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Equilibrium force

  1. Mar 18, 2005 #1
    the question is here

    so first I found the angle...
    [tex]tan\theta=\frac{O}{A}=\frac{0.5}{6}[/tex]
    [tex]\theta=4.7636416[/tex]

    I then needed to find the force of the hypotnuse...
    [tex]sin\theta=\frac{O}{H}[/tex]
    [tex]H=\frac{250N}{sin4.7636416}[/tex]

    [tex]H=3010.3987N[/tex]

    is this correct?
     
  2. jcsd
  3. Mar 18, 2005 #2
    I didn't bother going through your work, but I got the same answer, 3010.4 N. So I assume it's correct. Also, the distances in your problem were given to 2 significant digits, so realistically the answer is correct to only 2 sig dig.
     
    Last edited: Mar 18, 2005
  4. Mar 18, 2005 #3
    That looks good to me. Assuming all of your calculations are correct, that seems to be the correct method for solving this problem.
     
  5. Mar 19, 2005 #4
    it just seems amazing that if you applied only 500 newtons, you will get a resultant force of over 6 times of what you applied.

    how can that be?
     
  6. Mar 19, 2005 #5
    Mechanical advantage. You are applying force against a much larger angle than the force applied against the car. The difference in this angle will result in the force on the car being much greater.
     
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