Concurrent And Parrallel Forces 3

1. Feb 1, 2006

kgbwolf

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

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2. Feb 2, 2006

kgbwolf

Bump..................................................................................

3. Feb 2, 2006

lightgrav

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.

4. Feb 2, 2006

Valhalla

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.