Couple forces

[unknown_2]

Hey, i'm having some difficulty calculating couple moment. i understand that \sumM = \sumFd. here's an example:
http://i20.photobucket.com/albums/b219/twiztedazn/question.jpg
it asks to resolve the couple forces.

the answer is:
(4/5)(60lb) = 48lb
and
(40lb)(cos30)

\sumM = -(48lb)(4ft) + (40lb)(cos30)(8ft)

Now, i don't understand why "-(48lb)(4ft)" is negative. I'm not sure when to use a negative value. Can someone explain this to me?

Cheers

[stewartcs]

Hey, i'm having some difficulty calculating couple moment. i understand that \sumM = \sumFd. here's an example:
http://i20.photobucket.com/albums/b219/twiztedazn/question.jpg
it asks to resolve the couple forces.

the answer is:
(4/5)(60lb) = 48lb
and
(40lb)(cos30)

\sumM = -(48lb)(4ft) + (40lb)(cos30)(8ft)

Now, i don't understand why "-(48lb)(4ft)" is negative. I'm not sure when to use a negative value. Can someone explain this to me?

Cheers

It's a matter of convention. Typically, the clockwise direction is negative and counter-clockwise is positive. So if the applied force causes rotation in the clockwise direction, the moment is negative and vice-versa.

CS

[Andy Resnick]

I don't see where '8 ft' comes in. Shouldn't that be 'd ft'?