Moving lift in a shaft

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Suppose a lift is moving in a shaft upwards with a velocity a. Then why does the net acceleration of the lift become (a+g)? Life suppose a liquid of denisty ρ is kept at rest in a beaker. In this case as pressure changes only in horizontal direction so we can write:
dP/dx=0 and dP/dy=-ρg

But, soppose the beaker is accelerated and it has components of acceleration ax and ay in the x and y directions respectively, the above above equations in this case reduce to
dP/dx=-ρax
dP/dy=-p(g+ay)

Why does addition of accelerations in the y-direction happen? Shouldn't we subtract g from ay as both of them are in opposite directions?

I am posting two attachments. First one is for an accelerating cart with a bob hanging inside it.
Second is for an accelerating beaker with liquid inside it. I want to know why does the liquid incline from top left to down bottom when we accelerate the beaker towards the right. Is it analogous to cart situation in any way? If yes then how?
 

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  • #2
tiny-tim
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hi andyrk! :smile:
I want to know why does the liquid incline from top left to down bottom when we accelerate the beaker towards the right. Is it analogous to cart situation in any way? If yes then how?
yes

in both cases, you can draw a force diagram in the accelerating frame of reference

in both cases, the acceleration (of the bob or of the water) is zero

in both cases, there is the actual gravity, mg vertically downwards

in both cases, there is a fictional force, -ma, opposite to the rest-frame acceleration a

so there is a total force mg -ma, downward and to the left in both cases, which you can think of as being an "effective gravity" …

the effective gravity makes the bob point down and left, and make the water surface perpendicular to that :wink:
 
  • #3
collinsmark
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Suppose a lift is moving in a shaft upwards with a velocity a.
(Boldface mine.) I think you mean "acceleration," a.

Then why does the net acceleration of the lift become (a+g)?
Speaking strictly in terms of Newtonian mechanics (not relativity), It doesn't. :smile: The "net" acceleration is a, because it is implied that the net force acting on the lift is ma (if it was any different, the lift would be accelerating at some other rate besides a).

Newton's second law states,

[tex] m \vec a = \sum_n \vec F_n [/tex]
meaning an object's mass times its [net] acceleration is the sum of all forces acting on that object.

In our lift example, can you quantify all the forces involved that are acting on the lift? A free body diagram (FBD) might help. One of the forces has a magnitude mg (where m is the mass of the lift and its contents), pointing down. There is another force involved pointing up (presumably from the tension of the cable). What is the magnitude of this other force? (Hint: it's not ma.)

Life suppose a liquid of denisty ρ is kept at rest in a beaker. In this case as pressure changes only in horizontal direction so we can write:
dP/dx=0 and dP/dy=-ρg

But, soppose the beaker is accelerated and it has components of acceleration ax and ay in the x and y directions respectively, the above above equations in this case reduce to
dP/dx=-ρax
dP/dy=-p(g+ay)
Sounds reasonable so far.

Why does addition of accelerations in the y-direction happen?
It's still Newton's second law again, however this time it's distributed through the liquid.

Shouldn't we subtract g from ay as both of them are in opposite directions?
Don't confuse the net force, and thus the "a" in the net ma, with the individual forces involved.

In order for the beaker and contents to accelerate at [itex] \vec a = a_x \ \hat x + a_y \ \hat y [/itex], there must be some other force involved besides just [itex] -mg \ \hat y [/itex]. If you quantify that other force, I think the answer will become more clear.

I am posting two attachments. First one is for an accelerating cart with a bob hanging inside it.
Second is for an accelerating beaker with liquid inside it. I want to know why does the liquid incline from top left to down bottom when we accelerate the beaker towards the right. Is it analogous to cart situation in any way? If yes then how?
Yes, they are related. Can you see why they are now, after considering all the forces involved?

[Edit: tiny-tim beat me to it.]
 
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  • #4
Your formatting is very odd and hard to understand but I'm going to take a stab at your question. I think what you're really trying to solve for the lift going up is the force acting on it if there was a hypothetical person inside of it. If your lift was just sitting there the force acting on it would be just F=mg. Now imagine you're in that elevator, and it starts moving. You will feel the ground push up when it starts because the lift is trying to oppose the gravitational force, the force is greater than it was before. The total force is equal to the gravitational force plus the force of the lift going up so F=mg+ma. When you simplify you get F=m(g+a) and that's where you get your (g+a) from.

