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Newton's Third Law of Motion


The first law of motion simply says that the state of motion of an object can be changed only if a force acts on it.

The second law tells us how much change in the state of motion of an object can be brought about by a given force.

The third law tells us how forces are exerted.

We know that a horse pulls a cart and a hammer drives a nail, etc. We observe these actions very casually. But Newton carefully analyzed what he observed. In these examples, one object exerts a force and the other feels it - the horse exerts a force and the cart feels, it; a hammer exerts a force and the nail feels it.

But, is it really so simple? The third law of motion states that, whenever one body exerts a force on a second body, the second body exerts an equal and opposite force on the first, or to every action there is an equal and opposite reaction.

Action and Reaction Forces


The most important point to recognize is that the action and reaction forces act on different bodies. Action and reaction forces do not act on the same body. If they were to act on the same body, they would add up to zero.

No action can take place in the absence of a reaction.

Everyone who has played the game of tug-of-war knows from experience that he can exert a force only if there is somebody on the other end of the rope. No force can be exerted if the other end is free. One team provides the action force and the opposite team provides the reaction force.

Now the opposing team can be replaced by a wall by fixing the other end of the rope to the wall. The third law says that if you exert a force on the wall, the wall will exert an equal and opposite reaction force on you.

Now remove the rope and exert a force directly on the wall by pushing the palm of your hand against it. Now look at your hand. It is distorted a little and also a pains a little.

Only a force can distort the shape of an object. But you are not exerting any force on your hand; you are in fact exerting a force on the wall. Then which force has distorted your hand? It is the force exerted by the wall on your hand.

How can the wall exert a force on your hand? The wall can exert a force because it is elastic. It is because of elasticity that a non-living object can exert a force.

A spring, for example, has elasticity. We take a ball and press it against a spring fixed on a table. The spring is compressed by the force exerted on it. When the ball is released, it flies off due to the reaction force of the spring which is called into play as the spring begins to regain its original size. Similarly, when we press against the wall, the wall is compressed. The compression of the wall is very little and cannot be noticed. Due to this compression, a reaction force is called into play, which acts on the palm or on the ball which is being pressed against.

Some more examples of the third law are as follows

  1. When an ice-skater strikes against the wall of the skating-rink, he is pushed backwards. When he strikes against the wall, he exerts a force on it. The wall in turn exerts an equal and opposite reaction force on him.
  2. While swimming, a person pushes the water backwards and as a result he himself is pushed forward by the reaction force exerted by the water on him.
  3. The motion of rockets is a classic example of an action reaction pair. A rocket expels gases by exerting a force on them. The gases in turn exert an equal and opposite force on the rocket. A space vehicle can accelerate or change the direction of its motion just by firing rockets in the proper direction.
  4. We would not be able to walk if there were no reaction force. In order to walk, we push our foot against the ground. The earth in turn exerts an equal and opposite force. This force is inclined to the surface of the earth. The vertical component of this force balances our weight and the horizontal component enables us to walk forward (see Fig.).
Newton's Third Law of Motion
 

Two apparent paradoxes

  1. We push a wooden block of mass m against a solid wall with a force F as shown in the figure.

    According to Newton’s second law, the acceleration a is and thus, the block should accelerate and start moving. But from experience we know that the block will not move. What is wrong?

    The mistake is that in the equation
    F = ma, F is not just the force exerted on the object but the net force experienced by it. In addition to the applied force F, there is a second force F’ exerted by the wall on the block.
    The net force on the block is (F + F’). Now according to Newton’s third law F’ is equal and opposite of F, i.e. F’ = -F. The net force experienced by the block is F net = (F + F') =F+ (-F) = 0.

    Hence, the acceleration of the block is
  2. We place two blocks A and B of masses m1 and m2 respectively on a frictionless surface as shown in the given figure and apply a force F to A. The block B also experiences a force due to its contact with A.
    By Newton’s third law, the block B must exert an equal and opposite force F on block A. Thus the net force on A is
    F + (-F) = 0, so that

    Thus, we conclude that block A can never move, no matter how large a force F is applied to it which obviously is not true.

    This paradox arises because, we have assumed that the force F exerted on A is transmitted through it and is thus also applied to B. Newton’s laws do not say that if a force is applied to a body, then that body in turn must exert the same force on another body in contact with it.

    Let A exert a force F’ on B. From the third law B will exert a force –F’ on A. Thus, the net force on A is F - F’ and its acceleration a is given by

    F – F’ = m1a
    Since B is in contact with A, it will have the same acceleration a which is given by
    F’ = m2a
    Adding the two equations, we get
    F = m1a + m2a = (m1 + m2)a
    or
    i.e.
    which is what we expect.
Example
A man of mass 60 kg is standing on a weighing machine placed on the floor of a lift. The machine reads weight (w = mg) in newton. 
(i) What is the reading of the weighing machine when the lift is (a) stationary,

(b) moving upwards with a uniform speed of 10 m s-1

(c) moving downwards with a uniform acceleration of 5 m s-2, and

(d) moving upwards with a uniform acceleration of 5 m s-2?

 
(ii) What would be the reading of the weighing machine if the connecting rope of the lift suddenly breaks and the lift begins to fall freely under gravity?


Solution 

Mass of man (m) = 60 kg

Acceleration due to gravity (g) = 9.8 ms-2

(i) The man exerts a downward force on the weighing machine

The machine, in turn, exerts on him an upward reaction or Normal force which it measures.

(a) When the lift is stationary, 

The reaction (R) or Normal Force = magnitude of the weight of the man.

