A glass of water on an inclined plane

[Swap]

Homework Statement

A Glass of water is freely sliding down an inclined plane. The water surface has settled down after some initial disturbances. Which of the three cases is possible for the shape of the water surface:
1) It will remain parallel to the incline
2) The water surface will rise towards the direction of the acceleration parallel to incline
3) The water surface will rise away from the direction of acceleration parallel to incline.

The Attempt at a Solution

I think the answer is 3. It is because the water will experience a pseudo accln. away from the direction of the actual accln. parallel to the incline and another normal force acting on it perpendicular to incline. The resultant of this will produce a force that will cause the liquid surface to bend like it is stated in no. 3 as the liquid surface is always perpendicular to the force applied. But somebody said the answer is 1. So I'm not sure. ( N. B. I do not know the correct answer)

 

[Tusike]

Ask that person to explain exactly why he thinks so. I don't really know what is meant by "away from" or "towards" the direction of acceleration, so I can't decide between 2, and 3. All I can say is that if you in fact draw in the two accelerations of the water and add them as vectors, you will get that if, say, the cup's moving LEFT and down on the slope, the surface will look like some kind of / slash, perhaps a bit less steep.

 

[ehild]

The water surface never will be parallel with the incline. It is horizontal if the glass is steady or moves with constant velocity. Just try.
I tried to make my glass of water accelerate down on an incline and the water surface looked a bit higher at the back of the glass than at the front, that is the water level rose "away from the direction of acceleration"
I would try to explain the experiment with hydrostatic pressure and Pascal's law and would think of the forces acting to a volume element of water that moves with constant acceleration, together with the glass and the whole amount of water.

ehild

 

[Swap]

Yup I know I tried it experimenting it too. It gave me the same results. But the person was a professor and then when I asked the reason he said its some advanced topic, u won't know and didnt explain much. Thats why I was not sure.

 

[ehild]

You have learned about hydrostatic pressure and Pascal's law in fluids, don't you? If yes, I'll try to give some explanation if you are interested.

ehild

 

[Swap]

@ ehild :Yes I do know about it. Tell me how

 

[RoyalCat]

The water surface never will be parallel with the incline. It is horizontal if the glass is steady or moves with constant velocity. Just try.
I tried to make my glass of water accelerate down on an incline and the water surface looked a bit higher at the back of the glass than at the front, that is the water level rose "away from the direction of acceleration"
I would try to explain the experiment with hydrostatic pressure and Pascal's law and would think of the forces acting to a volume element of water that moves with constant acceleration, together with the glass and the whole amount of water.

ehild

A possible mental shortcut in this case would be to say that in hydrostatic equilibrium, the surface plane of the water will always be perpendicular to the gravitational acceleration.

Taking that notion into an accelerated reference frame, in which there is the d'Alembert force, [tex]\vec F_d = -mg\sin{\theta} \hat t[/tex] where [tex]\hat t[/tex] is the tangential direction going down the incline.

Taking the vector sum of this fictitious force and the force of gravity, we can find an equivalent gravity in this accelerated reference frame, and water, doing as water does, aligns itself neatly perpendicular to this direction. Make a small vector diagram and you'll immediately see how the water aligns itself.

A neat example of this method is if, let's say, you have some tank of water, and you rotate it about some axis, what shape will the water's surface take on?

 

[kuruman]

http://www.physics.umd.edu/lecdem/services/demos/demosc4/c4-12.htm

Cheers.

 

[Swap]

@ kuruman, thanks a lot buddy. I know my fault already. I assumed the initial water surface to be parallel to inclined plane but it should be horizontal.

 

[Dadface]

http://www.physics.umd.edu/lecdem/services/demos/demosc4/c4-12.htm

Cheers.

I was puzzling over this question and seeing the answer came as a surprise.Suppose the glass was filled to the brim and the slope got steeper,the increase being very slow and gradual so as to reduce any disturbance.When the slope reaches the the vertical the water surface will also be vertical.Ignoring frictional forces the glass and its contents would now be in free fall.Putting it another way we could fill a glass with water,put a cover over the top so that the water doesn't spill out,turn the glass through 90 degrees so the water surface is vertical and then drop the glass whilst simultaneously removing the cover.According to the answer above the water will stay in the glass.Of course the theory is ignoring other effects such as surface tension but it does seem odd.

