Acceleration of a liquid container

[koolman]

If we have a closed box filled with liquid . It is given a horizontal acceleration g/2 and a vertical acceleration g/2 .

 

Then I tried to find the angle of the free surface with the horizontal . It should be tan^(-1) 1/2 as slope is equal to tan^(-1)a(x)/g . But here there is vertical acceleration does it affect the slope .

 

Any help is appreciated

[swansont]

Why would there be an angle with the horizontal, other than the one it started with?

[imatfaal]

!

Moderator Note

Moved to homework help. Please make a start on the problem and show your workings and the members will guide you to the correct answer

[studiot]

This is clearly a wind up.

 

The free surface in a closed box, filled with liquid?

[koolman]

I think I have done correct

[swansont]

I think I have done correct

 

No, studiot's point (to which I was alluding) is valid: if the box is filled with liquid, it is not free to move around. Its angle will be that of the box. Why would the box tilt under the given accelerations? Something is missing from the problem.

[koolman]

No the question is complete

[DimaMazin]

If I understand something then we should define real acceleration.

a=[(g/2)²+(g/2)²]^1/2

a=g/2^1/2

am=gm_*sin(theta) because sin(theta)=cos(90-theta)

sin(theta)=1/2^1/2

theta=45 degrees

[swansont]

If I understand something then we should define real acceleration.

a=[(g/2)²+(g/2)²]^1/2

a=g/2^1/2

am=gm_*sin(theta) because sin(theta)=cos(90-theta)

sin(theta)=1/2^1/2

theta=45 degrees

 

what is gm?

 

There are two named accelerations, and they are of equal magnitude. There is no information given that would lead you to deduce any kind of rotation.

[DimaMazin]

 

what is gm?

 

There are two named accelerations, and they are of equal magnitude. There is no information given that would lead you to deduce any kind of rotation.

gm is just force not rotation. Read the OP question.

Edited December 22, 2015 by DimaMazin

[swansont]

gm is just force not rotation. Read the OP question.

 

I did. There is nothing called gm there. So I ask again: what is gm?

[studiot]

Even if we allow that either

 

1) The OP hasn't read the question properly and it actually reads

 

or

 

2) The question was badly posed and it should have read

 

"partly filled"

 

I would have expected some greater understanding of the question by the OP.

 

There are three accelerations to consider in total, not two.

 

When you consider all three the quoted figure is incorrect

 

 

It should be tan^(-1) 1/2 as slope is equal to tan^(-1)a(x)/g

 

I make the slope tan^-1(1).

 

Samagra, do you understand either the mathematics or the physics of what is going on? The problem can be solved as an exercise in applied mathematics or and exercise in physical understanding and physical reasoning.

 

DimaMazin, instead of butting heads with swansont, you are nearly there on the physical reasoning approach tidy up your work and complete the job.

 

The subject under discussion is known as relative equilibrium of liquids; googling this will reveal many learned articles and sample calculations for many circumstances.

Edited December 23, 2015 by studiot

[imatfaal]

I make the slope tan^-1(1).

 

Me too - the surface will eventually settle perpendicular to the resultant force

 

I presume your third acceleration you refer to is the one prior to the experiment which causes the liquid to form a free surface in the first place. I also wonder why the scenario doesn't use g/sqrt2 - which is an easily physically realisable frame rotation and makes the answer much more intuitive

[studiot]

 

I presume your third acceleration you refer to is the one prior to the experiment which causes the liquid to form a free surface in the first place.

 

Interesting way to put it, but yes the liquid is at all times subject to the third acceleration.

[DimaMazin]

 

I did. There is nothing called gm there. So I ask again: what is gm?

My calculation was wrong.

gm is gravitational force. I did liquid acceleration but the free surface isn't perpendicular to acceleration direction because gravitational force works there. We should take gm or simple 'g' for the calculation.

[swansont]

There is no gravitational force mentioned in the OP.

[studiot]

Without gravity the words horizontal and vertical have no meaning.

[koolman]

But there is gravity

[studiot]

 

samagra

But there is gravity

 

So what about a statement of the problem to include all the facts?

Followed by your revised solution?

 

All the facts include the assumption both imatfaal and I made that the vertical acceleration is upwards.

This is important since the answer is different if it is downwards.

 

You have been told that both gravity and the vertical acceleration play a part in this so

 

over to you?

[DimaMazin]

Let's consider triangle ABC

g + 0.5g=1.5g is AC

0.5g is CB

C is right angle

We have to find angle (90 - B).

 

tan(B)=1.5/0.5=3

B=71.56505 degrees

 

90-B = 18.4349488 degrees

Simpler we have to find angle A

tan(A)=0.5/1.5=0.333333333333333333333333333333333333

A=18.4349488 degrees

Edited January 5, 2016 by DimaMazin

[swansont]

Th acceleration in both the vertical and horizontal directions is g/2

[DimaMazin]

Th acceleration in both the vertical and horizontal directions is g/2

I know that. But there is antigravitational acceleration 'g' which is caused by absence of freefall.And which is vertical.

[swansont]

I know that. But there is antigravitational acceleration 'g' which is caused by absence of freefall.And which is vertical.

 

 

Huh?

[studiot]

There is something of a sign convention issue here, amongst all the other missing items the OP did not tell us.

 

 

It is given a horizontal acceleration g/2 and a vertical acceleration g/2 .

 

What does this mean?

 

Does it mean the vertical acceleration is in the same direction as gravity (downwards) since it has the same sign?

 

In which case the total vertical acceleration = g + g/2 and downwards is positive.

 

Or was it badly worded and in fact the additional vertical acceleration is upwards, which makes more sense, and numerically equal to -g/2.

 

With a downwards positive convention this makes the total vert accel = g - g/2

This is the calculation imatfaal and I used.

 

Or with an upwards positive convention the accel is g/2 - g = -g/2

[DimaMazin]

There is something of a sign convention issue here, amongst all the other missing items the OP did not tell us.

 

 

What does this mean?

 

Does it mean the vertical acceleration is in the same direction as gravity (downwards) since it has the same sign?

 

In which case the total vertical acceleration = g + g/2 and downwards is positive.

 

Or was it badly worded and in fact the additional vertical acceleration is upwards, which makes more sense, and numerically equal to -g/2.

 

With a downwards positive convention this makes the total vert accel = g - g/2

This is the calculation imatfaal and I used.

 

Or with an upwards positive convention the accel is g/2 - g = -g/2

We have only one variant here. 'g' is local acceleration which creates the horizon . I have made correct math for it.