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[HELP]- Determine the Resultant of the Force System.

  1. Nov 7, 2009 #1
    Hi, I am having a little problem understanding this exercise, need some route to follow.
    I attached a picture of the exercise.

    - The Problem Statement

    -Determine the Resultant of the Force System.

    - Relevant Equations

    M= FxD


    X(bar)= [tex]\Sigma[/tex]FXD/R=

    -I tried this which I think is part of the process, if not please set me in the right path.

    20 lb/ft x 8 ft= 160 lb

    Attached Files:

    Last edited: Nov 7, 2009
  2. jcsd
  3. Nov 8, 2009 #2

    Doc Al

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    Staff: Mentor

    What do you mean by the "resultant"? The sum of those forces?
  4. Nov 8, 2009 #3
    The statement is asking to determine the resultant of the force system,

    The resultant is the sum of the forces yeah, I need some guidance in that regard.

    [tex]\Sigma[/tex]F= 100 + 700 + 160 -, ect....

    then finish it off by finding the Moments, then Xbar,

    Is that how you approach this exercise?
    Last edited: Nov 8, 2009
  5. Nov 8, 2009 #4
    Help, don't give up on me :)
  6. Nov 8, 2009 #5

    Doc Al

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    Staff: Mentor

    The "resultant" just means just the sum of a set of forces. Are they just asking you to add up several forces? Which ones? The three that are shown?

    I'm guessing that the real question is "Find the support (or reaction) forces in the following system". Could that be it?

    If that's what you mean, then you are on the right track. Use ΣF = 0 & ΣTorque = 0.
  7. Nov 8, 2009 #6
    I don't think is asking for reactions or anything like that, it is asking for summation of the forces aka Resultant for the entire force system. But I believe that is just part of the process in order to get the solution of the exercise.

    Going by another exercise in the book with the same statement but different diagram, shows this as the process, not sure if it would work for this one also?.

    1. you add the forces to get the Resultant


    2. then you do the summation of the Moments( I NEED HELP WITH THIS ), which I believe is something like this

    [tex]\Sigma[/tex]M= -100lb( 8 ) - 700 lb(4 )

    3. then at the end you finish it by putting [tex]\Sigma[/tex]M/ over the Resultant

    to get the answer.
    Last edited: Nov 8, 2009
  8. Nov 8, 2009 #7

    Doc Al

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    You added the three forces shown?

    Moments about what point? A? What did you get?

    I don't quite understand the significance of this last step.
  9. Nov 8, 2009 #8

    Doc Al

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    I don't see where these numbers come from. It doesn't seem to match the diagram you attached.
  10. Nov 8, 2009 #9
    Sorry I got confused by another exercise I was looking at, read it again I edited with the proper numbers.
  11. Nov 8, 2009 #10

    Doc Al

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    Staff: Mentor

    What point are taking moments about?
  12. Nov 8, 2009 #11

    1. 3 forces shown???, the last force is 160lb isn't it? (20lb/ft x 8)?

    [tex]\Sigma[/tex]F= - 100 lb - 700 lb - 160 lb=...

    2. Read below for that one

    3. That last step is literally in the book, with a problem with the same statement BUT it does

    not have the uniformly line load, its just a Simple Beam type of exercise.
    Last edited: Nov 8, 2009
  13. Nov 8, 2009 #12
    I did that following this exercise I am looking at in the book, they multiply the forces

    with the distances they come down on.

    For example like if at Point A their is a Force of 300 lb coming down on the

    given distance of 8 ft, they say this is part of the process summation of the moments

    [tex]\Sigma[/tex]M= 300lb( 8 )
    Last edited: Nov 8, 2009
  14. Nov 8, 2009 #13

    Doc Al

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    The moment produced by a force depends on the point you are using to calculate the moment. For example, in your problem the moment of the 100 lb force about point A is 100*0 = 0, about the midpoint of the beam it would be 100lbs*8ft, and about the right end of the beam it would be 100lbs*16ft.
  15. Nov 8, 2009 #14

    I was about to post that lol, I seen this part in the book showing just that!.

    Why you didn't show the 700 lb Force in that example you gave?
  16. Nov 8, 2009 #15
    Ok so far with what you said we can assume this right, I hope this is the right path, if you see something wrong please point it out.

    [tex]\Sigma[/tex]F= -100 lb - 700 lb - 160 lb=...

    [tex]\Sigma[/tex]M= -100( 0 ) - 700 ( 8 )- 160 (16 )=....

    I am here waiting for your reply, if any other mentor is reading this please chime in on this Please, I need to understand how to do these types of exercises.
    Last edited: Nov 8, 2009
  17. Nov 8, 2009 #16


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    Homework Helper

    When you are applying forces to an isolated system, you have to identify the actions and reactions.
    If there is no linear or rotational motion in the system, then ΣF = 0 and ΣM = 0.
    In your post you have not taken into account the reactions.
    Your moment due to 160 lb is wrong, because the center of mass of 160 lb is not at 16 ft.
    Whether lb is mass or weight? If it is mass, then weight is mg.
  18. Nov 8, 2009 #17
    So it is 20lb/ft x 8 then? or 20lb/ft x 4?, you know I actually had it like that but changed it when the mentor brought up his advice in the last post.

    Yeah ΣF = 0 and ΣM = 0, thanks for clearing that up.

    Also is the path I am going there right?, other then what you pointed out, is all good right?.

    R=[tex]\Sigma[/tex]F= -100 lb - 700 lb - 160 lb= -960 lb

    [tex]\Sigma[/tex]M= -100( 0 ) - 700 ( 8 )- 160 (8 )= -6880 ft/lb

    After that^, the X/bar portion comes into the equation and the division of the [tex]\Sigma[/tex]M over the Resultant is the solution, I mean going by the book that's what it seems to say but I want to make before I move on.
    Last edited: Nov 8, 2009
  19. Nov 8, 2009 #18


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    Homework Helper

    Your ΣM is wrong. All the distances should be measured from the left end. So check the moment due to 160 lb. And about the moment due to the reaction of the roller?
  20. Nov 8, 2009 #19
    This is clearly a first year mechanics problem... you need to find the reactions at A and B. Try taking the moment at A or B in order to eliminate the reaction at that point from the equation. Also remember to use the centroid for a constant distributed load.
  21. Nov 8, 2009 #20
    Is for Statics 1 class, I am trying to follow the book and is pointless sometimes, it seems like I am learning more from you guys here then the book.
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