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Homework Help: Help Exam Tmr! Gr 11 Physics

  1. Nov 6, 2005 #1
    Help Exam Tmr! 112 Physics *FORCES*

    As you may have guessed from the title, I'm in a bit of a bind here. You see the website with the answers to the review questions I'm working on isn't working... and there's an exam tomorrow. Here's a few of the questions any help would be greatly appreciated.
    A 2.0kg fish is hooked on a line that rated to a max of 38 N. At one point, the fish pulls with 40 N of force (fish are weightless in water). What is the min acceration at which you must play out the line to avoid it from snapping?
    A 0.00021 kg spider is suspended from a thin strand of webbing. The greatest tension the strand can withstand without breaking is 0.002 N . What is the max acceleration with which the spider can safetly climb up the strand?

    ok so apparently i need to show my work so here goes:

    for the fish question:
    m= 2.0kg
    Fxrt= 40N
    a = ?

    Fnet canot be more than 38N soo..

    that doesnt seem right to me... awfully fast to make up for 2 N wouldnt you think?
    orr wait..

    Fnet needs to be -2? so then..

    a= -1m/s2


    See the problem with the spider question is this: the Fg of the spider ends up being the same as the max Force... but where Fg is also the net force i should be able to use F=ma... plug that in and you end up with 9.52m/s2... which seems to completely contradict that fact.?.?.?.
    I mean wouldnt the spider only be able to hang there with no acceration at all, unless that is he goes down instead of up... so maybe the answer is -9.52?
    Last edited: Nov 6, 2005
  2. jcsd
  3. Nov 6, 2005 #2
    :zzz: is anyone out there?
  4. Nov 6, 2005 #3


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    Homework Helper

    Force by the FISHING LINE (Tension) needs to be less than 38 N .
    Force by the water (acting on the fish) is known.
    Total Force acting on the fish (as your intuition had it) is 2N.
    total F on fish = (m_fish)(a_fish)

    The total Force acting on the spider looks very small,
    maybe zero depending on the exact "g" value there.
    so, T - mg = 0 = ma = 0
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