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Homework Help: Boat tugging a whale

  1. Jul 16, 2011 #1

    Femme_physics

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    Am getting stuck with 2 unknowns


    1. The problem statement, all variables and given/known data


    http://img40.imageshack.us/img40/5883/whaltboat.jpg [Broken]

    The 5.5-Mg humpback whale is stuck on the shore due to changes in the tide. In an effort to rescue the whale, a 12-Mg tugboat is used to pull it free using an inextensible rope tied to its tail. To overcome the frictional force of the sand on the whale, the tug backs up so that the rope becomes slack and then the tug proceeds forward at 3 m/s.
    If the tug then turns the engines off, determine the average frictional force F on the whale if sliding occurs for 1.5 s before the tug stops after the rope becomes taut. Also, what is the average force on the rope during the tow?


    3. The attempt at a solution
    As u can see, I get stuck with 1 equation and two unknowns. I was able to solve for N and a (hopefully I did it correctly) though.

    http://img121.imageshack.us/img121/3028/19084315.jpg [Broken]



    http://img339.imageshack.us/img339/7913/16323840.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 16, 2011 #2
    to find Tension (T)
    tension retards the ship
    so
    T=mass of ship * acc of ship(a)
    T= 12000*2
    = 24000 N

    now u can find Friction usin

    m*a = T -F

    F = T - m*a
    =24000 - 11000
    =13000 N
     
  4. Jul 16, 2011 #3

    I like Serena

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    Hey Fp! :smile:

    Can you make an FBD of the ship?

    Perhaps that will give you an extra equation?
     
  5. Jul 16, 2011 #4

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  6. Jul 16, 2011 #5

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    Good! :)

    Do you have enough information now to find the friction force?
     
  7. Jul 16, 2011 #6

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  8. Jul 16, 2011 #7

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    What?
    No victory dance? :confused:
     
  9. Jul 16, 2011 #8

    Femme_physics

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    I got it?!?? :biggrin:

    w000t! *victory dance!!!!*


    Wait, let's do it with calculus now!!!!

    Give me pointers, I'm sooooooooooo ready!
     
  10. Jul 16, 2011 #9

    Femme_physics

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    Hmm wait according to the manual

    F = 24[kN]
    T = 24 [kN]
     
  11. Jul 16, 2011 #10

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    Hmm, if F equals T, then....
    The resultant force on the whale is zero...
    That doesn't sound right. :confused:

    I know, I think they made a typo (a copy+paste error)! :smile:



    Ah, but then we would need Captain Calculus! Rawwwr!
    He got points and angles!

    [PLAIN]http://www.cantonschools.org/~lforastiere/00F5BEEB-0075833E.0/Captain%20Calculus.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  12. Jul 16, 2011 #11
    hmmm......sounds right

    i got it now: *ques asks for average force....average acceleration is zero so is average net force*

    i got the same answer using calculus
     
  13. Jul 16, 2011 #12

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  14. Jul 16, 2011 #13

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    Nope.
    Average acceleration is not zero, since it says in the problem: "sliding occurs for 1.5 s before the tug stops".
     
  15. Jul 16, 2011 #14
    but it stops after sliding 1.5 s...that means it retards which cancels the intiall acceleration and average becomes zero
     
  16. Jul 16, 2011 #15

    Femme_physics

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    Oh now you're just patronizing. Humphffh!! :grumpy: :tongue:

    So they're mistaken?

    LOL where'd you get that from?!? :rofl:

    Captain Calculus. I'm gonna use that haha!
     
  17. Jul 16, 2011 #16

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    The acceleration is never negative, so it can't cancel, nor can its average become zero.


    I believe that they are mistaken.


    I was googling for funny pointers on calculus when I found the book.
    Looking at it I found it *mildly enjoyable* :D


    And errrr.... I don't know how to do this problem with calculus....
     
    Last edited: Jul 16, 2011
  18. Jul 16, 2011 #17
    check this
     
  19. Jul 16, 2011 #18

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    Check what?
     
  20. Jul 16, 2011 #19
    the meanig of the sentence
     
  21. Jul 16, 2011 #20

    Femme_physics

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    Really? But I thought you studied physics with calculus since you took it at the university
     
  22. Jul 16, 2011 #21

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    That's not what I meant. ;)
    I could spew a whole lot of calculus formulas here, but it wouldn't be particularly useful for this problem.
     
  23. Jul 16, 2011 #22

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  24. Jul 16, 2011 #23

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    Ah, I see.
    They used another way to solve the problem.

    This is still not really calculus although it looks that way.
    They used calculus notation though.

    And they used a formula I haven't seen you using yet.
    The one about impulse and momentum.



    Hold on, I see it now.
    Sorry, I made a mistake earlier.

    The resultant force on the whale is (T-F), but this force will first accelerate the whale from zero to a certain speed, and after that it will decelerate the whale again, until it is back at zero speed.
    This also means that the average acceleration is indeed zero.

    On average this means that T and F are equal.
    So your answer must indeed be F = 24 kN.



    As for the formulas that were used, you have them on your list like this:

    attachment.php?attachmentid=37233&stc=1&d=1310825361.jpg

    If you want, I can explain how they were used.
     

    Attached Files:

    Last edited: Jul 16, 2011
  25. Jul 17, 2011 #24

    Femme_physics

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    Good morning, captain! :smile:


    And aha! it makes sense, then. :smile: Why did they use calculus notation though?

    That doesn't really make much sense to me.
     
  26. Jul 17, 2011 #25

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    Ahoy ahoy sailor! :smile:

    The calculus notation comes from the use of "impulse".

    Impulse is a varying force that is exerted over a small period of time (the jerking tug in this case).
    Typically it is unknown how the force varies exactly.
    In cases we would know the varying force F, we would use an integral to sum it up.
    That is [itex]J = \int F \cdot dt[/itex].

    However, when you work with impulse, you're usually only interested in the total effect, which is the change in momentum. The force F is taken as an average then. Hence the formula [itex]J = F \cdot \Delta t[/itex] on your formula sheet.

    Every now and then the calculus notation with an integral is still used to denote the impulse, although there is no intention of actually integrating.
     
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