Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Dr Amrutha on 22/04/2016 09:42:42
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It's zero [;)] But I don't know how scientifically.Can someone explain this to me?
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Imagine that you have a mass of 60kg, and are standing on a set of glass bathroom scales, in a stationary lift/elevator. Your mass of 60kg, acting under a gravity of 9.8 m/s2 pushes down on the scales with a Force of 60x9.8=588 Newtons, and the scales display that force as a weight in pounds, stones/pounds or kilograms force (depending on the country).
Now imagine that the lift enters free fall, Your mass of 60kg, acting under a gravity of 9.8 m/s2, accelerates downwards at 9.8m/s2 (as do the lift and the scales). Let's say that random air currents place the scales 1cm above the lift floor, and your feet 1cm above the scales as you fall. There is no pressure of your feet on the scales, and no pressure of the floor on the scales, so the scales will read zero*.
Hence, your weight, in a freely falling lift is zero.
But watch out - your weight reappears all at once to make up for it's temporary absence, when you reach the bottom of the lift well!
*Actually, the scales will read slightly less than zero, because the scales are calibrated to ignore the mass of the glass plate on the scales - only the glass plate now weighs nothing.
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Let's say that random air currents place the scales 1cm above the lift floor, and your feet 1cm above the scales as you fall. There is no pressure of your feet on the scales, and no pressure of the floor on the scales, so the scales will read zero*.
I do not understand how you relate pressure to zero weight in the above statements.Can you please elaborate more on that ?
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I do not understand how you relate pressure to zero weight in the above statements.Can you please elaborate more on that ?
Pressure is force/unit area so you can read force where Evan has written pressure without changing the sense of what he was saying.
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It's zero [;)] But I don't know how scientifically.Can someone explain this to me?
The weight of an object can be and is defined in two ways;
1) the force due to gravity on a body
2) the force required to hold a body at rest, i.e. to prevent it from falling, in a gravitational field.
Let's use the second definition since most of us think that way. When you're in free fall there are no forces other than gravity acting on your body. The force of gravity is not something you can feel when you're in free fall because it causes every single particle in your body to fall at exactly the same rate. In your frame of reference, i.e. the frame in free fall, there is, for all practical purposes, no gravitational field. In fact in general relativity there really isn't a gravitational field in a free fall frame. It's said to have been transformed away.
Was that helpful? [:)]
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Youngsters may never have experienced this, but anhyone ofer 50 has probably push-started a car at some time. Initially, you can feel the force you are applying to get the car moving, and you continue to provide a force to accelerate the car to the point where the engine starts. Suddenly, the car is moving at the same speed as you, and the push force is zero - you might even fall over.
In the lift situation, gravity is trying to accelerate you downwards but the floor is stopping you, just like the stationary car was opposing your push. As the lift accelerates, at some point the floor will be moving and accelerating at the rate that gravity demands, so once again you don't feel any force between you and the floor.
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What happens if you factor in air resistance? My limited knowledge of physics wants to tell me that the lift will experience drag at an increasing magnitude as it falls (as it accelerates). Meanwhile the person inside the lift is contained within their own little box of air, does not experience drag. So the person and the lift will initially accelerate at the same rate, but then the lift will decelerate while the person continues "falling". The difference in force will then, again be felt by the person's feet as "weight".
But, I could be just making all of that up.
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Sounds logical to me. The air in the lift would be moving with the lift and not affecting terminal velocity of passenger who would be trying to overtake the lift.