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Physics:finding torque equation in static equilibrium

  1. Sep 21, 2005 #1
    search keyword : " 80 W sin 13 = 40 W l sin 13 + 7 Fj cos 45 "

    look at the picture above the keyword you just searched for..
    i dont understand why you would add 7 Fj cos45 to the torque equation. I thought we only consider the forces perpendicular to the 'lever arm'. The Fj is on the pivot, so shouldnt we not take Fj into consideration for the torque equation? thanks
  2. jcsd
  3. Sep 21, 2005 #2
    Those diagrams could use some better clarification, where's the pivot point?
  4. Sep 21, 2005 #3
    well, im guessing that the pivot of this diagram is the little line, which would be the 'joint'.and the lever arm would be the long line part.
  5. Sep 21, 2005 #4
    I guess the pivot point is the point where the arrow with the 80 right next to it is pointing.In that case, [itex]F_j[/itex] does apply a torque about that point since it is not parallel to the smallest thick line. I guess thats the best way to illustrate that given that diagram?
  6. Sep 21, 2005 #5
    well.. what exactly are you suppose to include in your torque equation? i thought only tthe forces that are perpendicular to lever arm(the main base line) are only considered in the torque equation..
  7. Sep 21, 2005 #6
    Any force component that is perpendicular to the lever arm. If your level arm vector is [itex] \vec{r} [/itex] and your force vector is [itex] \vec{F} [/tex] then your torque equation would be

    [itex] \tau = \vec{r} \times \vec{F} [/tex] which evaluates to [itex] \tau = |\vec{r}||\vec{F}|\sin(\theta)[/tex] where [itex] \theta[/itex] is the included angle. If the vectors are completely parallel (or antiparallel) the included angle is 0 (or [itex] \pi [/itex]) and the sin of the angle will give 0, resulting in the torque being 0. Otherwise, the sin value scales the quantity to provide the correct amount of torque given the angle between the vectors.
  8. Sep 21, 2005 #7
    oh, i got it thanks a bunch
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