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Concurrent And Parrallel Forces 3

  1. Feb 1, 2006 #1
    No. 22: The force of friction of a mower is 20 lb. What force must the man (in fig. 6-23 )exert along the handle to push it at constant speed ? Hint: look up example 6 in this section. You can get up to 10 points for answering correctky this discussion before monday's class.


    Did i work this out correctly??

    M = Cos45 ^x MSIn45^y
    F= 20LB ^x 0LB ^y
    Mcos45+20=0 28.8sin45=20.3LB ^y
    -20/cos45= -28.8LB ^x

    attachment.php?attachmentid=6208&stc=1&d=1138814934.jpg
     

    Attached Files:

  2. jcsd
  3. Feb 2, 2006 #2
    Bump..................................................................................
     
  4. Feb 2, 2006 #3

    lightgrav

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    almost ... the man's Force IS 20 lb / cos(45) ,
    but that is along the diagonal handle, not in the negative x-direction.

    Either stick with components (-20, -20) lb , or fix your direction.
     
  5. Feb 2, 2006 #4
    Ya, you have issues with your directions. The friction and the direction of motion are going to be opposite directions and signs. Other than that your setup looks good.
     
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