# Newton's Third Law in an elevator

Some misconception I'd like cleared up.

From what I understand, when an elevator accelerates downwards with you, there must be a net force acting downwards on you. Since your true weight remains the same, the net force arises because the elevator exerts a smaller normal force on you which is matched from Newton's Third Law by an equal force downwards on the elevator (this downwards force is your apparent weight measured by weighing scales I believe). Is this correct? What causes you to not exert your full weight despite making contact with the elevator floor?

Edit: Thought about my first point a bit more and I understand where I went wrong so I removed it.

• You really need to carefully read through several chapters regarding Newton's laws and gravity, in a good high school physics book. Your current misconceptions imply that any answer you get here will be somewhat misinterpreted. Dec 23 '18 at 21:28
• Your body is accelerating downward at the same acceleration rate as the elevator. So there must be a net downward force on you. The downward force of gravity on you doesn't change. So the upward force on you from the elevator must be less. Dec 23 '18 at 23:51

Here is a simple explanation I used to understand it myself. When the elevator is accelerating downwards, it's moving away from you. I mean, of course, you're accelerating at $$9.81ms^{-2}$$ downwards, but lets say the elevator is accelerating downwards too at $$1ms^{-2}$$, then effectively, the weight you will measure from the weighing machine in the elevator is $$8.81ms^{-2}\times M$$.

So, what happens if the elevator were to accelerate downwards at acceleration of $$>9.81ms^{-2}$$? You would be essentially "levitating" from the POV of the elevator. Of course, you will end up crashing to the top of the elevator.

However, that explanation is probably not very scientific, but its really just for easy understanding. A more "physical" explanation of this would be: You are accelerating at $$1ms^{-2}$$ downwards (the acceleration of the elevator). However, without the elevator, you are accelerating at $$9.81ms^{-2}$$ downwards! Where is the missing upward $$8.81ms^{-2}$$? You have your answer.

• if the person with mass m is accelerated downwords $-g\cdot m$ than by third law the ground will give a force same in magnitude corresponding to $g\cdot m$. As the ground is now moving down with $-a$ we get another force, but do we take for that the mass of the accelerator M? is then the total force on the accelerator $m\cdot g - M \cdot a$ and the total force on the person $-g\cdot m + a\cdot M$? Jun 8 at 8:45

1) Gravity does. In general the acceleration gravity causes will be greater than the downward acceleration of an elevator, keeping you on the floor of the elevator.

2) Your 'weight' is the force you exert on the elevator floor, and is equal in magnitude to the normal force the elevator floor exerts on you. However, your weight does not stay the same when the elevator accelerates downwards. Your mass does. Your weight in an elevator accelerating downwards will be smaller than your weight when standing on a non-accelerating surface, due to newtons third law: $$F = ma$$. In this case:

$$F = ma = m(g-a_{elevator})$$ in the elevator going downwards, and simply $$F = mg$$ When standing on a non-accelerating surface.

What causes you to not exert your full weight despite making contact with the elevator floor?

This winds up being a slightly complicated interaction with four things to keep track of. First there is gravity, second there is the normal contact force with the floor which can be modeled as a very stiff elastic spring, third the acceleration of the elevator, and fourth your acceleration.

Now, gravity is a fixed force, and the mechanism of the elevator is constructed to deliver a safe acceleration regardless of the load as long as it is within the designed limits. So you can think of the acceleration of the elevator as being fixed also. So the only variable things are the elastic force and your acceleration.

If your acceleration exceeds the elevator’s acceleration then the floor will compress and the elastic force will increase. As the floor is represented as a very stiff spring the force increases very sharply with a very small compression. The sharply increased elastic force correspondingly reduces your acceleration until it matches that of the elevator.

If your acceleration is less than the elevator’s acceleration then the reverse happens. The sharply reduced elastic force increases your acceleration until it again matches that of the elevator.

Either way the floor deforms enough so that the normal force between you and the elevator is the exact value necessary so that your acceleration matches the elevator’s acceleration.