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Math Help - Pressure/Area word problem of a oil tank help?

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    Pressure/Area word problem of a oil tank help?

    A cylindrical oil storage tank 12 feet in diameter and 17 feet long is lying on its side. Suppose the tank is half full of oil weighing 85 lb per cubic foot. What's the total force on one end of the tank? The relevant relationships are that pressure where p is fluid density and h is height below the surface of the fluid, and where Area is the area of the surface under pressure p.

    I spent hours on this problem and got a force of around 6,000lbs but apparently that's not right! How would you do this problem? What should I have gotten?
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    Re: Pressure/Area word problem of a oil tank help?

    Let h be the distance below the surface of the liquid. You need to find an equation for the width of the tank as a function of h, call this w(h). Then take the integral from h=0, the surface of the liquid, to h=6, the bottom of the tank of dF, which is \int\, dF = \int \rho h\,dA = \int_0^6 \rho h w(h) \,dh.

    - Hollywood
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    Re: Pressure/Area word problem of a oil tank help?

    Quote Originally Posted by hollywood View Post
    Let h be the distance below the surface of the liquid. You need to find an equation for the width of the tank as a function of h, call this w(h). Then take the integral from h=0, the surface of the liquid, to h=6, the bottom of the tank of dF, which is \int\, dF = \int \rho h\,dA = \int_0^6 \rho h w(h) \,dh.

    - Hollywood
    I'm not sure how to find the width with this integral. When I try to integrate it I end up with some complicated number *w, So I know I'm doing it wrong. I'm not used to so many variables in one definite integral. How do you integrate it?
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    Re: Pressure/Area word problem of a oil tank help?

    It may be easier to do this using polar coordinates. You can set  h = R(1-sin \theta), w(h) =  2R\cos \theta and dh = R \cos \theta d \theta. So the integral becomes:

     F = \int_{-\pi/2} ^{\pi/2} 2 \rho R^3 (1-\sin\theta)\cos^2\theta d \theta
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    Re: Pressure/Area word problem of a oil tank help?

    Quote Originally Posted by ebaines View Post
    It may be easier to do this using polar coordinates. You can set  h = R(1-sin \theta), w(h) =  2R\cos \theta and dh = R \cos \theta d \theta. So the integral becomes:

     F = \int_{-\pi/2} ^{\pi/2} 2 \rho R^3 (1-\sin\theta)\cos^2\theta d \theta
    Okay it took me a while, but I think it helped to put it in those terms. When I integrated this I got that F=pi*p is this right? I could've gotten several things wrong so it might be off.
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    MHF Contributor ebaines's Avatar
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    Re: Pressure/Area word problem of a oil tank help?

    I get  \pi \rho R^3. Always check your units - your answer of  \pi \rho is not in units of force.
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    Re: Pressure/Area word problem of a oil tank help?

    Quote Originally Posted by ebaines View Post
    I get  \pi \rho R^3. Always check your units - your answer of  \pi \rho is not in units of force.
    Okay, so when it's in units of force as pi(p)R^3 how do I find the p and R variables? (That may be a stupid question, but I've never worked out this kind of problem before. At least not from a word problem, it's really confusing me)
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    MHF Contributor ebaines's Avatar
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    Re: Pressure/Area word problem of a oil tank help?

    You've been given the data that  \rho (density) is 85 pounds/ft^3 and R = 6 ft.
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    Re: Pressure/Area word problem of a oil tank help?

    Quote Originally Posted by ebaines View Post
    You've been given the data that  \rho (density) is 85 pounds/ft^3 and R = 6 ft.
    Oh, duh! I can be so blind sometimes. When I use that I get that F= 57,679. But is that for just one side of the tank? It seems high.
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    Re: Pressure/Area word problem of a oil tank help?

    I get 12,240 lb.

    - Hollywood
    Thanks from canyouhelp
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    Re: Pressure/Area word problem of a oil tank help?

    Quote Originally Posted by hollywood View Post
    I get 12,240 lb.

    - Hollywood
    I'm not sure how I'm integrating this one wrong, but 12,240lbs would make a lot more sense. I'll ask my professor to explain this one to me.
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    MHF Contributor ebaines's Avatar
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    Re: Pressure/Area word problem of a oil tank help?

    Quote Originally Posted by ebaines View Post
    I get  \pi \rho R^3.....
    This is not correct - my apologies but I realize I calculated the force as if the tank was full, not half full. The integral that should be evaluated is:

     \int _{-\pi/2} ^0 2 \rho R^3(1-\sin \theta) \cos^2 \theta d \theta

    which yields:

     F = 2 \rho R^3 \left(\frac {\theta} 2 + \frac { \sin \theta \cos \theta} 2 + \frac {\cos^3 \theta} 3 \right] _{- \pi/2} ^0  = 2 \rho R^3(\frac {\pi} 4 + \frac 1 3 )
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