Physics Question

Serendipity
Serendipity Posts: 6,975
edited September 2007 in The Clubhouse
Hi,

Was wondering if anyone here would know how to solve the following problem:

"A swimming pool measures a length of 5.80m, width 4.30m, and depth 3.20m. Compute the force exerted by the water against the bottom. Do not include the force due to air pressure. Take the free fall acceleration to be g= 9.80m/s^2 , and the density of water to be 1000kg/m^3. Also compute the force exerted by the water against either end. Do not include the force due to air pressure."

Any help would be greatly appreciated.

Thanks in advance!

P.S. This is for my "fluid mechanics" course.
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  • Ricardo
    Ricardo Posts: 10,636
    edited September 2007
    After 25 years or so I forgot all the formulas, but this seems a pretty basic thing....don't you have your book?? ;)
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  • billbillw
    billbillw Posts: 6,842
    edited September 2007
    Isn't that why you are in school? You tell us!
    For rig details, see my profile. Nothing here anymore...
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  • Serendipity
    Serendipity Posts: 6,975
    edited September 2007
    HTrookie wrote: »
    After 25 years or so I forgot all the formulas, but this seems a pretty basic thing....don't you have your book?? ;)

    I thought it was a basic question, but it doesn't appear to be. Spent hours yesterday trying to figure it out with a friend, to no avail. It's also labeled a "Challenge Problem" in the textbook.
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  • bobman1235
    bobman1235 Posts: 10,822
    edited September 2007
    I'm guessing you can't use the basic F=ma for this one?

    Force = mass * acceleration
    (where Mass = volume * density)

    Seems a bit too simple that way...
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  • Serendipity
    Serendipity Posts: 6,975
    edited September 2007
    I agree, seems WAY too simple that way...and the fact that it is a challenge problem. The professor said he would give 5pts extra credit for solving the question.
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  • jdhdiggs
    jdhdiggs Posts: 4,305
    edited September 2007
    Looks really straight forward to me....
    5.8*4.3*3.2=79.81 M^3*1,000Kg/M^3=79,810 Kg*9.8N/Kg= 782,000 N or 782 kN
    (3 sig figs as given)

    Did you write the whole question?

    Oh snap, I got it!!! Haha! That's just the pressure on the bottom, you need to calculate the pressure on the sides of the pool as well! Or I should say only on the sides... The wording is hard, not the problem...
    There is no genuine justice in any scheme of feeding and coddling the loafer whose only ponderable energies are devoted wholly to reproduction. Nine-tenths of the rights he bellows for are really privileges and he does nothing to deserve them. We not only acquired a vast population of morons, we have inculcated all morons, old or young, with the doctrine that the decent and industrious people of the country are bound to support them for all time.-Menkin
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  • bobman1235
    bobman1235 Posts: 10,822
    edited September 2007
    Ohhh, I missed the part whree it said compute the force on either end. I think what jdhdiggs said is true for the bottom though. No idea how to do the sides; I haven't taken physics in almost 10 years.
    If you will it, dude, it is no dream.
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  • strider
    strider Posts: 2,568
    edited September 2007
    Cool. I feel better now. I came up with the force on the bottom just like jdhdiggs, the force on the sides seemed a bit trickier.
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  • billbillw
    billbillw Posts: 6,842
    edited September 2007
    You'll need to dig into your Calculus to get the force on the sides.
    For rig details, see my profile. Nothing here anymore...
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  • jdhdiggs
    jdhdiggs Posts: 4,305
    edited September 2007
    Sides are easy, I just suck at metric. Pressure is just a linear function of depth and you just integrate it across the sides (Simple calc 1, hell trig works too)

    IIRC (LOOK THESE UP!!!)
    Roughly 9800 N/M^2 of depth Pressure at the bottom is 31,360 N/M^2, at the top it's 0 N/M^2. (Think triangles) Pressure on narrow end would be 67.4 kN and the long sides would be 90.9 kN. That's each side. So total pressure: 1,100 kN (1,098.854 kN before sig figs)
    There is no genuine justice in any scheme of feeding and coddling the loafer whose only ponderable energies are devoted wholly to reproduction. Nine-tenths of the rights he bellows for are really privileges and he does nothing to deserve them. We not only acquired a vast population of morons, we have inculcated all morons, old or young, with the doctrine that the decent and industrious people of the country are bound to support them for all time.-Menkin
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  • audiobliss
    audiobliss Posts: 12,518
    edited September 2007
    Ahhh, I remember problems like these. But integration was definitely Calc 2 for us, not Calc 1.

    But, alas, I've forgotten every bit of it.
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  • jdhdiggs
    jdhdiggs Posts: 4,305
    edited September 2007
    It's not really a calculas problem, it's trig 1. 1/2B*H then multply it be the length of the side and you're golden.
    There is no genuine justice in any scheme of feeding and coddling the loafer whose only ponderable energies are devoted wholly to reproduction. Nine-tenths of the rights he bellows for are really privileges and he does nothing to deserve them. We not only acquired a vast population of morons, we have inculcated all morons, old or young, with the doctrine that the decent and industrious people of the country are bound to support them for all time.-Menkin
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  • Serendipity
    Serendipity Posts: 6,975
    edited September 2007
    audiobliss wrote: »
    Ahhh, I remember problems like these. But integration was definitely Calc 2 for us, not Calc 1.

    But, alas, I've forgotten every bit of it.

    Same here, did integration in HS but forgot a lot of it now that I'm in college.
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  • Serendipity
    Serendipity Posts: 6,975
    edited September 2007
    jdhdiggs wrote: »
    Sides are easy, I just suck at metric. Pressure is just a linear function of depth and you just integrate it across the sides (Simple calc 1, hell trig works too)

    IIRC (LOOK THESE UP!!!)
    Roughly 9800 N/M^2 of depth Pressure at the bottom is 31,360 N/M^2, at the top it's 0 N/M^2. (Think triangles) Pressure on narrow end would be 67.4 kN and the long sides would be 90.9 kN. That's each side. So total pressure: 1,100 kN (1,098.854 kN before sig figs)

    Thanks for your help!
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  • jdhdiggs
    jdhdiggs Posts: 4,305
    edited September 2007
    Oh, and if I'm wrong its because I haven't even looked at that stuff for 11+ years... Good luck! ;)
    There is no genuine justice in any scheme of feeding and coddling the loafer whose only ponderable energies are devoted wholly to reproduction. Nine-tenths of the rights he bellows for are really privileges and he does nothing to deserve them. We not only acquired a vast population of morons, we have inculcated all morons, old or young, with the doctrine that the decent and industrious people of the country are bound to support them for all time.-Menkin
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