Concurrent And Parrallel Forces 3

[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

 

[kgbwolf]

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

 

[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.

 

[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.