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Sum of all forces zero,but net torque not zero?

  1. Sep 29, 2005 #1
    hiya

    Newton's law says if an object is not pushed or pulled upon, its velocity will naturally remain constant, or body will remain standing still

    But is this only true if the sum of all torques is also zero?

    If we pull on wheel (its axis attached to something so it doesn't fall down ) standing upwards with same force clockwise and counter-clockwise, but the distance vector is not the same for both torques, then for example clockwise torque will be greater and wheel will start moving clockwise, even though sum off all forces is zero ?

    thank you

    bye
     
  2. jcsd
  3. Sep 29, 2005 #2
    How do you define velocity of an object with many particles? You say the velocity of an object is equal to the velocity of its center of mass. Take your example, and you'll see that even though some particles were accelerated one way and others were accelerated the other way, the velocity of the center of mass, and thus the velocity of the object is unchanged. So Newtons law still holds. (lucky for us).
     
  4. Sep 29, 2005 #3
    So a sum of all forces equals zero for a body standing still or moving with constant velocity , but still net torque may not be equal equal to zero ?

    I though there was a law saying something to the effect : "if an object is not pushed or pulled upon, or is standing still, then the sum of all external and internal torques equals zero"?

    Do you have any idea what that is all about then?
     
  5. Sep 30, 2005 #4
    really need some explanation about this
     
  6. Sep 30, 2005 #5

    Doc Al

    User Avatar

    Staff: Mentor

    conditions for equilibrium

    It is certainly possible for the net force on an object to be zero, yet the net torque not be zero. That's why, for an extended body (as opposed to a point mass), there are two conditions for equilibrium:
    (1) The net force must be zero (translational equilibrium condition)
    (2) The net torque about any axis must be zero (rotational equilibrium condition)
     
  7. Sep 30, 2005 #6
    thank you for your help
     
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