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If you have a man in an elevator accelerating upwards, why does the elevator experience a normal reaction downwards rather than just the man's weight downwards.

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The man also has to accelerate upwards and so will have a net upward force on him.
So the force on the man due to the elevator (normal reaction) must be greater than the weight of the man.

The force on the man due to the elevator is equal in magnitude but opposite in direction to the force on the elevator due to the man and so is greater than the weight of the man.
This is just an application of Newton's third law.

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Draw a free body diagram for man and observe that $N-Mg=Ma$ Where $a$ is acceleration of lift upwards, $M$ is mass of the man and $N$ is normal force by the lift on man upwards.

Now this leads to $N=Mg+Ma$. Hence mathematically (as well as physically (Newton's third law)) lift feels normal force downwards as not only $Mg$(Weight of man) but more than that.

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  • $\begingroup$ So would I be right in saying that the lift feels the reaction force downwards due to N3L; lift exerts reaction force on man due to his weight => lift feels reaction force downwards as an equal and opposite reaction. $\endgroup$ – thisisaphysicsquestion Mar 27 '17 at 20:39
  • $\begingroup$ @thisisaphysicsquestion If lift was still, then the normal force would be exactly equal to the weight of the man. But here lift is accelerating. So normal force is due to weight of the man and the acceleration of the lift. $\endgroup$ – Error 404 Mar 28 '17 at 4:15
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The man's weight is the force of gravity on the man. It has nothing to do with the elevator. The only contact or interaction between the man and the elevator is at the bottom of the soles of his shoes. That's the only place a force occurs between the man and the elevator. The only force that the elevator experiences that is due to the man is the normal force.

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  • $\begingroup$ You're saying the elevator doesn't experience the man's weight? If you're including the weight in the normal force, you may want to make that clear. $\endgroup$ – electronpusher Mar 28 '17 at 0:17
  • $\begingroup$ @electronpusher I am responding to the language of the OP. He or she mentions that the elevator feels the man's weight. It does not. The elevator feels a normal force. In turn the man experiences normal force on his feet and weight. The magnitude of the normal force can be calculated from acceleration and weight, and that can be used (Newton's Third) to find the force, the normal force experienced by the elevator. $\endgroup$ – garyp Mar 28 '17 at 2:10
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In Newton's 3rd law there is a force/reaction pair. Those forces must be -Equal size -Opposite direction AND -Involve only two objects -Be of the same type of force.

For instance, the 3rd law pair of a gravitational force must be another gravitational force, not a normal force. A gravitational force is a non-contact force, whereas a normal force requires contact.

The normal force is between the floor and the person, the gravitational force is between the earth and the person. If the elevator is not accelerating then Fn = Fg, but the equivalence is NOT due to the 3rd law, it is because ∑F=0 in the 2nd law.

So the end result is:

Fg pulling down on the person is paired with Fg of the person pulling up on the earth and these two numbers are equal. Fn of the elevator pushing up on the person is paired with Fn of the person pushing down on the elevator. Since Fn on the person is greater than Fg on the person, then Fn from the person on the floor is greater than the person's weight.

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