Newton’s Third Law of Motion
Newton’s third law of motion was discovered and formulated, during the investigation of the fact that in all experiments it appeared that “when ever a body exerts a force on a second body, the second body always exerts a force on the first one”
Let us visualize and understand this phenomena with an experiment.
Suppose, we throw a stone on a surface of good strength; and the surface is made of glass, one finds it broken (the surface). From here one concludes that a force was exerted by stone on the surface and consequently it was broken.
Now, the question is, did that surface also exert a force on the stone. Just to know about it let us change our throwing object from stone to an egg of almost equal mass. Now, one throws this egg on the same surface of good strength with the same throwing force which he used for the stone. What happens? Obviously with your daily experience you know that the egg will be broken (And the damage to the surface will not be visible due to egg’s spoiling the observation).
This is only possible if there was a force acting on the egg at the time it hit the surface. In fact we can now conclude that there is a mutual force acting on the contact point of the surface and the object thrown. The breaking of either one (or may be both) depends on their ability to absorb forces without getting damaged (that is their strength) so in precise words:
Þ To every action there is always opposite and equal reaction, it is equivalent to say that mutual actions of two bodies upon each other are always equal and directed to contrary parts.
Þ The most important fact to notice here is that these oppositely directed equal action and reaction can never balance or cancel each other because they always act, on two different point (broadly on two different objects) For balancing any two forces the first requirement is that they should act on one and the same object. (or point, if object can be treated as a point mass, which is a common practice).
Illustrations
1. Suppose in figure 3.2 we put one more block of 5 kg mass adjacent to 10 kg and a force of 150 N acts as shown in the figure 3.3, then find the forces acting on the interface.
Sol. The combined acceleration of the two bodies when treated as one is
a= F/ (10+5) = 150 / 15 = 10m/sec²
So each one moves with a = 10m/sec² keeping their contact established.
Here you can feel that due to 150N force the body of 5kg feels as if it is being pushed by the 10 kg mass. There is a force acting on 5kg called R1, to oppose it by third law this body exerts a force R2 on 10 kg. The interface is as shown in figure 3.4.
Also, third law tells us that R1 = R2 in magnitude and is opposite in direction. (figure 3.5)
R1 = R2 = R
Here since 150 N force acts on the 10 kg mass and only R acts on the 5kg mass. For motion in 5 kg only R is responsible. We can write the initial equation as:
F = 150 = (10+5) a i.e., 150 = 10a + 5a
Here 10a is force experienced by 10 kg mass and 5a is experienced by 5 kg mass.
R = 5a a = 10m/sec2 => R = 50 N
Net force experienced by 10kg block is (150 – R) = 10a
150 – R = 10 × 10 = 100 N => R = 50 N
Therefore we get R = 50N for both blocks. Hence we find “action and reaction are equal and opposite”. Now net force on the body of 10 kg mass is 100 N & Net force on the body of 5kg mass is 50N and on the interface action and reaction are both equal and also are equal to force experienced by second body.