Need help finding the acceleration for this question

[Derpicus]

Homework Statement

A scientist in a special elevator in a research facility on Earth, where g = 9.8m/s^2 [down], is standing on a bathroom scale, calibrated in Newtons. The elevator can be given various vertical accelerations. The scientist weighs himself and finds that his weight has apparently doubled.What is the acceleration of the elevator?

Homework Equations

Fg = mg
Fnet = Fn + Fg
F = ma
W= mg

The Attempt at a Solution

Apparent Weight
g= 9.8/ms^2
W = -Fn
W= -m(a-g)
W= m(g-a)/m(g)
-W/m(g) = a

True weight
Fg = mg
Fg = m(9.81m^s^2)

Horziontal
Fnet = 0
Vertical
Fnet = Fn + Fg
ma = Fn + Fg
Fn = ma - Fg

But we still have three unknown variables. We don't know his true weight or his apparent weight; so I don't know how I can even begin to solve for the acceleration. I know in Physics I'm not supposed to rely on the examples in the text as I'm supposed to be learning about the concepts but in any question similar I've been given the mass + acceleration and in some cases gravity when it isn't 9.81 m/s^2.

 

[berkeman]

Homework Statement

A scientist in a special elevator in a research facility on Earth, where g = 9.8m/s^2 [down], is standing on a bathroom scale, calibrated in Newtons. The elevator can be given various vertical accelerations. The scientist weighs himself and finds that his weight has apparently doubled.What is the acceleration of the elevator?

Homework Equations

Fg = mg
Fnet = Fn + Fg
F = ma
W= mg

The Attempt at a Solution

Apparent Weight
g= 9.8/ms^2
W = -Fn
W= -m(a-g)
W= m(g-a)/m(g)
-W/m(g) = a

True weight
Fg = mg
Fg = m(9.81m^s^2)

Horziontal
Fnet = 0
Vertical
Fnet = Fn + Fg
ma = Fn + Fg
Fn = ma - Fg

But we still have three unknown variables. We don't know his true weight or his apparent weight; so I don't know how I can even begin to solve for the acceleration. I know in Physics I'm not supposed to rely on the examples in the text as I'm supposed to be learning about the concepts but in any question similar I've been given the mass + acceleration and in some cases gravity when it isn't 9.81 m/s^2.

It seems you are overthinking this, or I'm missing something. If his weight doubles, what total acceleration is he experiencing? How much more acceleration is that compared to just the acceleration due to gravity?

 

[cnh1995]

When the lift accelerates upwards, we feel our weight has increased due to increase in the normal force. From stationary frame(outside the lift),the net acceleation is upward, normal force is upward and weight is acting downwards. Draw FBD, write the equation and see.

 

[SteamKing]

Homework Statement

A scientist in a special elevator in a research facility on Earth, where g = 9.8m/s^2 [down], is standing on a bathroom scale, calibrated in Newtons. The elevator can be given various vertical accelerations. The scientist weighs himself and finds that his weight has apparently doubled.What is the acceleration of the elevator?

Homework Equations

Fg = mg
Fnet = Fn + Fg
F = ma
W= mg

The Attempt at a Solution

Apparent Weight
g= 9.8/ms^2
W = -Fn
W= -m(a-g)
W= m(g-a)/m(g)
-W/m(g) = a

True weight
Fg = mg
Fg = m(9.81m^s^2)

Horziontal
Fnet = 0
Vertical
Fnet = Fn + Fg
ma = Fn + Fg
Fn = ma - Fg
[/B]
But we still have three unknown variables. We don't know his true weight or his apparent weight; so I don't know how I can even begin to solve for the acceleration. I know in Physics I'm not supposed to rely on the examples in the text as I'm supposed to be learning about the concepts but in any question similar I've been given the mass + acceleration and in some cases gravity when it isn't 9.81 m/s^2.

Well, you've managed to thoroughly confuse yourself.

If the elevator is not moving, then the scale is going to read the scientist's weight, true?

There's a simple relationship between the mass of the scientist m, which does not vary, and his weight W, W = m ⋅ g, where g = 9.81 m/s²

Now, the scientist is still standing on the scale, but the elevator is moving. The scientist glances at the scale and finds that it reads 2 ⋅ W. Since m does not vary (the scientist has not grown an extra head or body in the meantime), what must the acceleration of the elevator be in order to produce this new reading on the scale? Remember, g = 9.81 m/s² is always present at the surface of the earth.

 

[Derpicus]

Well, you've managed to thoroughly confuse yourself.

If the elevator is not moving, then the scale is going to read the scientist's weight, true?

