1 acting downwards and (ii) tension T acting upwards. As the weight W1 is moving downwards, hence net force on the weight W1 is acting downwards.

Net downward force = (W1 – T)

But net force = Mass × Acceleration

Now consider the motion of weight, W2· The forces acting on W2 are: (i) its weight W2 acting downwards and (ii) tension T acting upwards. But the weight W2 is moving upwards, hence net force on weight W2, is acting upwards.

Net upward force = (T – W2)

But net upward force = Mass × Acceleration

Problem 5.56. Two bodies of weight 50 N and 30 N are connected to the two ends of a light inextensible string. The string is passing over a smooth pulley, Determine:

(i)                           The acceleration of the system, and

(ii)                        Tension in the string. Take g = 9.80 m/s2.

Sol Given:

Heavier weight, W1 = 50 N

Lighter weight, W2 = 30 N

Let a = Acceleration of the system, and

T = Tension in the string.

Fig. 5.26

(i) Using the equation (5.25) for acceleration,

= 2.45 m/s2 Ans.

(ii) Using equation (5.26) for tension in the string,

Problem 5.57. Two bodies of different weights are connected to the two ends of a light inextensible string, which passes over a smooth pulley. If the acceleration of the system is 3 m/s2 and bigger weight is 60 N, determine:

(i)                           The smaller weight, and

(ii)                        Tension in the string. Take g = 9.80 m/s2.

Sol. Given:

Acceleration, a = 3 m/s2

Bigger weight, W1 = 60 N

Let      W2 = Smaller weight, and

T = Tension in the string.

(i) Using equation (5.25); we get

Problem 5.58. A pulley whose axis passes through the centre O, carries load as
shown in Fig.
5.27. Neglecting the inertia of the pulley and assuming that the card is inextensible, determine the acceleration of the block A. How much weight should be added to or taken away from the block A, if the acceleration of the block A is required to be g / 3.0 downwards?

(AMIE Summer, 1981)

Sol. Given:

Bigger load, W1 = 700 N

Smaller load, W2 = 500 N

Let a = Acceleration of block A or the acceleration of the system.

Using equation (5.25),

How much weight should be added to or taken away from the block A (i.e., from bigger load 700 N) when acceleration of bigger load is g / 3.0 downwards.

Let W1* = Total weight of block A when acceleration is g/3.0

As W1*  is more than W1. Hence the weight must be added to the block A.

Weight added = W1* – W1 = 1000 – 700 = 300 N. Ans.

Problem 5.59. Determine the tensions in the strings and accelerations of blocks A and B weighing 150 N and 50 N connected by a string and a frictionless and weightless pulley as shown in Fig. 5.28.

(AMIE Nov., 1970)

Sol. Given:

Weight of block A = 150 N

Weight of block B = 50 N.

As the pulley is smooth, the tension in the string will be same throughout.

Let T = Tension in the string

a = Acceleration of block B.

Then acceleration of block A will be equal to half the acceleration of block B.

As the weight of block A is more than the weight of block B, the block A will move downwards whereas the block B will move upwards.

Fig, 5.28

Consider the motion of block B

The forces acting on block B are: (i) its weight 50 N acting downwards, and (ii) tension T acting upwards. As the block B is moving upwards, the net force on the block B is acting upwards.

Net force on block B = T – 50

Acceleration of block B = a.

But, Net force = Mass × Acceleration

Problem 5.60. A system of weights connected by strings passing over pulleys A and B is, shown in Fig. 5.29. Find the acceleration of the three weights, assuming weightless strings and ideal conditions for pulleys.

(AMIE Summer, 1977)

Sol. Given:

As the strings are weightless and ideal conditions prevails, hence the tensions in the string passing over pulley A will be same. The tensions in the string passing over pulley B will also be same. But the tensions in the strings passing over pulley A and over pulley B will be different as shown in Fig. 5.29.

Let T1 = Tension in the string passing over pulley A, and

T2 = Tension in the string passing over pulley B.

One end of the string passing over pulley A is connected to a weight of 15 N, and the other end is connected to pulley B. As the weight 15 N is more than the weights (6 + 4= 10 N), hence weight 15 N will move downwards, whereas pulley B will move upwards. The acceleration of the weight 15 N and of the pulley B will be same.

Let a = Acceleration of 15 N weight in downward direction.

Then acceleration of pulley B in upward direction is also equal to ‘a’.

The string passing over pulley B, has weights of 6 N and 4 N. Hence the weight 6 N will move downwards, whereas the weight 4 N will move upwards.

Let ‘a1‘ = Acceleration of 6 N downwards with respect to pulley B.

Then acceleration of weight of 4 N with respect to pulley

B = a1 in the upward direction.

Absolute acceleration of weight 4 N.

= Acceleration of 4 N w.r.t. to pulley B + Acceleration of pulley B.

= a1 + a (upwards)

(As both the acceleration are in upward directions, total acceleration will be sum of the two accelerations).

Absolute acceleration of weight 6 N

= Acceleration of 6 N w.r.t .. pulley B + Acceleration of pulley B

= (a1 – a) downwards.

(As a1 is acting downward whereas a is acting upward. Hence total acceleration in the downward direction = a1 – a).

Consider the motion of weight 15 N

The forces acting are: (i) 15 N downwards, and (ii) tension T1 upwards. As the weight 15 N is moving downwards, the net force is acting downwards.

Net downwards force = (I5 – T1).

Using, Net force = Mass × Acceleration

5.8.9. Analysis of the Motion of two Bodies Connected by a String when one
Body is Lying on a Horizontal Surface and other is Hanging Free

1. First case when the horizontal surface is smooth and the string is passing over a smooth pulley. Fig. 5.30 ‘how, the two Weight, W1 and W2 connected by a light inextensible string, passing over a smooth pulley. The weight W2  is placed on a smooth horizontal surface, whereas the weight WI is hanging free. The weight W1 is moving downwards, whereas the weight W2 is moving on smooth horizontal surface. The velocity and acceleration of W1 will be same as that of W2.

As the string is light and inextensible and passing over a smooth Pulley, the tensions of the string will be same on both sides of the pulley.

Let T = Tension in the string

a = Acceleration of the weight W1 and also of W

Problem 5.66. Two bodies of weights 40 N and 15 N are connected to the two ends of a light inextensible string, which passes over a smooth pulley. The weight 40 N is placed on a smooth inclined plane, while the weight 15 N is hanging free in air. If the angle of the plane is I5°determine:

(i)                           acceleration of the system, and

(ii)                        tension in the string. Take g = 9.80 m/ s2.

Sol. Given:

Weight placed on inclined plane, W2 = 40 N

Weight hanging free in air, W1 = 15 N

Angle of inclination, θ = 150

Acceleration due to gravity, g = 9.80 m/s2

Let a = Acceleration of the system

T = Tension in the string.

The inclined surface is smooth. Hence the acceleration and tension are obtained by using equations (5.31) and (5.32).

(i) Using equation (5.31) for acceleration,