Calculating Couple Moment - Explaining Negative Values

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SUMMARY

The discussion centers on calculating couple moments in mechanics, specifically addressing the use of negative values in moment calculations. The formula \(\sum M = \sum Fd\) is applied, with an example yielding a moment of \(-(48 \text{ lb})(4 \text{ ft}) + (40 \text{ lb})(\cos 30)(8 \text{ ft})\). The negative sign indicates a clockwise rotation, which is a standard convention in mechanics. Clarification is sought regarding the distance used in the calculations, with a suggestion that '8 ft' should be replaced with a variable 'd ft'.

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unknown_2
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Hey, I'm having some difficulty calculating couple moment. i understand that \sumM = \sumFd. here's an example:
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
 
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unknown_2 said:
Hey, I'm having some difficulty calculating couple moment. i understand that \sumM = \sumFd. here's an example:
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
 
I don't see where '8 ft' comes in. Shouldn't that be 'd ft'?
 

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