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Homework Help: Really clueless = Minimum Mass Tension Force Question

  1. Jul 13, 2010 #1
    1. The problem statement, all variables and given/known data

    A fisherman yanks a fish straight up out of the water with an acceleration of 2m/s^2 using very light fishing line that has a "test value" of 50 pounds. the fisherman, loses the fish as the line snaps.
    What is the minimum mass of the fish?

    The answer given is mass=18.9kg

    No additional information is given, and it's just a question by it self, no other values attached.

    2. Relevant equations

    3. The attempt at a solution

    I converted pounds into kg since they asked for mass as solution
    They asked for 3 Sig Figs, so 50 pounds =22.72kg
    I'm assuming the acceleration is 2m/s^2 all the way
    but I'm not sure how that get me the 18.9 kg answer

    I'm seriously clueless with this question, about both the concept and how the answer is gotten. Been trying to figure it out for at least 3 hours, really need your help Q__Q
    Please and Thank you!
  2. jcsd
  3. Jul 13, 2010 #2
    When a fisherman yanks a fish straight up out of the water, the acceleration of 2m/s^2 should be added to gravity acceleration (9.81m/s^2).

  4. Jul 13, 2010 #3

    Thanks for the reply, Bartek!

    I plugged the new acceleration in, and played around with different equations
    I've got many answers, but none that's 18.9kg
    F= ma
    = (22.72)(11.8)
    = 222.656 N
    How should I continue next?
  5. Jul 13, 2010 #4
    No! When line was tested a=g. So Fmax is (22.72)*9.81, not 11.81 (during test the mass of 50 pounds hanging motionless). F_max is a maximum line tension.

    When fisherman yanks a fish, this Fmax should be equal to (m_fish)*(2+9.81). You can calculate m_fish.


    50 pounds is equal to 22,6796185 kg, How did you got 22.72?
    Last edited: Jul 13, 2010
  6. Jul 13, 2010 #5
    Hahah I have no idea how did I get 22.72, probably 1kg = 2.2 lbs

    Thank you so much ♥ Finally it's solved~! :D

    BTW, I have another tensions concept question

    Two tug-of-war teams are at opposite ends of rope. Newton's third law says that the force exerted by team A will equal the force that team B exerts, how can either team win the tug-of-war?

    Does the player's mass and friction play a role?
  7. Jul 13, 2010 #6
    Imagine both teams and a rope as a one rigid body. What forces exert this "body" horizontally? When it will be moved?
  8. Jul 13, 2010 #7
    Friction force exerts on the body horizontally, it'd move when the forces on one side is greater than the other? But what's going to trigger that unbalanced force?
  9. Jul 13, 2010 #8
    muscles :biggrin:. Both teams are pushing the ground... so, ground pushing the teams. Stronger win.
  10. Jul 13, 2010 #9

    Thank you thank you ♥
    I can finally hand in my work and sleep in peace :D

    PS, are you teaching physics? or just doing for fun?
  11. Jul 13, 2010 #10
    Last edited by a moderator: Apr 25, 2017
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