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Homework problem Leaking faucet

  1. Jan 5, 2005 #1
    Homework problem "Leaking faucet"

    Problem : Leonardo da Vinci constructed an apparatus consisting of two vertical boards hinged together on one side and covered with blotting paper on the inside faces. A leaking water tap lets drops fall down between the boards at presumable equal intervals of time. When a sting is suddenly pulled, the boards are clapped together and the position of the drops on the blotters can be inspected. Suppose that the vertical boards are 3 feet long and the leaking faucet is at such height and the rate of the dropping is so regulated that when a drop is just coming out of the faucet, the next or second drop is just at the top of the boards and the sixth drop is at the bottom of the boards, while drops 3, 4 and 5 are between the boards. Calculate the number of drops leaving the faucet per second.

    I'm stumped on this. I figured there is probably some simple approach I'm missing. But when I try to figure it out I always end up needing some other piece of information that I don't have and can't find out how to get ( like the distance from the faucet to the top of the boards).

    Any help would be greatly appreciated, thanks.
     
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  3. Jan 5, 2005 #2

    Andrew Mason

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    Use: [tex]h = \frac{1}{2}gt^2[/tex]

    The trick here is to realize that there are 5 equal (single drop spacing) time intervals between the tap and the bottom and 1 such interval between the tap and the top of the board. So:

    [tex]\Delta h_{1-6} = 3 + \frac{1}{2}g\Delta t^2 = \frac{1}{2}g(5\Delta t)^2[/tex]

    Work out the time interval between drops from that. That will allow you to determine the number of drops/sec.

    AM
     
    Last edited: Jan 5, 2005
  4. Jan 5, 2005 #3

    BobG

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    Yes, you need the distance to the top of the board. But, you can find that.

    I take it you managed to find the amount of time it takes for the 6th drop to reach the bottom of the boards, except it was in relation to x, the distance between the faucet and the top of the boards.

    [tex]s_f=s_i + v_i t - \frac{1}{2} a t^2[/tex]

    or

    [tex]x-3 = 0 + 0 - 16 t^2[/tex]

    During that time, how many drops came out of the faucet? That enables you to find the amount of time between each drop.

    One drop just reaches the top of the board as the next drop is about to leave the faucet. In other words, the amount of time to reach the top of the boards is equal to the amount of time between drops. Using this, you can solve for the distance between the faucet and the top of the board and actually put a number to the time. The rate the faucet is dripping (the number of drops per second) is equal to the reciprical of the time between drops.

    Edited to put everything in the right direction.
     
    Last edited: Jan 5, 2005
  5. Jan 5, 2005 #4
    Measure the distance from the faucet to the boards then write back the measurements.
     
  6. Jan 5, 2005 #5

    BobG

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    There are 5 equal time intervals (I started to do this myself until I took a second to think about it). 6 dots - how many intervals between them.
     
  7. Jan 5, 2005 #6

    Andrew Mason

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    Right you are. I have changed it.

    I am a little confused with your equation. Shouldn't your equation be: 3 = 0 + v14t +8gt2 where v1=gt ? or 3 + x = 0 + 0 +12.5gt2 ?

    AM
     
  8. Jan 5, 2005 #7

    BobG

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    Oops. To be technically accurate, I started out with the idea that the direction is negative (down) and half way through I switched to common sense in that you don't really care about the direction since it's all one direction (multiply both sides by -1 and don't worry about it).

    So, yeah, either acceleration should be negative with 'x' being a negative value, or leave everything positive with '3 + x' on the left side.
     
  9. Jan 5, 2005 #8
    Thanks for the help. I understand where you got [tex]x-3 = 0 + 0 - 16 t^2[/tex] from, but I'm still not completely sure on how to get the distance [tex]x[/tex] (from the faucet to the boards) from this. I know you guys explained it, but I didn't quite get it. From what I understand, you are trying to tell me that since all the [tex]t's[/tex] are equal I can use this to find the x... but I'm not exactly sure on how to use this fact to do so.
    Thanks for your time.
     
  10. Jan 5, 2005 #9

    BobG

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    Actually, it's the fact that the 't' required for a drop to reach the bottom of the boards can be defined in terms of x:

    [tex]-\frac{x-3}{16}=(t_6)^2[/tex]

    If you divide by 5, you have the time between each drop, which, in this case, happens to correspond with the amount of time it takes a drop to reach the top of the boards (or to fall a distance of x)

    [tex](t_1)^2 = -\frac{x-3}{80}[/tex]

    (multipying the denominator by 5 to divide the time into 5 intervals)

    t_1 is the amount of time required to fall x distance. Then you use the same basic distance equation, except for the drop 2 instead of drop 6. And, in this case, the distance is x instead of x-3.
     
  11. Jan 5, 2005 #10

    Andrew Mason

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    You have to use: [itex]12gt^2=3[/itex]. g is the acceleration due to gravity. You can't do the analysis without g.

    It is quite simple: The total distance is x+3 and we know that is equal to [itex].5g(5t)^2[/tex] where 5t is the total time. We also know that [itex]x=.5gt^2[/itex] where t is the time between the first interval. Substitute for x and solve for t.

    AM
     
  12. Jan 5, 2005 #11

    Andrew Mason

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    The correct equation is:

    x+3 = .5g(5t)^2. You are missing the g and your 16 should be 12.5.

    Your explanation is confusing poor nineyes. You can't divide by 5 like that to get t1. Since when is (5x)^2/5=x^2?

    AM
     
  13. Jan 5, 2005 #12
    In reply to Andrew Mason Post #10:
    I recieved t = .354 sec. This is only for one interval so I just do 1drop/.354 sec = 2.83 drops / sec.
    Was that the right way to go about it?
     
  14. Jan 5, 2005 #13

    Andrew Mason

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    How did you get your answer? It is way too high - that would mean there was over 1.5 seconds between the 6 drops and things drop at 16 feet in one second! So you know something is wrong. Since you are using feet you have to use g as 32 feet/sec^2.

    [tex]12gt^2=3 \rightarrow t = \sqrt{\frac{1}{4g}} =\sqrt{\frac{1}{128}} = .089 sec[/tex]

    So there are 1/.089 = 11.3 drops /sec.

    AM
     
  15. Jan 5, 2005 #14
    I see what I did wrong, I made a math error and divided by .5g (I forgot about the ... + 3, which when I look at it makes me wonder how I came up with that, I must have been distracted doing something else.) Thanks for the help!
     
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