Thanks for posting the site address kuruman.It looks like there's some good stuff on there and when I get some time I will have a browse.

 

[Dadface]

@ kumaran, thanks a lot buddy. I know my fault already. I assumed the initial water surface to be parallel to inclined plane but it should be horizontal.

Did you look at all of the pictures?When the glass was sliding its surface became parallel to the plane.

 

[Swap]

@ dadface : Yes, I did. Thats why I was able to point out my mistake. Accn. to figure the initial water surface is horizontal.

 

[ehild]

The water surface never will be parallel with the incline.

Shame of me... I knew that it becomes more parallel with the incline as in steady state, but I did not think that it becomes really parallel. :( Yes, it is like free fall...

ehild

 

[Tusike]

I'm sorry, I just can't believe that until someone gives me some kind of proof written down.

So far, I had a glass of water go down a gentle slope, and it was nowhere near parallel with the incline. I thought that fair enough, but perhaps you are right, and the only reason it didn't get parallel was because of the friction; so now I'm building a small car for the cup using my old metal erector set:) There will still be some friction, but it's supposed to get really close to parallel. We'll see. But still, if it does get parallel, I'd be really happy for a written proof, I just don't see how it can' happen.

 

[Tusike]

Yeah OK I am stupid. All you need to do is simply draw down the forces acting on the water... There's mgsin(alpha) pulling it parallel to the incline, -mgsin(alpha) acting as the inertial force exactly in the opposite direction. These two cancel each other out, so all we have left is mgcos(alpha) which is pulling the water perpendicular to the incline. And as we all now, the surface of the water is perpendicular to the forces acting on it, so with all that combined, we get that the surface will indeed be parallel with the incline.

(My experiment with the car also led to this conclusion... It was almost parallel, but that's only because of the friction. If the inertial force is less than mgsin(alpha), we would also get near parallel.)

 

[kuruman]

Yeah OK I am stupid. All you need to do is simply draw down the forces acting on the water... There's mgsin(alpha) pulling it parallel to the incline, -mgsin(alpha) acting as the inertial force exactly in the opposite direction. These two cancel each other out, so all we have left is mgcos(alpha) which is pulling the water perpendicular to the incline. And as we all now, the surface of the water is perpendicular to the forces acting on it, so with all that combined, we get that the surface will indeed be parallel with the incline.

(My experiment with the car also led to this conclusion... It was almost parallel, but that's only because of the friction. If the inertial force is less than mgsin(alpha), we would also get near parallel.)

Think about what you are saying here. If these two "canceled each other out" the acceleration down the incline would be zero and the water would move with constant velocity in which case the level of the water would be horizontal.

The easiest way to see what is going on is to transform to the non-inertial frame of the water. In that frame, the effective acceleration is the sum of two vectors, (a) g vertically down and (b) gsinθ directed up the incline. The sum of the two is the effective acceleration of the water and is gcosθ directed perpendicular to the incline.

Corollary: A cork floating in the water with, say, half its volume exposed in zero acceleration, will float with more than half the volume exposed when the water is accelerating down the incline.

 

[Tusike]

the effective acceleration is the sum of two vectors, (a) g vertically down and (b) gsinθ directed up the incline. The sum of the two is the effective acceleration of the water and is gcosθ directed perpendicular to the incline.

There's mgsin(alpha) pulling it parallel to the incline, -mgsin(alpha) acting as the inertial force exactly in the opposite direction. These two cancel each other out, so all we have left is mgcos(alpha) which is pulling the water perpendicular to the incline. And as we all now, the surface of the water is perpendicular to the forces acting on it, so with all that combined, we get that the surface will indeed be parallel with the incline.

I really don't see the difference between these two :) Since g vertically down can be split into mgsin(alpha) and mgcos(alpha), both of us are talking about the same forces...
I would like to point out, however, that when talking about the surface of the water, one most only take into account the forces that act upon the whole volume of the water.

