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Homework Help: Force couples - Finding resultant couple

  1. Jun 26, 2009 #1
    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]

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

    2. Relevant equations
    [tex]\vec{M_{C}} = \vec{r} \times \vec{F} = \vec{F}d[/tex]
    [tex]\vec{M_{R}} = \vec{M_{1}} + \vec{M_{2}} + \vec{M_{3}} + . . . [/tex]

    3. The attempt at a solution

    I found the perpendicular distance in between the lines of action, [tex]d[/tex], 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 [tex]\vec{M_{C}} = \vec{F}d[/tex] 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. jcsd
  3. Jun 26, 2009 #2
    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. :smile:

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

    Are there any flaws in this approach?

    Thanks in advance,

  4. Jun 28, 2009 #3


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    Did you get the right answer yet, or not?
  5. Jun 29, 2009 #4
    Sorry nvn, yes I did for both parts.

    Thanks for asking,

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