Moment of force exerted by water on tube

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Homework Statement


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Homework Equations



Bernoulli's equation : ##P + \frac{1}{2}ρv^2 + ρgh ## = constant

The Attempt at a Solution



The pressure at the interface is ##P_{int} = P_0 + ρgH## .

Applying bernoulli's equation between point at the interface of two liquids and exit of the tube .

##P_{int} + ρg(\frac{H}{2}+\frac{H}{4}) ## = ## P_0 + \frac{1}{2}(2ρ)v^2##

##v = \frac{\sqrt{7gH}}{2}## .

This doesn't give correct answer .

.What is the mistake ?

Thanks
 

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  • #2
TSny
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Applying bernoulli's equation between point at the interface of two liquids and exit of the tube .

##P_{int} + ρg(\frac{H}{2}+\frac{H}{4}) ## = ## P_0 + \frac{1}{2}(2ρ)v^2##
Did you use the correct density on the left?
 
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  • #3
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Ok .

It should be 2ρ . It gives ##v = \frac{\sqrt{5gH}}{2}## which again seem incorrect .
 
  • #4
TSny
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It should be 2ρ . It gives ##v = \frac{\sqrt{5gH}}{2}## which again seem incorrect .
Check to see if the 2 should be inside the square root.
 
  • #5
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Sorry . It should be ##v = \sqrt{\frac{5gH}{2}}## . Is this the speed at which the fluid flows in the tube ?
 
  • #6
TSny
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Sorry . It should be ##v = \sqrt{\frac{5gH}{2}}## . Is this the speed at which the fluid flows in the tube ?
You calculated ##v## for a point at the exit of the tube. Can you give a reason why this is also the speed at other points inside the tube?
 
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  • #7
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This will not be the speed at other points in the tube .

If I need to find vertical force exerted by the tube , should I be looking at speed of the fluid at the point where tube bends ?
 
  • #8
TSny
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This will not be the speed at other points in the tube .
Why not? Does the "continuity equation" tell you anything?

If I need to find vertical force exerted by the tube , should I be looking at speed of the fluid at the point where tube bends ?
Yes.
 
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  • #9
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Why not? Does the "continuity equation" tell you anything?
Right . The speed would be same in the entire tube length = ##v = \sqrt{\frac{5gH}{2}}## .

Now calculating the vertical force exerted by fluid on the tube . For that I considered the volume of fluid in the tube i.e from the hole in the container till the exit of the tube .
The vertical forces on this parcel of fluid are Fv , force due to tube and pressure force ##P_0 A## at the exit . The rate of change of vertical momentum of fluid would be ##ρAv^2## .

Doing force balance in vertical direction ,taking upwards as positive , ##P_0 A + F_v = -ρAv^2## i.e ##F_v = -ρAv^2 - P_0 A ## ??
 
  • #10
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Now calculating the vertical force exerted by fluid on the tube . For that I considered the volume of fluid in the tube i.e from the hole in the container till the exit of the tube .
The vertical forces on this parcel of fluid are Fv , force due to tube and pressure force ##P_0 A## at the exit . The rate of change of vertical momentum of fluid would be ##ρAv^2## .

Doing force balance in vertical direction ,taking upwards as positive , ##P_0 A + F_v = -ρAv^2## i.e ##F_v = -ρAv^2 - P_0 A ## ??
I think that's OK, except that in the statement of the problem ρ stands for the density of the other fluid, not the density of the fluid in the tube.
 
  • #11
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Yes , it should be 2ρ .

But , then the moment of the force exerted by liquid will have a P0 term whereas none of the options have P0 ??
 
  • #12
TSny
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Yes , it should be 2ρ .

But , then the moment will have a P0 term whereas none of the options have P0 ??
You'll need to consider the atmospheric pressure acting on the tube over the region shown below.
 

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  • #13
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You'll need to consider the atmospheric pressure acting on the tube over the region shown below.
But this pressure is not acting directly on the fluid ?
 
  • #14
TSny
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Right. You are ultimately interested in the net torque on the tube. (I believe this is what the problem is actually having you calculate). I will let you think about this.
 
  • #15
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If I consider (tube+fluid flowing through it) as system , then since the system is in vertical equilibrium , the force due to atmosphere at the top of the bend is balanced by the pressure force at the exit .

Is this a valid reason why P0A should not appear in the expression for vertical force and ##F_v = -ρAv^2## only and not ##F_v = -ρAv^2 - P_0 A## ?
 
  • #16
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You are ultimately interested in the net torque on the tube. (I believe this is what the problem is actually having you calculate). I will let you think about this.
Net torque on the tube = net force on the tube in vertical direction x moment arm

net force on the tube in vertical direction = vertical force on the tube due to fluid . So , again I have come to the point whether I should use ##F_v = -ρAv^2 - P_0 A## or ##F_v = -ρAv^2## ?

In the book also , in one of the sample problems it is mentioned that vertical force due to liquid exerted on the tube = ##F_v = -ρAv^2## , but then in post#9 I have obtained that ##F_v = -ρAv^2 - P_0 A## o_O. So which one is correct ??

Please let me know which expression for the vertical force is correct before I think about torque .
 