By the way, there's something a bit off with our reasoning. In the frame of reference of the accelerating water, out of the two forces we mentioned, (a) and (b) acting upon the water, (a), being g vertically down isn't quite true, since mgsin(alpha) is the reason for the acceleration. (for example, if a car is accelerating to the left, in the frame of reference of the accelerating car, would you say there's a force acting on an object in the car to the left? no...). So actually (a) should be reduced to mgcos(alpha), (b) is the inertial force, -mgsin(alpha), and if we want to get the resultant force perpendicular to the incline, there must be some other force that's exactly mgsin(alpha). Perhaps the side of the cup? But is that a force acting on the whole volume of the water? I think so! Any ideas, or where did I go wrong?

 

[kuruman]

You reasoning is confused because you think of two frames of reference simultaneously.

Case I: In the inertial frame of reference (at rest with the incline)

The forces acting on the sliding mass are
1. Gravity
2. Normal force

Gravity can be broken into two components
1. Perpendicular to the incline that is canceled by the normal force
2. Parallel to the incline in the down direction that is canceled by nothing.

Conclusion: The mass accelerates relative to the inertial frame down the incline with acceleration gsinθ.

Case II: In the accelerating frame (at rest with the sliding mass)

The forces acting on the sliding mass are
1. Normal force (perpendicular to the incline)
2. Apparent force of gravity (also perpendicular to the incline but in the other direction).

Conclusion: The mass it at rest in this frame but has a weight that is other than m*9.81 m/s².

In the accelerated frame, ma appears on the other side of Newton's Second Law with a change in sign as a "pseudo-force." Think about it, if you are an observer in the accelerated frame you are at rest in that term by definition. Therefore any acceleration is interpreted by you as a force.

 

[RoyalCat]

You reasoning is confused because you think of two frames of reference simultaneously.

Case I: In the inertial frame of reference (at rest with the incline)

The forces acting on the sliding mass are
1. Gravity
2. Normal force

Gravity can be broken into two components
1. Perpendicular to the incline that is canceled by the normal force
2. Parallel to the incline in the down direction that is canceled by nothing.

Conclusion: The mass accelerates relative to the inertial frame down the incline with acceleration gsinθ.

Case II: In the accelerating frame (at rest with the sliding mass)

The forces acting on the sliding mass are
1. Normal force (perpendicular to the incline)
2. Apparent force of gravity (parallel to the incline but in the direction opposite the acceleration down the incline in the inertial frame).

Conclusion: The mass it at rest in this frame but has a weight that is other than m*9.81 m/s².

In the accelerated frame, ma appears on the other side of Newton's Second Law with a change in sign as a "pseudo-force." Think about it, if you are an observer in the accelerated frame you are at rest in that term by definition. Therefore any acceleration is interpreted by you as a force.

Just a small correction.

 

[Tusike]

I realized how I was wrong after I left so yeah never mind about that.

However, I'm still confused about one thing:
(1) In the accelerated frame of reference, the two forces acting on the whole volume of water, as we said, are gravity pulling down and the pseudo--force or inertial force of -mgsin(alpha), - meaning it's directed uphill. Add these two together, you get the overall force of mgcos(alpha), which is why the surface is parallel to the incline.

(2) In the inertial frame of reference, the only force I can think of that is acting upon the whole volume of the water is g pulling down. What am I forgetting about in this frame of reference? We should still get that the surface is parallel to the incline, not that it stays horizontal...

 

[kuruman]

Just a small correction.

Thanks for pointing this out, RoyalCat. I meant to say that there is an apparent force up the incline and the "real" vertical force of gravity that combine to give a resultant perpendicular to the incline.

(2) In the inertial frame of reference, the only force I can think of that is acting upon the whole volume of the water is g pulling down. What am I forgetting about in this frame of reference? We should still get that the surface is parallel to the incline, not that it stays horizontal...

You are forgetting that the water is accelerating. As I said earlier, it is easier to figure this out in the non-inertial frame. I will try to explain this from the point of view of the inertial frame in a quasi hand-waiving manner.

Let's accept the "plumb-bob" principle (I made it up) that says that A pendulum hanging straight down (no oscillations) shows the direction of local gravity, i.e. which way is "down" at a particular point in space. Let's also accept that water surfaces (no waves) are locally perpendicular to the direction that is "down".

Now consider a cart with a plumb-bob attached to its ceiling. Put it on an inclined plane so that it accelerates down the incline with acceleration gsinθ. Draw a free body diagram of the hanging mass, apply Newton's second Law and find the angle of the string relative to the plane of the incline. It should be 90^o.

 

[Tusike]

OK thanks I have it now.