# Moment of force exerted by water on tube

1. May 11, 2016

### Vibhor

1. The problem statement, all variables and given/known data

2. Relevant equations

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

3. 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. May 11, 2016

### TSny

Did you use the correct density on the left?

3. May 11, 2016

### Vibhor

Ok .

It should be 2ρ . It gives $v = \frac{\sqrt{5gH}}{2}$ which again seem incorrect .

4. May 11, 2016

### TSny

Check to see if the 2 should be inside the square root.

5. May 11, 2016

### Vibhor

Sorry . It should be $v = \sqrt{\frac{5gH}{2}}$ . Is this the speed at which the fluid flows in the tube ?

6. May 11, 2016

### TSny

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?

7. May 11, 2016

### Vibhor

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. May 11, 2016

### TSny

Why not? Does the "continuity equation" tell you anything?

Yes.

9. May 11, 2016

### Vibhor

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. May 11, 2016

### TSny

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. May 11, 2016

### Vibhor

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. May 11, 2016

### TSny

You'll need to consider the atmospheric pressure acting on the tube over the region shown below.

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13. May 11, 2016

### Vibhor

But this pressure is not acting directly on the fluid ?

14. May 11, 2016

### TSny

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. May 11, 2016

### Vibhor

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. May 11, 2016

### Vibhor

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 $F_v = -ρAv^2 - P_0 A$ . So which one is correct ??

Please let me know which expression for the vertical force is correct before I think about torque .

Last edited: May 11, 2016
17. May 11, 2016

### TSny

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.

Anyway, see what you get. Hopefully, others will chime in.

Last edited: May 11, 2016
18. May 11, 2016

### Vibhor

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 .

Last edited: May 11, 2016
19. May 11, 2016

### TSny

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.

Last edited: May 11, 2016
20. May 11, 2016

### Vibhor

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 $F_v = -(2\rho) A v^2$ ??

Last edited: May 11, 2016