There's a simple relationship between the mass of the scientist m, which does not vary, and his weight W, W = m ⋅ g, where g = 9.81 m/s²

Now, the scientist is still standing on the scale, but the elevator is moving. The scientist glances at the scale and finds that it reads 2 ⋅ W. Since m does not vary (the scientist has not grown an extra head or body in the meantime), what must the acceleration of the elevator be in order to produce this new reading on the scale? Remember, g = 9.81 m/s² is always present at the surface of the earth.

So based on your help; and the person above we know when a person is stationary has true weight and if an increase in Fn causes a increase in weight. So I'm going to say the acceleration would be a= 9.81 m/s<sup>2. I'm still confused on what formula I should be using.

Would I exclude mass in the formula? Since I understand the free body Diagram my Fn counteracts the Fg then with the addition of acceleration the [up] direction is greater which would case his apparent weight to be double.

But like I said I'm not sure how I'm supposed to play with my formula to go about solving for acceleration. Like I feel like I'm missing out on something fundamental and I'm sorry that I have to ask for another hint but I'd greatly appreciate it.</sup>

 

[Derpicus]

It seems you are overthinking this, or I'm missing something. If his weight doubles, what total acceleration is he experiencing? How much more acceleration is that compared to just the acceleration due to gravity?

His acceleration would be double that of the gravity. Which would be 9.81 x 2 = 19.62m/s^2

So would I really just write a = 2(9.81m/s^2) ??

 

[cnh1995]

His acceleration would be double that of the gravity. Which would be 9.81 x 2 = 19.62m/s^2

So would I really just write a = 2(9.81m/s^2) ??

When going up, net acceleration of the man is same as that of the elevator.
So,
ma=F_Normal-mg
So, F_normal=ma+mg
If the man sees his weight doubled, then the normal reaction should also be twice the normal reaction when the lift is moving up with constant velocity(i.e.a=0), which is N_a=0=mg.
Using this, you don't get a=2g.

 

[RedDelicious]

I think you're getting confused because you don't understand your free body diagram. It will also help to keep track of your signs.

If an object is resting on a table, the normal force will be the force pushing upward on the object and, because the object is at rest, it must be equal and opposite to the gravitational force (otherwise it would be accelerating, not resting). So for an object resting on a surface, the net force on it is zero and we have

F = Fn + Fg = 0 ( case when the object isn't accelerating)

Now what happens if you give that object a net acceleration? F is no longer zero. F is then the force acting on it, which we know as ma. So if we go back to our relationship.

F = Fn + Fg = ma

Solving for Fn gives

Fn = F - Fg

Do you know what each of the terms represent? That is, do you know which term is his acceleration, which term is his apparent weight, and which term is his actual weight? Furthermore, how is his acceleration relevant to the elevator's acceleration? If you get confused, think about the more simple resting book example and how they were connected in that case and what must have changed.

 

[cnh1995]

https://lh5.googleusercontent.com/proxy/fWecjLEoPAxBElhgAQw1aLH-iajef_ymX_kl5YzUCnXriVTVz9_FXy6HobyMGysBxwi5UyId2uQMsHYSrX-j4B4MRu6ZxXTON_gLff4-VKLv3znZlPAcoqYjVXCcjvRo_rSOk4Dg62Y7ow=w456-h320-nc

When going up, net acceleration of the man is same as that of the elevator.
So,
ma=FNormal-mg
So, Fnormal=ma+mg
If the man sees his weight doubled, then the normal reaction should also be twice the normal reaction when the lift is moving up with constant velocity(i.e.a=0), which is Na=0=mg.

When accelerating upwards,
ma=W_apparent-mg
W_apparent=ma+mg
Since W_apparent=2mg,
ma=2mg-mg=mg
That means, a=g

What is the acceleration of the elevator?

It should be g instead of 2g.

 

[Derpicus]

https://lh5.googleusercontent.com/proxy/fWecjLEoPAxBElhgAQw1aLH-iajef_ymX_kl5YzUCnXriVTVz9_FXy6HobyMGysBxwi5UyId2uQMsHYSrX-j4B4MRu6ZxXTON_gLff4-VKLv3znZlPAcoqYjVXCcjvRo_rSOk4Dg62Y7ow=w456-h320-nc

When accelerating upwards,
ma=W_apparent-mg
W_apparent=ma+mg
Since W_apparent=2mg,
ma=2mg-mg=mg
That means, a=g

It should be g instead of 2g.

Thanks for going above and beyond to help me; I'm currently re-reading all of the chapter to make sure I understand the fundamental concepts that I feel like I've been lacking since my Free body diagram has been incorrect and this problem really brought it to light. I know it's not much but I just want you to know I'm really grateful for the help you and everyone else has given me.

That picture really helped me grasp the concept of the elevator. I felt comfortable finding apparent and true weight when given 2 know variables then plugging them into a formula but the doubling of mass when only given acceleration really just didn't click. I know I'm not the brightest so I really appreciate the extreme amount of help you guys provided.