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| Newton's Third Law of Motion |
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| Newton's third law states that to every action, there is always an equal (in magnitude) and opposite (in direction) reaction. |
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 The two forces act on different bodies. |
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 If two bodies are interacting and if the force on body 2 due to body
1 is
then the force on body 1 due to body 2 is  |
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 Newton's third law is applicable whenever the bodies are at rest or in motion or are at a distance from each other. |
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 We can conclude that a single isolated force is not possible. |
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| Consider a simple system in which two blocks are connected by a string which passes over a light, frictionless pulley as shown below. |
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| Let m1 accelerate upward with an acceleration 'a'. Hence, m2 accelerates downward with an acceleration 'a'. The forces on m1 are as shown. |
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| M1a = T-m1g |
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| M2a = m2g-T |
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| a (m1 + m2) = (m2 - m1) g |
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| T (m1 + m2) = 2m1 m2g |
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 In a lawn sprinkler when water sprays out of the nozzles, a backward force is exerted on the nozzle as a result of which it starts rotating. Consequently, water is sprayed in all directions. |
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 Consider a block of weight w resting on a horizontal surface. The block exerts a downward force of w on the ground. Similarly, the ground also exerts an upward force of magnitude w on the block. |
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 When we walk we push the ground backwards, at an angle, with a
force
. The ground pushes us with a force  |
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 In order to swim, a person has to push the water backward in order to go forward. |
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 When a ball strikes a wall, it bounces backwards due to the reaction which the wall exerts on the ball. |
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 For a ship or submarine to go forward, it has to push the water backward. |
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| A good illustration of the ideas discussed in this chapter is the motion of a cart pulled by a horse. Suppose the cart is at rest when the driver whips the horse. The horse pulls the cart and the cart accelerates forward. The question posed is as follows. The horse pulls the cart with a force F1 in the forward direction. From the third law of motion, the cart pulls the horse by an equal force F2=F1 in the backward direction. The sum of the forces is, therefore, zero. Why then, should the cart accelerate forward? |
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| F |
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| = F |
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| = F |
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Try to locate the mistake in the argument. According to our scheme, we should first decide the system. We can take the horse as the system or the cart and the horse taken together as the system. Then the forces on the cart should be listed and the forces on the horse should not enter the discussion. The force on the cart is F1 in the forward direction and the acceleration of the cart is also in the forward direction. How much is this acceleration? Take the mass of the cart to
be Mc. Is the acceleration of the cart 
in forward direction? Think carefully. |
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Let us now try to understand the motion of the horse. This time, we have to consider the forces on the horse. The forward force F1 by the horse acts on the cart and it should not be taken into account when we discuss the motion of horse. The force on the horse by the cart is F2 in the backward direction. Why does the horse go in forward direction when whipped? The horse exerts a force on the cart in the forward direction and hence the cart is accelerated forward. But the cart exerts an equal force on the horse in the backward direction. Why is the horse not accelerated in backward direction? (Imagine this situation. If the cart is accelerated forward and the horse backward, the horse will sit on the cart kicking out the driver and the passengers.) Where are we wrong? We have not considered all the forces acting on the horse. The road pushes the horse by a force P which has a forward component . This force acts on the horse and we must add this force when we discuss the motion of the horse. The horse accelerates forward if the forward component F of the force P
exceeds F2. The acceleration of the horse is  We should make sure that all the forces acting on the system are added. Note that the force of gravity acting on the horse has no forward component.
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Going back to the previous paragraph, the acceleration of the cart may
not be  . The road exerts a force Q on the cart which
may have a backward component f'. The total force on the cart is
F1 - f'. The acceleration of the cart is then 
in the forward direction. The forces f and f' are self
adjustable and they adjust their values so that  . The
acceleration of the horse and that of the cart are equal in magnitude and direction and hence, they move together. So, once again we remind you that only the forces on the system are to be considered to discuss the motion of the system and all the forces acting on the system are to be considered. Only then,  |
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| Cart System |
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| (indicating only forces experienced by cart) |
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| f is the functional force and it is always in the direction opposing the motion. Hence, it is directed backward since the motion is forward. |
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| Net force on cart = F1-f1 |
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| Mass of cart = mc |
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| Horse system (indicating only forces experienced by horse) |
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| f is the horizontal component of the reaction experienced by the horse's leg when it applies an inclined thrust on the ground. Hence, it is in the forward direction. |
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| Net force on horse = f-F1 |
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| Mass of horse = mH |
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| Since the horse and cart move together, acceleration of cart must be equal to acceleration of the horse. |
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