scale and elevator question

[grayfalcon89]

I'm in first year of physics in my high school but I really don't get this concept. Say I'm on scale and going up on the elevator with a positive acceleration. Then my force is

 

F = m(a+g)

 

But my teacher told me that it's different if the elevator is goin' up and slowin' down or if the elevator is goin' down and speedin' up. I do not get those differences in terms of Force. Can anyone explain it? Example will be nice. Thnx!

[Mr Skeptic]

Think about this: what would the scale say if you were in free fall (ie, falling at g)

[iNow]

When driving in a car, do you feel pushed against the seat more when a) the car is speeding up, or b) the car is slowing down?

 

Your being pushed against the seat of the car is roughly the same as being pushed against the scale (which is pushed against the elevator floor).

[swansont]

I'm in first year of physics in my high school but I really don't get this concept. Say I'm on scale and going up on the elevator with a positive acceleration. Then my force is

 

F = m(a+g)

 

But my teacher told me that it's different if the elevator is goin' up and slowin' down or if the elevator is goin' down and speedin' up. I do not get those differences in terms of Force. Can anyone explain it? Example will be nice. Thnx!

 

 

Slowing down or speeding up is the "a" in that equation, and a is a vector, so it has a direction (sign) that may or may not be the same as g.

 

If you go up at a constant speed, a=0, and you read your usual weight (i.e the "normal" normal force) on the scale.

[MrMongoose]

But both accelerations are implicitly in the same direction as g, so the force on the scale should be the same in both cases...

[swansont]

But both accelerations are implicitly in the same direction as g, so the force on the scale should be the same in both cases...

 

In general that's not true. g is always down, but you can accelerate in either direction.

[MrMongoose]

But "going up and slowing down" implies a downward acceleration, as does "going down and speeding up".

[swansont]

But "going up and slowing down" implies a downward acceleration, as does "going down and speeding up".

 

Which are both different from going up with a positive acceleration (or can be, depending on your coordinate system), which was the first scenario in the OP. And the difference is in the sign of the external acceleration with respect to g. (I'm pretty sure from your latest post that you aren't comparing the same situations that I am.)

[MrMongoose]

OH. I thought he was comparing the second and third cases to each other rather than to the first case... I need to work on my English!

[swansont]

OH. I thought he was comparing the second and third cases to each other rather than to the first case... I need to work on my English!

 

Don't worry about that too much. That was my first reaction, too.

[grayfalcon89]

@Mr Skeptic: Hmm... Falling in free fall... I'm actually not that sure. I thought it would increase but then by F = ma, a is constant in free fall (isn't it just 9.80 m/s^2)?

 

@iNow: I get that example... It's like because of my inertia right? I mean, inertia is not a force but that's still why I have that force on other side (the one that pushes me back to the seat).

 

@swansont: I watched a video on F = ma yesterday. It says a and F have to be in same direction or something... Anyway, I said that but I"m not sure if that's related to anything. As with what you said, so if my direction of movement and acceleration is opposite, I subtract it from 9.80 m/s^2?

 

Thank you!! And sorry for bad explanations on problem.

[swansont]

@swansont: I watched a video on F = ma yesterday. It says a and F have to be in same direction or something... Anyway, I said that but I"m not sure if that's related to anything. As with what you said, so if my direction of movement and acceleration is opposite, I subtract it from 9.80 m/s^2?

 

It's true that a and F have to be in the same direction. But your scale measures the normal force, N. So if the only things happening are that you are accelerating and standing on the scale, two forces will add to give the net force.

 

[math]\vec{F} = m\vec{a} = \vec{N} + m\vec{g}, so \vec{N} = m(\vec{a}-\vec{g})[/math] (if a is up the signs will be opposite and that introduces another "-" sign, so you get a+g as scalars; you feel like you weigh more when accelerating up)

 

In the case of freefall, a and g are the same, so the net is zero — you feel weightless