The beaker and the cart in your other example are the exact same scenario. The objects in the example are simply experiencing a force due to the acceleration of the cart/beaker.
 
  • #5
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The total force is equal to the gravitational force plus the force of the lift going up so F=mg+ma.
WHY? (By total force I assume you mean the total force on the body in the lift)
And if we do this and apply newton's laws, we say that Net Force is mass times acceleration
So, F=mg+ma=ma...so mg=0? Got confused here! I think the net force on the body in the lift should only be ma and not m(a+g).. the m(a+g) force would come as follows(according to me):
In order to accelerate the body with an acceleration of a upwards, we first need to over come the force of gravity. So mg (upwards comes into place. and then we need to accelerate it upwards so thats where ma comes in (upwards) Adding them we get m(g+a) which would also be equal to the normal reaction force. But then I have another doubt that we know that Normal reaction is a reaction force between two objects. So as N=m(g+a) here, so does that mean the body in the lift exerts a force of m(g+a) on the floor of the lift?

Don't confuse the net force, and thus the "a" in the net ma, with the individual forces involved.

In order for the beaker and contents to accelerate at [itex] \vec a = a_x \ \hat x + a_y \ \hat y [/itex], there must be some other force involved besides just [itex] -mg \ \hat y [/itex]. If you quantify that other force, I think the answer will become more clear.


Yes, they are related. Can you see why they are now, after considering all the forces involved?

[Edit: tiny-tim beat me to it.]
You mean to say that we firstly need to overcome the force of gravity so that where the mg omes from and then we need it accelerating upwards so thats where ma comes from?

hi andyrk! :smile:


yes

in both cases, you can draw a force diagram in the accelerating frame of reference

in both cases, the acceleration (of the bob or of the water) is zero

in both cases, there is the actual gravity, mg vertically downwards

in both cases, there is a fictional force, -ma, opposite to the rest-frame acceleration a

so there is a total force mg -ma, downward and to the left in both cases, which you can think of as being an "effective gravity" …

the effective gravity makes the bob point down and left, and make the water surface perpendicular to that :wink:
Can you explain it without using pseudo forces? Pseudo forces is a made up concept firstly. So it doesn't actually answer it completely logically.
 
  • #6
collinsmark
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So as N=m(g+a) here, so does that mean the body in the lift exerts a force of m(g+a) on the floor of the lift?
Yes, that's right. :smile: Of course in this context we assume that the m is the mass of something inside the lift, and not necessarily the entire lift itself and everything in it.

If you were to place a simple, bathroom scale in the lift, between the floor of the lift and something heavy on top of it with mass m, the force the scale would read m(g + a). That's what simple, bathroom scales really do: they measure the normal force.

You mean to say that we firstly need to overcome the force of gravity so that where the mg omes from and then we need it accelerating upwards so thats where ma comes from?
Essentially, yes. That's what I mean.

Let's go back to defining m as the mass of the lift itself and everything in it.

You've initially defined a as the acceleration of the lift and everything in it, and that acceleration, a, is pointed up (assume "up" is positive). We know that the force of gravity is mg, pointed down (down is negative). But there's another force, T, the tension on the cable pulling the lift up. Calculate the magnitude and direction of of the tension, T.

ma = -mg + T

The same idea applies if we then use the "something on the bathroom scale, on the floor of the lift" scenario. But instead of calculating the tension "pulling" on something, we calculate the force of the bathroom scale "pushing" on something.
 
  • #7
tiny-tim
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hi andyrk! :smile:

(just got up :zzz:)
Can you explain it without using pseudo forces? Pseudo forces is a made up concept firstly. So it doesn't actually answer it completely logically.
yes:

if the real forces are F, and the acceleration of the vehicle (etc) is a and and the acceleration of the bob (etc) is b,

then in the accelerated frame of reference newton's second law for the bob is

F - ma = mb

(because relative acccelerations add)

but in the rest frame, it is

F = m(b + a) :wink:
 

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