Force couples - Finding resultant couple

[KEØM]

Homework Statement

The steel plate shown supports six 2 in. - diameter idler rollers mounted on the plate as shown. Two flat belts pass around the rollers, and rollers A and D are adjusted so that the tension in each belt is 10 lb. Determine (a) the resultant couple acting on the plate is a = 8 in., (b) the value of a so that the resultant couple acting on the plate is 480 lbin. clockwise.

It is problem 3.70 in the image.

http://img5.imageshack.us/img5/5109/problem3701.th.jpg

Here also is a link to the pdf: "[URL title="download file from Jumala Files"]http://jumalafiles.info/showfile2-46942995970229528183533325984723086/problem370.pdf [/URL]

Homework Equations

\vec{M_{C}} = \vec{r} \times \vec{F} = \vec{F}d
\vec{M_{R}} = \vec{M_{1}} + \vec{M_{2}} + \vec{M_{3}} + . . .

The Attempt at a Solution

I found the perpendicular distance in between the lines of action, d, for each couple. For the force couple that is horizontal the distance is 16 in. if a is 8 in. and for the angled couple i chose to find the distance between points F and C. Using the Pythagorean theorem if found this distance to be about 22.6 in. Using the right hand rule I found that both moments are clockwise.

Now using \vec{M_{C}} = \vec{F}d and summing all of the moments I find the resultant moment to be 386.3 lbin. which isn't right. What am I doing wrong here?

 

[KEØM]

I figured out part a after looking at the diagram for a while and realizing the significance of the diameter of the pulleys. The points on the diagram that I was measuring from are not located on the lines of action but are a pulley's radius away.

For part two I will set the sum of both of my moments equal to -480lb*in and then put their d's in terms of a and solve for a.

Are there any flaws in this approach?

Thanks in advance,

KEØM

 

[nvn]

Did you get the right answer yet, or not?

 

[KEØM]

Sorry nvn, yes I did for both parts.

Thanks for asking,

KEØM