Resultant Force on a Conical Flask Filled with Liquid

[leena19]

Homework Statement

A conical flask contains a liquid of density $$\rho$$ to a height 'h'.
If the area of the base of the flask is A and V is the volume of the liquid in the flask,then the resultant force acting on the curved surface is,
1) h$$\rho$$gA - V$$\rho$$g horizontally
2) h$$\rho$$gA - V$$\rho$$g vertically upwards
3) h$$\rho$$gA - V$$\rho$$g vertically downwards
4) h$$\rho$$gA + V$$\rho$$g vertically upwards
5) h$$\rho$$gA + V$$\rho$$g vertically downwards

Homework Equations

Pressure and upthrust

The Attempt at a Solution

since the horizontal components of the forces acting on the curved surface cancel off,(only the downward vertical components add on)so the answer can't be (1) ?
The force acting on the base of the flask is ,
force due to the pressure of the liquid - the weight of the fluid?
h$$\rho$$gA - V$$\rho$$g downwards, so is this the resultant force acting on the curved surface also?

Thank you

 

[tiny-tim]

Hi leena19!

(have a rho: ρ )

A conical flask …
If the area of the base of the flask is A and V is the volume of the liquid in the flask, …
…
since the horizontal components of the forces acting on the curved surface cancel off,(only the downward vertical components add on)so the answer can't be (1) ? …

Yes, it can't be (1).

I don't understand what shape this conical flask is, or which way up it is.

(or did you mean comical flask?)

Can you please describe it?

 

[leena19]

Hello sir,
A conical flask is a flask that we use to store and mix chemicals.
It's got a conical base(not comical and a narrow cylindrical neck.
Here's a picture of it,
http://www.advtechind.com/psaw/images/Conical-Flask.jpg"

 

[tiny-tim]

A conical flask contains a liquid of density $$\rho$$ to a height 'h'.
If the area of the base of the flask is A and V is the volume of the liquid in the flask,then the resultant force acting on the curved surface is …

It's got a conical base(not comical and a narrow cylindrical neck.
Here's a picture of it,
http://www.advtechind.com/psaw/images/Conical-Flask.jpg"

Hi leena19!

hmm … I'm still confused

I'm guessing that your diagram is wrong, and that the flask has a flat bottom, and that the question is asking about the force on the sloping surface …

calculate the force of the flat bottom on the water, and then use the fact that the water isn't moving, so the total forces on it must be zero, to find the force from the sloping surface.

 

[leena19]

Urm...I'm fairly certain the image on the link is exactly how a conical flask looks like.It's not a flat bottom flask,but I also think the flask has a somewhat flat bottom(it's not entirely round bottomed either) :)
so,

and that the question is asking about the force on the sloping surface …

Yes,I too think they're asking for the resultant force on the sloped surface,

Force on the bottom,due to water = weight of water + h$$\rho$$gA
V$$\rho$$g + h$$\rho$$gA downwards?

Therfore force on water due to the base = V$$\rho$$g + h$$\rho$$gA upwards?

And due to the equilibrium of forces,
the force on the slope = V$$\rho$$g + h$$\rho$$gA downwards ?
So is the answer (5) ?

THANK YOU

 

[tiny-tim]

Hi leena19!

(what happened to that ρ I gave you? )

Force on the bottom,due to water = weight of water + h$$\rho$$gA
V$$\rho$$g + h$$\rho$$gA downwards?

Noooo …

Hint: suppose the flask was a cylinder (with vertical sides) …

what would the force on the bottom be then?

 

[ideasrule]

Hi leena19!

(what happened to that ρ I gave you? )


Noooo …

Hint: suppose the flask was a cylinder (with vertical sides) …

what would the force on the bottom be then?

Another hint: water pressure exists because the water has weight, not in spite of it.

 

[leena19]

Another hint: water pressure exists because the water has weight, not in spite of it.

Yes,I forgot.Thanks

(what happened to that ρ I gave you? )

Thanks for the rho tiny-tim,but,since I've used $$\rho$$ in all the choices given,i decided to stick with it,but if you insist...

Hint: suppose the flask was a cylinder (with vertical sides) …

what would the force on the bottom be then?

It would be Ahρg which is equal to Vρg ?
so if the above equation were true,
the force acting on the bottom would be R(due to curved surface) + Ahρg ?
I'm still unsure about this.

 

[tiny-tim]

Hi leena19!

It would be Ahρg which is equal to Vρg ?

Yes, and they're both equal to mg, the weight of the water.

So your formula would give 2mg, which is obviously twice too much …

can you see that you're double-counting?

You can't count both the weight and the force from the pressure …
and anyway the weight (of the water) doesn't act on the glass … it only acts on the water!

Try again!

 

[turin]

... anyway the weight (of the water) doesn't act on the glass … it only acts on the water!

You're blowin' my mind, tiny-tim.

 

[tiny-tim]

You're blowin' my mind, tiny-tim.

we goldfish know a lot about water! ​

 

[HallsofIvy]

My first thought would be that, since the flask is not moving, the total force on any part of it is 0! But now my question is, "exactly what do you mean by 'resultant force'?" The result of what?

 

[tiny-tim]

… But now my question is, "exactly what do you mean by 'resultant force'?" The result of what?

Hi HallsofIvy!

She means the resultant or total of the force on all the sloping parts of the flask.

 

[leena19]

My first thought would be that, since the flask is not moving, the total force on any part of it is 0! But now my question is, "exactly what do you mean by 'resultant force'?" The result of what?

Hi HallsofIvy!

She means the resultant or total of the force on all the sloping parts of the flask.