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  • #17
TSny
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Good question. I think that they probably expect you to use just ##F_v = -\rho Av^2##. If you include the term ##-P_0A##, then you will get an extra upward force exerted on on the tube by the fluid of ##+P_0A##. But this is cancelled out by the external atmospheric pressure acting on the tube as indicated in my drawing. If I use ##F_v = -\rho Av^2## I do get one of the specific answers listed. [EDIT: ##\rho## should be ##2\rho## in these expressions.]

They also seem to be neglecting the force due to the weight of the fluid in both the horizontal and vertical sections of the tube. If I include this, then I do not get one of the specific answers. Of course, there is always (D), none of the above. o_O

Anyway, see what you get. Hopefully, others will chime in.
 
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  • #18
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I think that they probably expect you to use just ##F_v = -\rho Av^2##
That gives option A) which is indeed the correct answer . But this expression is used everywhere , rather I should say it is used as a standard result . But I am really surprised by the omission of ##P_0A## term .I am still not convinced why the downward force on the bend as indicated in your drawing should cancel the upward ##P_0A## force .
 
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  • #19
TSny
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That gives option A) which is indeed the correct answer . But this expression is used everywhere , rather I should say it is used as a standard result . But I am really surprised by the omission of ##P_0A## term .I am still not convinced why the downward force on the bend as indicated in your drawing should cancel the upward ##P_0A## force .
The ##P_0A## term that you included is a force on the fluid. Then, when you solved for the vertical force, ##F_v##, that the tube exerts on the fluid, you got ##F_v = -(2\rho) A v^2 - P_0A##. Thus, by the third law, the fluid exerts the same force, but upwards, on the tube. But the atmospheric pressure acting externally on the tube has an unbalanced force contribution downward (shown in my figure) that happens to equal ##P_0A##. This downward force on the tube from the atmosphere cancels the part, ##P_0A##, of the upward force on the tube from the fluid. So, the net effect is to just use ##F_v = -(2\rho) Av^2## and forget about the unbalanced atmospheric pressure acting on the tube. At least that's how I see it.

I'm still confused as to why we can neglect the weight of the fluid in the analysis. It seems to me that it contributes the same order of magnitude as the other forces.
 
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  • #20
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The ##P_0A## term that you included is a force on the fluid. Then, when you solved for the vertical force, ##F_v##, that the tube exerts on the fluid, you got ##F_v = -(2\rho) A v^2 - P_0A##. Thus, by the third law, the fluid exerts the same force, but upwards, on the tube. But the atmospheric pressure acting externally on the tube has a an unbalanced force contribution downward (shown in my figure) that happens to equal ##P_0A##. This downward force on the tube from the atmosphere cancels the part, ##P_0A##, of the upward force on the tube from the fluid. So, the net effect is to just use ##F_v = -(2\rho A) v^2## and forget about the unbalanced atmospheric pressure acting on the tube. At least that's how I see it.
Nice analysis !

Considering (tube+fluid flowing through it) as system and using the fact that the system is in horizontal equilibrium , there will also be a net atmospheric pressure acting on the tube horizontally (at the bend) acting leftwards , but this will be balanced by pressure force PA at the hole of the container ??

Likewise would it be correct to say that the horizontal force exerted by fluid on the tube is ##F_v = -(2\rho) A v^2 - PA## , where ##P## is the pressure at the hole of the container but , net horizontal force on the tube will be ##F_v = -(2\rho) A v^2## ??
 
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  • #21
TSny
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Considering (tube+fluid flowing through it) as system and using the fact that the system is in horizontal equilibrium , there will also be a net atmospheric pressure acting on the tube horizontally (at the bend) acting leftwards
Yes.
but this will be balanced by pressure force PA at the hole of the container ??
Well, the atmospheric force will be ##P_0 A## which will be different in magnitude from the force ##PA##.

Likewise would it be correct to say that the horizontal force exerted by fluid on the tube is ##F_v = -(2\rho) A v^2 - PA## , where ##P## is the pressure at the hole of the container
I would write this with positive signs to indicate a direction to the right (and maybe a subscript h instead of v).
but , net horizontal force on the tube will be ##F_v = -(2\rho) A v^2## ??
As mentioned above, the forces ##PA## and ##P_0A## will not cancel. So, the "net" horizontal force on the tube would be ##F_h = (2\rho) A v^2 +(P - P_0)A##. Since the tube remains at rest, there will actually need to be another horizontal force to balance this "net" force. This would come from the tank where the tube is attached to the tank.
 
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  • #22
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So, the "net" horizontal force on the tube would be ##F_h = (2\rho) A v^2 +(P - P_0)A##
This is net horizontal force on the tube OR horizontal force on the tube due to fluid ??
 
  • #23
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This is net horizontal force on the tube OR horizontal force on the tube due to fluid ??
It's the sum of the horizontal force on the tube due to the fluid and the horizontal force on the tube due to the atmosphere.
 
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  • #24
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Ok .

And horizontal force on the tube due to fluid will be ##(2\rho) A v^2 +PA## ??
 
  • #25
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Ok .

And horizontal force on the tube due to fluid will be ##(2\rho) A v^2 +PA## ??
Yes, that's how I see it anyway. (Where's Chestermiller when we need him? Maybe I'll send him a note and see if he will take a look.)
 
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