Water pressure in a tank

  • #26
Regla
@Nidum is trying to lead you past the integration step so that you do not have to evaluate that integral.

The key is that you are being asked to evaluate ##\int_0^h \rho g L \ x \ dx##

The ##\rho g L## factors out and all that's left is evaluating ##\rho g L\ \int_0^h x \ dx##

That integral should be easy to evaluate. It should give the same result that @Nidum is jumping to: The force on a side is the average pressure on a side multiplied by the [wetted] area of that side.
thanks for showing me how I was supposed to form the integral, my only question is what do you mean that ρgL factors out?
 
  • #27
jbriggs444
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thanks for showing me how I was supposed to form the integral, my only question is what do you mean that ρgL factors out?
The ##\rho##, the ##g## and the ##L## are all constant terms for our purposes. They do not depend on x. They are simply a constant multiplier on the value. As long as you have a constant multiplier on the function you are integrating, for example the k in:
$$\int_0^h k\ f(x)\ dx$$
then you are allowed to move that constant outside of the integral and instead write
$$k\ \int_0^h f(x)\ dx$$
Moving the constant out of the integral is what I call "factoring it out". It seems that Wiki even has a page on this principle: https://en.wikipedia.org/wiki/Constant_factor_rule_in_integration
 
  • #28
Regla
The ##\rho##, the ##g## and the ##L## are all constant terms for our purposes. They do not depend on x. They are simply a constant multiplier on the value. As long as you have a constant multiplier on the function you are integrating, for example the k in:
$$\int_0^h k\ f(x)\ dx$$
then you are allowed to move that constant outside of the integral and instead write
$$k\ \int_0^h f(x)\ dx$$
Moving the constant out of the integral is what I call "factoring it out". It seems that Wiki even has a page on this principle: https://en.wikipedia.org/wiki/Constant_factor_rule_in_integration
oh that makes sense.
 
  • #29
Regla
The ##\rho##, the ##g## and the ##L## are all constant terms for our purposes. They do not depend on x. They are simply a constant multiplier on the value. As long as you have a constant multiplier on the function you are integrating, for example the k in:
$$\int_0^h k\ f(x)\ dx$$
then you are allowed to move that constant outside of the integral and instead write
$$k\ \int_0^h f(x)\ dx$$
Moving the constant out of the integral is what I call "factoring it out". It seems that Wiki even has a page on this principle: https://en.wikipedia.org/wiki/Constant_factor_rule_in_integration
did I enter everything correct?
 

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  • #30
jbriggs444
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did I enter everything correct?
You are supposed to be integrating from 0 to h, not from 0 to 4. You are supposed to be integrating x, not 4. And what @haruspex is trying to get you do to is to express the result in symbolic form. You should wind up with an expression involving ##\rho##, ##g##, ##h## and ##L##.
 
  • #31
Regla
You are supposed to be integrating from 0 to h, not from 0 to 4. You are supposed to be integrating x, not 4.
the whole problem was stated as a volume of 64 dm3... so I thought that since 43=64 doesn't that mean h=4?
 
  • #32
jbriggs444
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the whole problem was stated as a volume of 64 dm3... so I thought that since 43=64 doesn't that mean h=4?
You are trying to find h. It will not be equal to 4.

h in this context refers to the height (or depth) of the water in the tank. The tank is not full.
 
  • #33
Regla
You are trying to find h. It will not be equal to 4.
like this?
Screenshot (116).png
 
  • #34
jbriggs444
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like this?
That appears to be a correct result. However, as has been suggested previously, it would be better to leave it in symbolic form and to avoid replacing ##\rho##, ##g## and ##L## with numeric values:

$$F_{side} = \rho g L \frac{h^2}{2}$$

Now can you write down a formula for the total force on the bottom of the tank in terms of ##\rho##, ##g##, ##L## and ##h##?
 
