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Pitch and wrench

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data
    please replace the force-couple system of the previous problem with a wrench. Please state the pitch of the wrench and the coordinates of the point at which the wrench axis intercepts the xy plane


    2. Relevant equations



    3. The attempt at a solution

    ok there are 3 forces going in different directions at different points. the force-couple system they are referring too consists of the moment about point D (MD) and the resultant force (R) if I understand correctly.

    MD = <2376.1, 950, -710>
    R = <238.1, -46.9, 237.7>

    note: 1 force created no moment because it was alligned with point D

    pitch (p) = (R MD)/(magnitude of R^2) = 3.054

    that much seems easy, im confident that i did that much correctly. I was trying what the book told me to do for the rest and I keep getting answers with infinity and unmatching answers. they said M1 = pR. so I did pR to get M1.
    then they said: M1 + rXR = MD
    so i need to figure out vector r. so i set rXR = MD - M1.
    if i set r = <x,y,z> and plug that in then i get answers with infinity (for x,y, and z)
    they ask for the xy axis so on my next attempt i set z = 0 and plugged in r = <x,y,0> then i got answers that didnt match...
    any idea what I am doing wrong?

    p.s.
    i also tried just doing rXR = MD. i had the same results

    OK i just relooked the part where i put in r = <x,y,0>
    i came up with value for x and y but in the 3rd equation it doesnt add up to 0 like it should. it is off by .0722

    is this a problem? or is it only a matter of signifigant figures? because if its just SF then i think i actually did the problem correctly. just let me know if you see anything wrong with the concept of what i explained
     
    Last edited: Nov 8, 2008
  2. jcsd
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