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More advanced FBD problems-need some tips!

  1. Oct 28, 2005 #1
    More advanced FBD problems--need some tips!

    Okay, just got assigned 3 more FBD-based problems with which I not only have to do the normal calculations, but also use various computer programs of choice for additional calculations, plots, etc.
    As such, these are a bit more advanced than the FBD problems I've done this far, and I would GREATLY appreciate some guidance/tips on successfully solving each one.

    1)http://www.ihostphotos.com/show.php?id=182814" [Broken]
    2)http://www.ihostphotos.com/show.php?id=182816" [Broken]
    3)http://www.ihostphotos.com/show.php?id=182817" [Broken]

    I get the basic concepts, but each of these problems is a little more complicated (or, moreso, just are seeming that way right now) than I'm used to.
    THANKS! Any and all suggestions welcome, as I start trying to work on these buggers...
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 28, 2005 #2
    your page is to big to see you need to shrink it
  4. Oct 28, 2005 #3
    I know...it's fixed now^
    Last edited: Oct 28, 2005
  5. Oct 29, 2005 #4
    The sizes of the pages are fixed now. Anybody have tips/clues as to how to start and find each one? Don't want answers, just a running start and direction.
  6. Oct 29, 2005 #5
    Still haven't gotten very far. Anyone have any tips on resolving the geometries of each? That seems to be what's throwing me, to start with.

    1) Can't figure out what/how gets combined to find the distance "a", to start with?
    2) Angles....how to express theta as a function of M, W, and I?
    3) Similar to two, but now how to express the Tension as a function of theta?

    This probably isn't too difficult, but I'm just having trouble thinking how to start/layout the steps, especially for the ones with angles.
  7. Oct 30, 2005 #6


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    Geometries? I'd use the rectangular coordinates that houses are built on.

    1) 0,0 at B; tan(theta) = 2R/a = (Z_p)/(X_p), L - sqrt(4R^2 + a^2)
    it might be useful to take Moments around the z-line of action of P
    at the x-line of action of the Pipe's F_N (beneath left side of a)
    You know B_x already , so that will get you B_z
    But there aren't any WRONG places the sum Moments around ...

    2) at least it's a 90 degree angle! and frictionless ...
    try taking Moments around the place where the two F_N intersect
    did you sum F_x and F_z first?

    3) this one's pretty easy. Moments around A, gives you T.
    back-substitute into Sum F_x and F_z, to get reactions at pin.
  8. Oct 30, 2005 #7
    Okay, THANKS--that should get me going!
  9. Oct 31, 2005 #8
    Okay, finally got a chance to get back to these problems.

    For #1:
    a) Determining distance "a" as a function of known length "l" and the radius of tube "r":
    You said: "0,0 at B; tan(theta) = 2R/a = (Z_p)/(X_p), L - sqrt(4R^2 + a^2)"

    So, should I simplify down to the point that I get a "a=" with L & R on the on the right?

    And with this:
    "it might be useful to take Moments around the z-line of action of P
    at the x-line of action of the Pipe's F_N (beneath left side of a)
    You know B_x already , so that will get you B_z"

    So, Bx = 0? Then how do I get Bz? Similarly, what is the z-line of action of P at the x-line of action of the Pipe's Normal Force--and how do I get moments from that?

    Sorry to be so confused.....it doesn't seem too hard but having so many unknowns is making me lost.
    Last edited: Oct 31, 2005
  10. Nov 1, 2005 #9
    Anybody? Am I just trying to simplify down that expression for the first part of #1, or what?
  11. Nov 1, 2005 #10
    I'm still confused with #1.:rolleyes:

    From this:
    Okay. So, to determine the distance a as a function of the known length L and radius r, in order to hold the rod in equilibrium with force P, am I taking this:

    2R/a = L - sqrt(4R^2 + a^2)

    and simplifying down so a is alone on one side basically? What about the (Pz) and (Px) mentioned above or the tan(theta)? I don't really need to include those for part a), do I?
  12. Nov 1, 2005 #11
    Should I FIRST be finding the total moments and total forces?

    There's just something about this "easy" problem I'm not getting....:mad:
  13. Nov 1, 2005 #12


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    Sum of Forces in the x direction = 0 (this *IS* statics, no?)
    there's a P in the -x direction, anf there's no friction in the pipe.
    So the Bx must cancel Px.
  14. Nov 1, 2005 #13

    Yeah, it's Statics alright...too bad I tend to forget the simple things.

    What about finding the factor a as a function of L and R? Am I headed the right direction with that or no?
  15. Nov 1, 2005 #14


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    I think the best way to involve parameters like "a" and "R"
    is just to write the torque equations. They'll be messy enough,
    you don't need to resort to involved geometry (Pythagoras, trig)
    to include the important parameters.

    If some problem asks you to write an important quantity
    as a function of an UNimportant variable,
    tell them to walk off a cantilever.
    (an important variable will show up all by itself
    as you write the statics Force component and torque eq'ns)

    by the way, once you know Bz, then Fz=0 gets you az.
    but you have to know how far down the force P is applied.
    That's the only reason to do this geometry work.
    Last edited: Nov 1, 2005
  16. Nov 1, 2005 #15

    Summing the forces:
    FB=0 -> (-Px + Bx)i + (Py + By - Ay)j ?

    Did I forget the sin and cos?
    Last edited: Nov 1, 2005
  17. Nov 1, 2005 #16


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    The Force P is horizontal, so Py = 0.
    Don't forget the weight of the rod.
  18. Nov 1, 2005 #17
    After doing the sum of forces, got:
    Ay= -49jN
    By= 49jN

    Then finding the moments about B, as an example, would this be right?
    MA=0 -> [(L sin(theta))j x -Pi] + [Byj x (sqrt(4R^2 + a^2))i]

    Last edited: Nov 1, 2005
  19. Nov 1, 2005 #18


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    I'm pretty sure 10 kg weighs 98N (-j) here, so By + Ay = 98N .
    otherwise ... looks like these are moments around A ...
    where is the 98N thru c.o.m.?
    Last edited: Nov 1, 2005
  20. Nov 1, 2005 #19
    Yeah, I had:
    F=0 -> (By-Ay-W)j=0
    (By-Ay)j = Wj
    (By-Ay)j = 98jN
    By = 49jN
    Ay = -49jN

    Okay, does look like I'm FINALLY getting somewhere on this problem. But, to once again go back to part a), how would I actually write out the equation where a is a function of L and R? Like in the very beginning, what I was talking about, or what?

    The big point of this assignment (3 problems) is to resolve certain components into functions of other components, and then plot that data using a couple given ranges and stuff.
  21. Nov 1, 2005 #20
    Okay, got this total moment:

    (P*L*sin(theta) - By*cos(theta)*sqrt(4R^2 + a^2) + (W*L/2)*cos(theta))k N-m = 0

    I have By as calculated previously, but do I really get out of this total moment? Is it how I get my a = function of.... value?
    Last edited: Nov 1, 2005
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