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Homework Help: [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

    R=[tex]\Sigma[/tex]F=

    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
     

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

    Doc Al

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    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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    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

    [tex]\Sigma[/tex]F=

    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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    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

    rl.bhat

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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

    rl.bhat

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    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.
     
  22. Nov 8, 2009 #21
    Ok so the ΣF is correct...

    but how is ΣM wrong?, you threw me in a loop there because I was following the book

    when I did that portion, see like I said before sometimes this book is not worth 150 bucks!.


    Ok so from what you said there, it needs to be like this?(keep in mind I am trying to use

    the book also, it says something about using 1/2 of the distance in the Uniformly Load

    line)


    Is it this?, or not even close( next post will feature my 2nd try )

    [tex]\Sigma[/tex]M= -100( 0 ) - 700 ( 8 )- 160 ( 4 )= -6240 ft/lb

    help me out here, if I get this part done the problem is pretty much done, since the X/bar portion is easy.
     
    Last edited: Nov 9, 2009
  23. Nov 9, 2009 #22
    I found this similar exercise in the book but the diagram is mirrored, the line load is on the left and the rest of the beam is on the right.

    I am trying to use that though is not the same type of problem. it is helping me understand how to get the moments and reactions.

    I used it and I got this for mine, not sure is if correct though, just me trying to learn how to get the moments/reactions, can you check it out for me please.


    [tex]\Sigma[/tex]M= -100( 16 ) -700(8) - 160( 4 )=
     
    Last edited: Nov 9, 2009
  24. Nov 9, 2009 #23

    rl.bhat

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    What do you mean by the ratio moments/reaction? Through out the discussion you have not taken the reaction of forces on the two supports. But you expecting the ratio of moments/reaction! Write down the exact wording of the problem in the book.
     
  25. Nov 9, 2009 #24

    The problem I am using in the book is just an example, in order to understand how to take the reactions of the two supports. Read my previous post and see what I came up with using this example, clearly a whole new type of equation came out of it.

    The problem I am trying to get done here is just a sheet, with the statement as stated in the picture, teacher told us to get used to these types of problem, clearly the reactions factor flew over my head but now I got an understanding about it.

    btw I appreciate all the help, also I am sorry if this problem that you probably got done already is taking me years to finish, but you know I am trying to do my best to understand this, that is why I am here 2am( NYC) in the morning trying to get it done, hopefully I don't pass out, if I do then I will be back tomorrow.
     
    Last edited: Nov 9, 2009
  26. Nov 9, 2009 #25

    tiny-tim

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    Hi Dellis! Thanks for the PM. :smile:

    (just got up :zzz: …)
    Yes, it's not asking for the reaction force (though, of course, the main reason for finding the resultant is so that you can then easily find the reaction force! :wink:)

    I think the "resultant" means finding the single force and the point of application of that force, that would have the same effect as the given forces.

    So you not only must find the magnitude and direction, F, of the force, but you also need to know the distance, d, along the beam at which the force must be applied.

    So you just draw this force on the diagram, and then use the ordinary equations and the moment equations. :wink:
     
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