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Simple Problems

  1. Oct 24, 2006 #1
    1) Why doesn't the coefficient of friction have any units?

    2) A rop can withstand 750 Newtons. If 2 people in opposite directions pull with a force of 500 newtons will the rope break?

    3) In a tug-of-war Newtons 3rd law states that any action causes an equal yet opposite reaction. Explain how a tug-of-war match can be won using this law. I put because one side may loosen their grip, or weigh less and therefore have less friection on their side, or may be weaker so thery cannot possible apply the force the other side can without sliding.

    4) A fisherman uses a line that has a test value of 50 pounds. If he pulls the fish upwards at 2m/s/s how much does the fish weigh at least? I put 25 lbs.
     
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  3. Oct 24, 2006 #2

    PhanthomJay

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    See comments above in red.
     
  4. Oct 24, 2006 #3
    so the rope wouldn't break, i donno how but I got about 40 lbs. for the fish, but why duznt the coefficient have units?
     
  5. Oct 24, 2006 #4

    PhanthomJay

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    Why doesn't the rope break?

    That fish problem has messed up units, the line strength is given as 50 pounds, but you give the acceleration in meters/s/s? Did you mean a = 2ft./s/s? If not, best to convert pounds to newtons before solving. At any rate, use newton's 2nd law and set the rope strength equal to the line tension to solve for the max weight. Show work.

    What's the formula for the friction force?
     
  6. Oct 24, 2006 #5
    Think about the equation of friction: F = mu N. F and N have units of newtons. Solve for mu and ask yourself what the units would be.

    Sometimes, I think it's easier to deal with something more familiar. So, if I drive 4 miles on thursday, and 8 miles on Friday, I drove twice as far. What are the units of 'twice'?

    Dorothy
     
  7. Oct 24, 2006 #6
    so it's just a ratio right?
     
  8. Oct 25, 2006 #7
    well if the other end of the rope were attached to a wall, and you pulled at 500 N, and its breaking point was 750, would it break? or if would u only have to pull at half of 750 N in order to break it?. Isn't it the same concept since the wall exerts 500 N back, so does the other team. Is is the magnitude of the smallest direction of force that determines the break?
     
    Last edited: Oct 25, 2006
  9. Oct 25, 2006 #8
    Also about the units in the formula. The coeffiecient of friction is a constant that is derived from an ODE (which you may or may not know). The result from the ODE is that the constant (the coefficient of friction) is unitless.
     
  10. Oct 25, 2006 #9
    I came up with F-222.754=2m, but now what do I do? Do i divide 222.754 by 9.8 and use that as the mass? If so the answer would be 27 kg. And yes the question says 2m/s/s but also 50 lbs.
     
  11. Oct 25, 2006 #10

    PhanthomJay

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    Whether it's 2 people pulling on the rope with a force of 500 each, or one pulling on it at 500 while it is attached to a wall at the other end, yes, correct, tension in rope is the same. What is that tension? If it's less than 750, the rope will hold.
     
  12. Oct 25, 2006 #11

    PhanthomJay

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    it's 222 - F = 2m, where F is the weight, mg, = to 9.8m. Now solve for m.
     
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