# Force couples - Finding resultant couple

1. Jun 26, 2009

### KEØM

1. The problem statement, all variables and given/known data
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 [Broken]

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

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?

Last edited by a moderator: May 4, 2017
2. Jun 26, 2009

### KEØM

I figured out part a after looking at the diagram for a while and realizing the signifigance 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?

KEØM

3. Jun 28, 2009

### nvn

Did you get the right answer yet, or not?

4. Jun 29, 2009

### KEØM

Sorry nvn, yes I did for both parts.