Perceived Weight In Elevator

[MedPR]

As far as I know, this is a pretty classic example used to help explain Force so hopefully a lot of you are familiar with it.

TBR does an equation derivation to show your "apparent weight" when you are in an elevator that is accelerating upwards.

S-W=ma, where S is the force of the scale pushing up on the person and W is the person's weight (mg). TBR says the net force is the force "up" minus the force "down"

S=W+ma

Substitute for W/g for m (since W=mg)

S=W(1+a/g)

So, when in an elevator accelerating upwards, your apparent weight is greater than if you were standing flat on the ground.

TBR then goes on to say that, if the cable in the elevator were cut, and you were now in a free-falling elevator, you would have S=W(1-g/g)

I understand that the acceleration is now equal to g, since there is no "upward" force, but how would you derive that equation, since there is no "upward" force, what is your original equation when accelerating downwards? (The original equation when accelerating upwards was S-W=ma)

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[chiddler]

Why don't you write it as the following:

F = mg + ma = m (g + a)

for upward elevator movement.

For downward,

F = mg - ma = m ( g - a )

and when a = g, Fnet = 0 which explains the weightlessness and zero gravity experienced during freefall.

If you're asking how I found the equations, it's because force down is gravity of course. Force up can be thought of as a rope attached to the person that pulls on them. If you're accelerating at 2 m/s^2, then the rope is pulling (2)(your weight). You can add this value to the weight if you're moving up, or subtract if moving down.

Is this what you are asking?

 

[Pisiform]

Why don't you write it as the following:

F = mg + ma = m (g + a)

for upward elevator movement.

For downward,

F = mg - ma = m ( g - a )

and when a = g, Fnet = 0 which explains the weightlessness and zero gravity experienced during freefall.

If you're asking how I found the equations, it's because force down is gravity of course. Force up can be thought of as a rope attached to the person that pulls on them. If you're accelerating at 2 m/s^2, then the rope is pulling (2)(your weight). You can add this value to the weight if you're moving up, or subtract if moving down.

Is this what you are asking?

👍👍

However, what they did is:

Since the object is accelerating downward:

Fnet = ma is down

ma = mg - S

mg = mg - S ( a = g because object is in free fall)

S = mg - mg

S = W - W

S = W - W (g/g) ( i don;t know why they put g/g since g/g will be 1. I think its to make the equation look like previous

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one (S=W(1+a/g)) so that it can easily be compared)

S = W (1 - g/g)

 

[grapepopsicle]

I like to think of it like this: when the elevator is accelerating upwards (say at 2 m/s^2), it has to overcome both your weight (due to gravity) and provide enough force to be give an upward acceleration of 2. If you are accelerating down, the elevator does not have to provide as much force because it doesn't have to completely cancel out the acceleration due to gravity.

 

[deleted393595]

1. Draw force vectors
2. Set up basic equations
3. Substitute in given quantities
4. Solve