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What is the force on the exerted edge of the cap

  1. Mar 5, 2008 #1
    1. The problem statement, all variables and given/known data
    To remove a bottle cap, a student exerts a force of 40 N on the opener. What is the force on the exerted edge of the cap by the opener?

    http://img225.imageshack.us/my.php?image=22965751hd6.png
    2. Relevant equations

    sigma(Tccw) = sigma(Tcw)
    t=r*d

    3. The attempt at a solution

    t= 40*.35
    t= 40*.30
    +
    = 26 n*m

    i know its supposed to be in newtons but im not sure how to calculate force when torque is involved
     
    Last edited: Mar 5, 2008
  2. jcsd
  3. Mar 5, 2008 #2

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    Since the torque is the same, force*dist is the same in both cases.
     
  4. Mar 5, 2008 #3
    im still confused
    i got torque for the whole bottle opener (t=40*.3) and i got torque for the opener coming in contact with the edge of the lid (t=40*.25)
    the answer is 240 N but i come no where close to it :cry:
     
  5. Mar 5, 2008 #4

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    Torque of 40 N is acting at a dist of 30 cm from the point = torque of F acting at a dist of 5 cm from the pint =>
    F*0.05 = 40*0.30 =>
    F = 240. (in N)
     
  6. Mar 5, 2008 #5
    argh i got it, I was thinking the start of the handle was a pivot so i was backwards in thinking for terms of distance
    thanks for clearing it up :)
     
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