I admit the original question was not in english,but the teacher translated it for us &as tiny-tim stated ,I'm sure we were asked to find the resultant force acting on the sloping surface.

So your formula would give 2mg, which is obviously twice too much …

can you see that you're double-counting?

Which formula are you referring to?
Is it,

the force on the slope = Vρg + hρgA downwards ?
I understand why this is wrong,cause as you said ,
Vρg = hρgA =mg,
Therfore for the above equation I get 2mg?
But if you were referring to this,

the force acting on the bottom would be R(due to curved surface) + Ahρg ?

I don't understand how 2mg is obtained,here.

Here's the link to the diagram I drewto try to understand the problem,but I'm still very confused,

http://img140.imageshack.us/img140/1028/force.png"

The forces(acting on water)Fx and Fx' cancel off,the Fy,Fy' vertical forces add on,and the Fz,Fz' forces form a resultant force(of $$\sqrt{2Fz}$$ downwards,maybe?
It is these downward forces on water or these forces on the slope that we're supposed to find ,right?
If this was true,then I guess we're left with only 2 choices (2) or (4) cause the foces on the slope have to be upwards?
But I really don't know how to get the magnitude of this resultant force.
Taking R as the resultant force,
I get R + Ahρg , but Ahρg = Vρg,
so I don't get an equation?
I referred my notes on mechanics/hydrostatics ,but I still can't figure it out. :(

the weight (of the water) doesn't act on the glass … it only acts on the water!

by glass are you referring to the sloped surface or the base,or the whole flask in general? & how does a force acting downwards not act on the base of the flask or the glass?
I'm really trying to understand,but I'm still very confused.

Thank you

 

[tiny-tim]

Yes, you're very confused.

Try it for a cylindrical flask first (instead of a conical flask) … what is the force from the water on the flat bottom of the flask?

Now lean the sides in a bit (or out! ), so it becomes a bit conical … what do you expect will happen?

 

[leena19]

Try it for a cylindrical flask first (instead of a conical flask) … what is the force from the water on the flat bottom of the flask?

h"ρgA =mg=Vρg

Now lean the sides in a bit (or out! ), so it becomes a bit conical … what do you expect will happen?

When we lean it in,
The pressure would increase,cause,
1)if the volumes of the liquid in the 2 flasks(cylindrical & conical)are constant,due to the leaning in of the sides,the pressure would increase as the height of the liquid level increases,so the force(F)on the bottom increases
2) Also due to action=reaction or something (I think?)there will be an extra force (R)acting from the sloped walls on the water which would inturn act on the bottom,
So,
F = h ρgA + R
R is what we need to find? so R= F- hρgA ?
But what is F?

All the students in my class do chemistry,so the teacher showed us what a real conical flask was,with the sides leaned inwards,
but anyway with the sides bent outwards,wouldn't the force on the bottom of this flask be less than that of a cylindrical flask?

I hate to sound like this sir,but can you please tell me how the Vρg fits in(all the choices given includes it)?I've tried and tried for hours and hours,but I still can't figure this out(it must be very easy,cause afterall it's only a 2 minute MCQ)

 

[tiny-tim]

Hi leena19!

(just got up :zzz:)

(oh, and stop calling me "sir" … I'm only a little goldfish!

though my friend sir isaac the newt is a "sir"! )​

h"ρgA =mg=Vρg

That's right!

When the sides are straight, the force is only mg.

When we lean it in,
The pressure would increase,cause,
1)if the volumes of the liquid in the 2 flasks(cylindrical & conical)are constant,due to the leaning in of the sides,the pressure would increase as the height of the liquid level increases,so the force(F)on the bottom increases

All the students in my class do chemistry,so the teacher showed us what a real conical flask was,with the sides leaned inwards,
but anyway with the sides bent outwards,wouldn't the force on the bottom of this flask be less than that of a cylindrical flask?

Yes, more height (and the same area of contact at the bottom) means more pressure on the bottom: the force becomes more than mg.
And less height (that's with the sides bent outwards) means less than mg.
Only the height matters (and the area of contact, of course).

2) Also due to action=reaction or something (I think?)there will be an extra force (R)acting from the sloped walls on the water which would inturn act on the bottom,
So,
F = h ρgA + R
R is what we need to find? so R= F- hρgA ?
But what is F?

No, this is double-counting.

The water acts on the bottom once.

You're saying that the sloping glass acts on the "sloping top" of the water, and that that force is somehow transmitted to the bottom.

Well, yes it is, but that extra force is already included in the pressure at the bottom.

No force acts on the bottom of the glass except for the pressure.

I hate to sound like this sir,but can you please tell me how the Vρg fits in(all the choices given includes it)?

Yes, the Vρg is mg, and it fits in in the force equation for the water …

the water isn't moving, so the total forces on it must be zero

there are 3 forces: the reaction ρghA (up) from the bottom, the reaction R (unknown) from the sloping top, and the gravitational force on the water, -mg = -Vρg.

So R + ρghA -Vρg = 0.

 

[leena19]

Thanks tiny-tim.I think I may have finally got it.

So R + ρghA -Vρg = 0.

I hope by this, you mean,
$$\uparrow$$ hρgA - Vρg - R = 0
therefore R= hρgA - Vρg ,vertically upwards,which would mean answer no.(2) ?

Thank you

 

[tiny-tim]

Hi leena19!

I hope by this, you mean,
$$\uparrow$$ hρgA - Vρg - R = 0
therefore R= hρgA - Vρg ,vertically upwards,which would mean answer no.(2) ?

That's right!

(btw, your R is minus my R, because my R was the force of the glass on the water )