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Torque a problem from Serway

  1. Sep 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Im having exam about Torque on monday and im trying to solve problems about it I stumbled with this problem, I cant solve it. I think im missing a fundamental concept about torque because of the fact that I dont understand how this problem was solved.


    2. Relevant equations



    3. The attempt at a solution

    The problem is solved, I just need someone to tell me why when doing the sum of torques a 2000 cos 65 was used . I belive the torque should be the perpendicular component of the force to the rod or for the uniform boom. I wrote instead 2000 sin of 25 . any help please?
     

    Attached Files:

  2. jcsd
  3. Sep 3, 2011 #2
    please anyone? and why is the tension T divided as tx and ty in the sum of torques? T already has 90 degrees in respect to the axis , so shouldnt the torque of T just be T multiplied by its lenght?
     
  4. Sep 3, 2011 #3

    PeterO

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    Torque = Force x Radius of operation

    You either take the component of the force at right angles to the lever F.cos65
    OR
    You take the effective radius of operation R.cos65

    SO either F x R.cos65 or F.cos65 x R giving F.R.Cos65

    EDIT: they have done a vertical and horizontal for the second part
     
  5. Sep 3, 2011 #4

    PhanthomJay

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    Gold Member

    Yes, you did find the torque of T the easy way...Serway first broke up T into its x and y components then summed torques of each.....to give the same result. But ultimately you need to break up T into its components anyway to get the reaction forces. Remember the 2 ways to find torques...Frsin theta or F times perpendicular distance from line of action of F to pivot point.

    Also note that AB sin25 = AB cos 65 ....
     
    Last edited: Sep 3, 2011
  6. Sep 4, 2011 #5
    The calculator wasent working well for some reason AB sin25 wasent the same as AB cos 65 i had to reset it. thanks a lot for your time .
     
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