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Suppose i push a box with my hand which has a mass=1kg. I apply a force of 2 newtons and it will accelerate at 2m/s^2. But, according to the third law, the box applies a force on me, which is also equal to 2 newtons, in the opposite direction which will cause ME to accelerate in the backward direction. But this is not the case, i am accelerating forward ( my hand ) but that force ( the reaction of the box ) is forcing me to move backward, and that force is equal to 2 newtons. So i MUST apply a force larger than 2 newtons to move forward, but i am not doing this, i am still applying a force of 2 newtons and accelerating forward.

Why?

PS- I know the forces donnot cancel out as they act upon different objects.

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  • $\begingroup$ The force you apply is the force you apply, but the resulting movement is less than the distance that you move your arm forward because the center of mass of your body moves backwards a little. It is this lapse in movement that you perceives as "inefficient" rather than a change in force. If you calculate this formally then you end up with what is known as the "rocket equation", which basically tells you that rockets incur an exponentially rising "fuel cost" as we try to boost the payload to ever higher velocities. $\endgroup$ – CuriousOne Jan 23 '16 at 6:54
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The answer is simply that there is some other force that keeps you away from accelerating backwards. A simple example here: imagine yourself on a frictionless icerink. If you push someone, then it is true that you and the other person accelerate in opposite directions. However, when you are standing on earth and pushing a box, the box receives your force and accelerates forward, you receive the backward force from the box plus the forward static friction from the ground. The two forces on you cancel out and leave you static!

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  • $\begingroup$ So if i push the box when i am standing on a frictionless surface, then will i just fall? $\endgroup$ – Aaryan Dewan Jan 23 '16 at 6:19
  • $\begingroup$ Well if you push the box in x-direction you would just move in the opposite direction (-x). I don't see how you can fall by pushing a box. $\endgroup$ – Zhengyan Shi Jan 23 '16 at 6:21
  • $\begingroup$ As i stated i am standing on a frictionless surface, then the reaction of the box will cause me to fall down! Isn’t it? $\endgroup$ – Aaryan Dewan Jan 23 '16 at 6:23
  • $\begingroup$ in which direction do you push the box... I am confused. $\endgroup$ – Zhengyan Shi Jan 23 '16 at 6:24
  • $\begingroup$ if i push the box in the left direction and then it will push me in the right one, if i am standing on a frictionless surface, will i fall down? ( as there is no force that will cancel the reaction of the box ) ? $\endgroup$ – Aaryan Dewan Jan 23 '16 at 6:50
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Same force is applied to both you and the box( seperate objects) The behaviour of each of these objects then is given according to the Second law...where for each body this force may be a part of all the forces acting on the body. In case of a friction less ice rink(assuming) the distance moved by Centre if mass of each object will be inversely proportional to their masses.

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