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What is the net torque on this object?

  1. Jun 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Make the following into a web address; it wouldn't let me post one.
    img161.imageshack.us/img161/3728/physicspv5.pngimg161.imageshack.us/img161/3728/physicspv5.b3825b8665.jpg[/url]
    What is the net torque on this object?

    2. Relevant equations
    torque=(radius)(force)(sin(
    [tex]\theta[/tex]))

    3. The attempt at a solution
    I know that the answer is found by doing the following:
    .95(10N)-.22m(50N)-1.5m(3.0N)(sin40)
    but I don't understand why. Like, I get that it's plugging each section into the above equation, but I don't get why the values should be subtracted.

    Thank you!
     
  2. jcsd
  3. Jun 3, 2008 #2
    The link for the image doesn't work for me. However, my guess is that some of the torques are negative because they oppose the other torque. If your question is why the first is positive and the last two negative instead of the first negative and the last two positive then the answer is because there is a sign convention, called the right hand rule, with the cross product. E.g. the reason that i x j = k instead of -k is purely a convention (where i is a unit vector in the x direction, j a unit vector in the y direction, and k a unit vector in the z direction).
     
  4. Jun 3, 2008 #3
    take out the spaces:
    h t t p : / / s138.photobucket.com/albums/q275/charliethedogg/?action=view&current=physics.png

    Some of the torques are opposing each other. However, it doesn't fit with the location of the negative signs.
     
  5. Jun 3, 2008 #4
    Are you familiar with the right hand rule? If into your computer screen is the negative direction and out is the positive direction then all of the signs are correct.
     
  6. Jun 3, 2008 #5
    I'm positive that we weren't taught the right hand rule..
     
  7. Jun 4, 2008 #6

    alphysicist

    User Avatar
    Homework Helper

    Hi BrooklynBees,

    If you were not taught the right hand rule, then the usual convention is that clockwise torques are considered to be negative and counter-clockwise torques are positive. The equation in your post then indicates that the torque from the 10 N force is counterclockwise, and the torques from the other two forces are clockwise. Can you see how that matches the diagram?
     
  8. Jun 4, 2008 #7
    Thanks. I get it now.
     
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