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Fluid problem

  1. Nov 23, 2007 #1
    1. The problem statement, all variables and given/known data
    hey guys i'm kinda new to this forum and was wondering if you could help me on this fluid question i was having;

    Fig. 15-47 shows a stream of water flowing through a hole at depth h = 17 cm in a tank holding water to height H = 43 cm. (a) At what distance x does the stream strike the floor? (b) At what depth should a second hole be made to give the same value of x? (c) At what depth should a hole be made to maximize x?

    [​IMG]
    i dont know if u guys are gonna get the picture but its just a can with a height of H and a hole on its right side with the distance h from the top, and the water spills a distance of x away from the can


    2. Relevant equations
    well i tried to use bernoulli's equation for part a, but am stuck on part b and c


    3. The attempt at a solution
    well i used a combination of bernoulli's equation, liquid continuity equation and also simple projectile motion to find that th water falls 0.42 metres away from the can, but really have no clue what to do for part b and c.....if u guys can help it would be really awsome
     
  2. jcsd
  3. Nov 23, 2007 #2
    [tex]\rho g h=\frac{1}{2}\rho v^2[/tex]

    [tex]v=\sqrt{2gh}[/tex]

    [tex]H-h=\frac{1}{2}gt^2[/tex]

    [tex]x=vt=\sqrt{2gh}\sqrt{\frac{2(H-h)}{g}}[/tex]

    [tex]x^2=4hH-4h^2[/tex]

    As you can see that we have quadratic equation which has two values of h.

    for max x you need to just differentiate x to get h

    [tex]\frac{dx}{dh}=0[/tex]
     
  4. Nov 23, 2007 #3
    thanks....i didn't understand how you got the quadratic formula part, but i knew how to do c...just u kinda needed b to do c=P thanks again
     
  5. Nov 23, 2007 #4
    What I did is just

    [tex](x)^2=\left(\sqrt{2gh}\sqrt{\frac{2(H-h)}{g}}\right)^2[/tex]
     
  6. Nov 24, 2007 #5
    oo ic thanks....but i had one more question;
    In Fig. 16-41, a stick of length L = 1.6 m oscillates as a physical pendulum. (a) What value of distance x between the stick's center of mass and its pivot point O gives the least period? (b) What is that least period?
    [​IMG]

    would i do the same thing that you did in the other question but with the formula of
    T=2[tex]\pi[/tex][tex]\sqrt{I/mgh}[/tex]
    and then square both sides?
    if i do i get T^2 is equal to a bunch of stuff over h and i want to find the derivitave of the period so i can find h at the max and min?
     
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