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Homework Help: Two questions about Newton's law

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data
    9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s^2 using very light fishing line that has a "test" value of 22N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish?

    A. m > 1.5 kg

    15. A 75 kg theif wants to escape from a third story jail window. Unfortunately, makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the theif use this "rope" to escape? Give quantitative answer.

    A. a (downward) >_ 2.2 m/s^2

    2. Relevant equations
    9. F = ma
    15. F = mg

    3. The attempt at a solution

    9. F = ma → m = F / a = 22 / 4.5 = 4.9 kg
    This is what I thought. However, the answer is 1.5 kg. I don't understand what I did wrong.

    15. I don't think the thief can get out of the jail. Howevery, there is the "ununderstandable" answer.
     
  2. jcsd
  3. Feb 6, 2010 #2
    Your first solution seems right to me. Maybe it was a miscalculation by your teacher, typo in the solutions, or it hinges on what the definition of a "test value" is for fishing equipment.

    For the second problem, I'd say to fold the rope, making each part of the fold carry half the weight.
     
  4. Feb 6, 2010 #3

    ideasrule

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    You forgot to account for gravity.

    I also don't understand what answer the teacher is looking for. Maybe using two ropes in parallel?
     
  5. Feb 7, 2010 #4

    ehild

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    The force on the rope is less than the weight of the man if he slides down with a certain acceleration.

    He can do it by grabbing the rope at his breast and letting his body fall, and grabbing the rope with the other hand and so on. You can try it.

    The forces acting on the man are gravity and the tension of the rope. This tension can not exceed 58g.
    If the downward acceleration of the man is a, its mass is m,

    mg-T=ma, 75a>=75g-58g, a>=g(1-58/75)=2.2 m/s^2.
     
  6. Feb 7, 2010 #5
    I still don't understand what to do for number 9. I drew Free Body Diagram, and I realized that there are three kinds of forces: mg(down), ma(up), and T(up; along with ma(up)). I found those three forces, but I don't know what to do next.
     
  7. Feb 7, 2010 #6
    for problem(9) .. you got the answer and you want the explanation ..

    first you did great when you decided to draw the free body diagram, but why didnt you just substitute the numbers you have?

    its just simply applying F(net) = ma >> T-mg = ma >> 22 - 10*m = 4.5m >> 22 = 14.5m
    which leads you to m = 1.51 , then the mass of the fish should be greater than this value in order for the fishing line to snap .. :)
     
  8. Feb 7, 2010 #7
    Thank you so much~!
     
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