  • #35
Regla
That appears to be a correct result. However, as has been suggested previously, it would be better to leave it in symbolic form and to avoid replacing ##\rho##, ##g## and ##L## with numeric values:

$$F_{side} = \rho g L \frac{h^2}{2}$$

Now can you write down a formula for the total force on the bottom of the tank in terms of ##\rho##, ##g##, ##L## and ##h##?
Like this? or does the 4 have to be an x?
Screenshot (118).png
 
  • #36
jbriggs444
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Like this? or does the 4 have to be an x?
You can include "L" or you can include "4". Using both is nonsense. As I had suggested in #34, the correct equation is:

$$F_{side} = \rho g L \frac{h^2}{2}$$

You had previously come up with a formula for the force on the base:

I've got the formula for the base of the tank bF=(ro)gxL2
When you wrote that you were apparently using the variable "x" to denote depth. Let us use "h" for depth and reformat that result:

$$F_{base} = \rho g L^2 h$$

Now then. We have formulas for the force on the side and for the force on the base. The problem statement tells us the ratio of those two forces. Can you write down an equation expressing that fact?
 
  • #37
Regla
Now then. We have formulas for the force on the side and for the force on the base. The problem statement tells us the ratio of those two forces. Can you write down an equation expressing that fact?
Fbase/Fside=3 you mean something like this? ρgL2h/ρgL(h2/2)=3
 
  • #38
jbriggs444
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Fbase/Fside=3 you mean something like this? ρgL2h/ρgL(h2/2)=3
Yes. You should be able to simplify that significantly.
 
  • #39
Regla
Yes. You should be able to simplify that significantly.
so how do we find the volume or height of the water level? since that only tells the ratio.
 
  • #40
jbriggs444
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More calculating. Fewer questions.
 
  • #41
Regla
More calculating. Fewer questions.
so what does the integral in post #34? h? the height of the water level? or the number that I have to put in the post #37 formula?
but we still get the h in the integral.
 
  • #42
jbriggs444
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In #37 you had an equation:

Fbase/Fside=3 you mean something like this? ρgL2h/ρgL(h2/2)=3
Solve that equation for h. Or at least simplify it as I had suggested in #38.
 
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  • #43
Regla
More calculating. Fewer questions.
I get h=0.00991988 or 0.01
 
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  • #44
Regla
In #37 you had an equation:



Solve that equation for h. Or at least simplify it as I had suggested in #38.
if I try to find the pressure to the bottom and side by putting in h I get always the same thing as 2. (ρgL2h/ρgL(h2/2)) It's a bit counter intuitive since I was stated that that should be 3, is the formula for the side equation right?
 
  • #45
Nidum
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ρgL2h/ρgL(h2/2)=3

Let's do a tidy re-arrangement of that equation :

ρgL2h = (3/2)ρgLh2

the ρg's cancel so :

L2h = (3/2)Lh2

Divide both sides by hL :

L = (3/2)h

Can you go from there ?
 
  • #46
Regla
ρgL2h/ρgL(h2/2)=3

Let's do a tidy re-arrangement of that equation :

ρgL2h = (3/2)ρgLh2

the ρg's cancel so :

L2h = (3/2)Lh2

Divide both sides by hL :

L = (3/2)h

Can you go from there ?
that simplifies a lot of things, thanks, but when I calculate h (if L=40) it I get h=26.6667 or 27. the problem is that if I try to calculate the pressures they don't equal to the statement give in the equation (Pbase/Pside=3)
Pbase=(1000*9.8*402*27)/40*40=264600
Pside=(1000*9.8*40*(272/2)/40*27=132300
264600/132300=2. It doesn't match with the statement given. I feel like there is something wrong with the side force equation.
 
  • #47
Nidum
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L is not 40 . Simple numerical mistake I think ?
 
  • #48
Regla
L is not 40 . Simple numerical mistake I think ?
If I know that the volume of the tank is 64 dm3 43=64 since that 4 is dm I change it into cm which is 40cm, isn't that right?
 
  • #49
Nidum
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Cube root of 640 is roughly 8.62

8.62cm x 8.62 cm x 8.62 cm = 640 cm3 .

.
 
  • #50
Regla
Cube root of 640 is roughly 8.62

8.62cm x 8.62 cm x 8.62 cm = 640 cm3 .

.
even if I change L with 8.62 and h changes to roughly 5.75 I still get 2 when I calculate the Pressures in the way stated in post #46
 

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