1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    [​IMG]
     

    Attached Files:

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

    lightgrav

    User Avatar
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Concurrent And Parrallel Forces 3
  1. Concurrent forces (Replies: 3)

  2. Concurrent forces (Replies: 5)

Loading...