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Fishing and torque

  1. Feb 23, 2005 #1
    The fishing pole in the picture makes an angle of 20 degrees with the horizontal. What is the torque exerted by the fish about an axis perpendicular to the page and passing though the fisher's hand?

    Is this question basically asking about how much torque there is?

    T=rFsinX
    T=2m*100N*sin37
    T=120Nm

    is this all that the question is asking for?
     
  2. jcsd
  3. Feb 23, 2005 #2
    torque = Fd sin(theta)
    theta is the angle between the force and the fishing pole...
    your theta is not quite correct, otherwise seems great
     
  4. Feb 24, 2005 #3
    is theta 123 degrees?
     
  5. Feb 24, 2005 #4
    Yes, it is 123 degrees. Optionally you could take 37 + 20 = 57 degrees. They have the same [tex]sin[/tex].
     
  6. Mar 2, 2005 #5
    does the fact that the the fisher has a 20 degree angle with the hoizon? Will that make a difference in theta being 123 degrees?
     
  7. Mar 2, 2005 #6
    Isn't the 20 degrees a completely different angle than the 123 degree angle?
     
  8. Mar 2, 2005 #7
    yes but I'm wondering if I should take into account the 20 degrees above the horizontal... I shouldn't right? it shold just be...

    T=rFsinX
    T=2m*100N*sin123
    T=167.7Nm
     
  9. Mar 2, 2005 #8

    learningphysics

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    Your answer is correct. But I don't understand what you mean by "taking the 20 degrees into account". If that angle wasn't 20, then theta wouldn't be 123.

    For example if the angle was 30 instead of 20, then you'd be taking sin 113.
     
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