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Homework Help: Suspended Fish

  1. Feb 23, 2010 #1
    I have the solution for the problem. I just don't understand HOW this formula was derived. It came from another student, and nowhere in the text has our teacher taught us how do something like this..

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

    http://loncapa2.physics.sc.edu/res/sc/gblanpied/courses/usclib/joneslib/jonespictures/jonesf4_43.gif

    What is the force exerted by the string on the wall at point A in the figure if the suspended fish has mass m = 0.56 kg?

    2. Relevant equations

    m * g * (Square root of 3)


    3. The attempt at a solution

    .56 * 9.81 * (square root of 3) = 9.52 N .. but why would you use the square root of 3 here? 30 degrees?
     

    Attached Files:

    • fish.jpg
      fish.jpg
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    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Feb 23, 2010 #2

    kuruman

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    Sorry, we cannot help until you attach a picture showing the situation.

    Also, m * g * (Square root of 3) is not an equation. What is it supposed to mean?
     
  4. Feb 23, 2010 #3
    Sorry, I linked an image from the site but I guess it didnt' show up. Can you see the attachment now?

    and I'm not sure where he got the mg(sqr3) from, I think he may have told me the latter part of another equation that was used and I just don't know the first part.
     
  5. Feb 23, 2010 #4

    kuruman

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    Yes, the attachment is there. You need to say that the sum of all the forces in the x direction is zero and the sum of all the forces in the y direction is zero. Can you write

    1. An expression for the sum of all the forces in the x-direction
    2. An expression for the sum of all the forces in the y-direction

    ?
     
    Last edited: Feb 23, 2010
  6. Feb 23, 2010 #5

    collinsmark

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    ('Had to jump in here -- that picture of the fish is great :rofl:)

    Hello kboykb,

    Use kuruman's advise on how to go about this problem. The advise is sound. Since you know that the fish is not accelerating, the sum of all the forces in a given direction must be zero. So you can just sum them up as kuruman advises, knowing the net force in any direction is zero.

    By the way, if you are wondering where in the heck a [tex] \sqrt {3} [/tex] fits in, here are a couple useful trig identities. You can check these using the Pythagorean theorem if you wish. Again though, follow kuruman's advise on how to solve this problem, but in the process you may find these identities useful.

    [tex] sin(30 ^{\circ}) = \frac{1}{2} [/tex]

    [tex] cos(30 ^{\circ}) = \frac{\sqrt{3}}{2} [/tex]
     
  7. Mar 16, 2010 #6
    I know I shouldn't revive a long gone thread. But I keep getting so caught up in the homework that I forget to show my appreciation.

    So to Kuruman & Collins, thanks for the help :)
     
  8. Jun 15, 2010 #7
    I am also trying to solve this problem, but the mass of the fish is .23kg. You say break down into sum of forces in each direction. Would it be something like:

    x-
    cos30 + cos180 = -.13
    y-
    sin30 + sin180 = .5

    and I'm not sure where to go from here.
     
  9. Jun 15, 2010 #8

    collinsmark

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    You need to sum all the forces in each direction, such that they sum up to zero (in a particular direction), like Kurman says. It's more than just summing up sines and cosines of angles alone.

    In other words, you need to take the tensions on each string into account too, as well as the gravitational force on the fish.:wink: A free body diagram is a good place to start.
     
  10. Jun 15, 2010 #9
    Ok so I've tried to break them down and equal to zero. I'm letting T1 be AB, T2 BC, and T3 B-fish.

    T1x = -T1 cos 0 = -1 T1
    T1y = T1 sin 0 = 0 T1
    T2x = T2 cos 30 = .87 T2
    T2y = T2 sin 30 = .5 T2
    T3x = 0
    T3y = -T3 = -.23

    T1x + T2x + T3x = 0
    -1 T1 + .87 T2 + 0 = 0
    T1 = .87 T2

    T1 y + T2 y + T3 y = 0
    0 T1 + .5 T2 + -.23 = 0
    0 T1 + .5 T2 = .23
    0 (.87 T2) + .5 T2 = .23
    T2 = .46

    T1 = .87 (.46)
    T1 = .4002

    so using an example I found, I got to tension 1 being .4002N which isn't right. I'm not sure what I did wrong and what I can do now to fix it.
     
  11. Jun 15, 2010 #10

    collinsmark

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    ".23kg" is in units of kg, a measurement of mass. Forces (in the SI system) have units of Newtons. :smile::wink::tongue:
     
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