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Sum of torque's question

  1. Nov 6, 2009 #1
    1. The problem statement, all variables and given/known data

    The nut is now placed in a nutcracker with only one lever, as shown, (Intro 3 figure) and again friction keeps the nut from slipping. The top "jaw" (in black) is fixed to a stationary frame so that a person just has to apply a force to the bottom lever. Assume that [tex]F_2[/tex] is directed perpendicular to the handle.


    2. Relevant equations

    [tex]\Sigma\tau=0[/tex]

    3. The attempt at a solution

    I'm not sure if the sum of the torques includes [tex]F_{2}D[/tex] and/or [tex]F_{2}d[/tex]. I though that it was the sum of the torque of the smaller handle + the torque of the larger handle + the normal force. Since counter clockwise is positive, it would be [tex]\tau_{nut}-\tau_{small}-\tau_{large}=0[/tex] and solve for [tex]F_{2}[/tex], but that's not right.
     

    Attached Files:

  2. jcsd
  3. Nov 6, 2009 #2

    stewartcs

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    What is the problem asking you to find?

    CS
     
  4. Nov 6, 2009 #3
    I think that the torques are two.The one from the force on the large part and the one from the nut.correct me plz if i am wrong
     
  5. Nov 6, 2009 #4
    the jaw is fixed so the torque at it does not count
     
  6. Nov 6, 2009 #5
    The question asks to find [tex]F_{2}[/tex]. Since its the max torque just to begin to break the shell, the sum of all the torques would be 0. So wouldn't that mean that the normal force of the shell at distance [tex]d[/tex] equals the torque of the little lever plus the torque of the larger lever?
     
  7. Nov 7, 2009 #6
    I believe that the torque due to the shell on the small lever should be equal to F2*D
     
  8. Nov 7, 2009 #7
    if it is just to start break the F2*D>=Torque from the nut
     
  9. Nov 9, 2009 #8

    stewartcs

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    Is this a continuation of another problem? If so, what else are you give (or did you find)?

    The torque on the hinge would be equal to T = F_2 x D since it is the only force being applied. The perpendicular force acting on the nut would then be, F_nut = T / d since the torque at the hinge is the same.

    CS
     
  10. Nov 9, 2009 #9
    the correct answer according to our online physics homework is [tex]\frac{F_{n}d}{D}[/tex]
     
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