R = mg = 60 × 9.8 = 588 N

Reading of weighing machine = 588 N

(b) When the lift is moving with a uniform speed, it has no acceleration of its own.

Hence, the reading of the weighing machine will still be 588 N.

(c) When the machine is moving downwards with an acceleration a = 5 m s-2 

Force F = ma acts downwards. But the reaction R = mg acts upwards

Hence the effective reading of the weighting machine will be

Reff = R - F = mg – ma = m (g - a) = 60 × (9.8 – 5 ) = 288 N

(d) when the weighing machine is moving upwards with an acceleration a = 5 m s-2 

Force F = ma and the reaction R = mg both act in the same direction (upwards).

Hence, the reading of the weighing machine will be
Reff = R + F = mg + ma = m (g + a) = 60 × (9.8 + 5 ) = 888 N

(ii) If the lift is falling freely under gravity, than a = g and the downward F = mg. Since the upward reaction R is also equal to mg, the reading of the machine will be

Reff = mg – mg = 0

Example

A helicopter of mass 1500 kg is rising vertically upward with a uniform acceleration of 5 m s-2. If the mass of the persons in the helicopter is 500 kg, find the magnitude and direction of the

(i) Force exerted by the persons on the floor of the helicopter, 

(ii) Action force exerted by the helicopter (with persons in it) on the surrounding air, and

(iii) Reaction force exerted by the surrounding air on the helicopter and the persons in it.


Solution  
Mass of helicopter (M) = 1500 kg

Mass of persons (m) = 500 kg

Acceleration of helicopter or of persons (a) = 5 m s-2 (vertically upward)

Acceleration due to gravity (g) = 9.8 m s-2 (vertically downward)

(i) Since the helicopter (and the persons in it) is rising upwards, the force exerted by the persons on the floor will

Feff = m (g + a) = 500 × (9.8 + 5) = 7400 N

The direction of the force is vertically downwards.

(ii) The action force exerted by the helicopter and the persons in it on the surrounding air is

F = M (g + a) + m (g + a)

F = (M + m) (g + a)

F = (1500 + 500) × (9.8 + 5.0) = 29600 N

The direction of the force is vertically downwards.

(iii) From Newton’s third law, the reaction force R exerted by the surrounding air on the helicopter with persons in it, is equal and opposite to the action force F calculated above.
Hence R = 29600 N (vertically upwards).

Example

A rope which can withstand a maximum tension of 400 N is hanging from a tree. If a monkey of mass 30 kg climbs on the rope, in which of the following cases will the rope break: the monkey

(i) climbs up with a uniform speed of 5 m s-1

(ii) climbs up with a uniform acceleration of 2 m s-2

(iii) climbs up with a uniform acceleration of 5 m s-2

(iv) climbs down with a uniform acceleration of 5 m s-2 and

(v) falls down the rope almost freely under gravity. Take g = 10 m s-2 and neglect the mass of the rope.


Solution  

Mass of monkey (m) = 30 kg, 

g = 10 m s-2

If the monkey climbs up the rope with a uniform acceleration a, the tension in the rope is
T = m (g + a)

(i) Since the speed of the monkey is uniform, a = 0.

Hence T = mg = 30 x 10 = 300 N

(ii) In this case, a = 2 m s-2.

Hence T = m (g + a) = 30 x (10 + 2) = 360 N

(iii) In this case, a = 5 m s-2

Hence T = m (g + a) = 30 x (10 + 5) = 450 N

(iv) If the monkey climbs down the rope with a uniform acceleration a, the tension in the rope will be

T = m (g – a)

Here a = 5 m s-2

Hence T = 30 x (10 – 5) = 150 N

In this case a = g. Hence the tension in the rope will be almost zero, Since the rope can withstand a maximum tension of 400 N, the rope will break only in case (iii) for which the tension in the rope is 450 N which is greater than the breaking tension.



Example

A man standing on a platform lifts a block of mass 25 kg in two different ways (a) and (b) as shown in the figure. If mass of the man is 60 kg, find the action exerted by the man on the platform in cases (a) and (b). If the platform yields to a normal force of 800 N, in which of the two cases will the platform yield?


Solution  

Mass of block (m) = 25 kg. Mass of man (M) = 60 kg.

Force needed to lift the block is Fb = mg = 25 x 9.8 = 245 N

Force exerted on platform by man is

Fm = weight of the man

     = Mg = 60 x 9.8 = 588 N

(i) If the man lifts the block as shown in the above figure (a), he has to exert action in a direction opposite to his weight. In other words, he has to overcome his own weight first and then apply extra effort to lift the block. Hence the action on the platform will be

Fp = Fm + Fb = 588 + 245 = 833 N

(ii) If the man lifts the block as shown in the figure (b), he has to exert action in the direction of his weight. In other words, his weight helps him to exert action. Hence, in this case, the action on the platform will be

Fp = Fm - Fb = 588 - 245 = 343 N

It is given that the platform can withstand a maximum normal force of 800 N. Since the action in case (a) (which is 833 N) exceeds 800 N, the platform will yield in this case. The man should adopt the method shown in the figure (b) to lift the block.

Alternative Method

In the first case, the man has to exert a force, which should be equal to the sum of the weight of the man and the weight he is supposed to lift.

R = m1g + m2g = g (m1 + m2) = 9.8 × (25 + 60) = 833 N

Second case:

R = m2g - m1g = g (m2 - m1) = 9.8 × (60 - 25) = 343 N
 





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