Statics is this problem easy or hard?

  1. Femme_physics

    Femme_physics 2,555
    Gold Member

    Statics....is this problem easy or hard?

    Yes, I know, easy and hard are relative statements, but I mean easy or hard compared to the other stuff I posted here-- if anyone remembers.

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


    A uniform beam, AB, whose length is 2 meters and mass is 10 kg is supported in its tip, A, by a smooth vertical wall and at its other tip B - he's tied to the wall by a wire BC, as depicted in the drawing. At what distance, AC, you tie the distance to the wall so the beam is at equilibrium? As well, calculate the tension on the wire (T), and the reaction force (R) of the wall at point A


    [​IMG][/URL]

    3. The attempt at a solution

    I just want to ask you guys for ideas. I can't find the angles of the damn thing. There's only one triangle in this diagram whose angles and lengths I can find-- that's AB as the hypotenuse of a right triangle whose other angles are 45 degrees. Without angles, what am I to do? I tried parallelograms... nothing. Moments won't get me anywhere as I don't have enough info with respect to the angles...you might say that without angles I got nothing... so no parallagram....no Pythagoras...no tricks? Help.... a lot of folks in my class are struggling with that problem...
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,041
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    Hi Dory! :smile:
    So far as the geometry is concerned, I don't see what the difficulty is :confused: … you add AC to the bottom bit, and you have the side of another right-angled triangle.

    Draw the forces on the diagram

    (why haven't you done so already?????)

    you know that the moments on the beam must add to zero (about any point), so look at the forces on the diagram, and decide which point do you think you should take moments about to make it easiest?
     
  4. Femme_physics

    Femme_physics 2,555
    Gold Member

    Re: Statics....is this problem easy or hard?

    Holy s***! I didn't see that. I add AC to the bottom bit!! That's friggin' brilliant :D That's how I discover AC right? I'll do at home... at work now....once I got the geometry I'll solve it :)
     
  5. Femme_physics

    Femme_physics 2,555
    Gold Member

    Re: Statics....is this problem easy or hard?

    In order to add AC to the bottom bit, I'd have to assume that the horizontal imaginary line connecting B to the wall is equal to AC...how do I know that for sure?
     
  6. tiny-tim

    tiny-tim 26,041
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    Sorry, Dory, I have no idea what you're talking about :confused:

    if you draw that horizontal line (BD say),

    then you have two right-angled triangles, BCD and BAD.
     
  7. Femme_physics

    Femme_physics 2,555
    Gold Member

    Re: Statics....is this problem easy or hard?

    Right, but what good are these triangles? I mean, I can find out everything about the smaller right triangle, but the bigger right triangle-- I only have its BD and no angle! So, 1 length and no angle...still stuck. Or am I missing something.
     
  8. tiny-tim

    tiny-tim 26,041
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    Dory, I get the impression that you don't like drawing forces on diagrams. :redface:

    You need to get used to this!

    Draw the forces on the beam (only) …

    how many are there?

    what do they look like?

    do they cross? at how many points? and where? :wink:
     
  9. Femme_physics

    Femme_physics 2,555
    Gold Member

    Re: Statics....is this problem easy or hard?

    There are 3! Na (wall's normal), Bc (rope pulling) and mass x gravity (split in middle)...there, tiny-tim..you know I'll always make diagrams for you if you only ask! :) [sorry, can't use the scanner this late so quality sucks]

    [​IMG]

    Uploaded with ImageShack.us

    I just wasn't sure how it helps me see the picture....should I build a triangle with these forces you suggest and I have the angles?

    Hmm, let me see... okay.... I tried building all sorts of triangles.... I know one angle is 45 degrees but it's not a right triangle...don't have the other two....
     
    Last edited: Feb 1, 2011
  10. tiny-tim

    tiny-tim 26,041
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    I know! :biggrin:

    ok, now you have 3 lines of force …

    they can meet in 3 points, or in one point …

    which is it? :wink:
     
  11. Borek

    Staff: Mentor

    Re: Statics....is this problem easy or hard?

    Completely OT, but I really think you should post this picture here.
     
  12. Femme_physics

    Femme_physics 2,555
    Gold Member

    Re: Statics....is this problem easy or hard?

    (Thanks Borek I did :) )

    Tiny-tim, I've already built that triangle in the picture. Do see it? It's a right triangle with the hypotenuse being BC. Problem is, I only got the fact it's a right triangle. The rest of the angles aren't given to me. Pleassssssse can you throw me a bigger bone?
     
  13. tiny-tim

    tiny-tim 26,041
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    Dory, why did you not answer my last post? :confused:

    ok, now you have 3 lines of force …

    they can meet in 3 points, or in one point …

    which is it? :wink:
     
  14. Femme_physics

    Femme_physics 2,555
    Gold Member

    Re: Statics....is this problem easy or hard?

    That one point is between C and B... I THINK. Technically you can move vectors around so I'm not sure as to significance of where they meet. I do see the significance of builing a triangle with them. Those 3 points are C, B, and the straight line I've drawn in red that connects B to the wall.
     
  15. tiny-tim

    tiny-tim 26,041
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    No!

    You can move free vectors (like velocity) around (that's how we make vector triangles of velocities),

    but you can't move bound vectors (like force) around …

    a bound vector (I'm not sure that's the correct name, btw :redface:) has a line which is part of the vector

    a force applied to the edge of something does not have the same effect as a force applied to the centre.
    Suppose a body in equilibrium has exactly 3 forces acting on it, and suppose that the three lines of force meet in 3 different points …

    choose one of the points, and take moments (torques) of forces about that point …

    what do you find? :wink:
     
  16. Femme_physics

    Femme_physics 2,555
    Gold Member

    Re: Statics....is this problem easy or hard?

    First off I used my better trigonometry to find AC = 1.41

    Now I used the Law of Sines to find out that the angle between CB and AB is 24.11.

    Now, being very excited and feeling close to the solution, I do this:

    Sum of all moments on A = 0 ; Tsin(24.11) x 2 - mgcos(45) x 1 = 0

    Then I see that the result of T is way off to what T should be ... then I get a myocardial infraction. :(

    Is my equation correct?
     
  17. tiny-tim

    tiny-tim 26,041
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    hmm …

    i] how did you find AC = 1.41 (= √2) ?

    ii] how did you get 24.11° ?

    iii] how about answering my last question?
     
  18. Femme_physics

    Femme_physics 2,555
    Gold Member

    Re: Statics....is this problem easy or hard?

    I used parallelograms to find out AC = 1.41

    See here

    [​IMG]

    :D

    I got 24.11 degrees by knowing the following information:

    [​IMG]

    To answer your last question I thought I did by doing sum of all moments on A! That's one of the points where the vectors meet. I don't mean to disapppoint you, I'm trying! See, I'm scanning and writing and ****...it's a tough problem!
     
  19. tiny-tim

    tiny-tim 26,041
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    Now you've swapped over A and C !

    Are you trying to confuse me? :biggrin:
    But I don't understand how you knew it was a parallelogram. :confused:
    But only one force (on the beam) goes through A (either A :rolleyes:) …

    Try the point where the normal force and the weight of the beam meet. :smile:
     
  20. Femme_physics

    Femme_physics 2,555
    Gold Member

    Re: Statics....is this problem easy or hard?

    Oops, I swapped A and C by accident! Heh.

    How did I know it was a parallelogram? I just drew parallel triangles :) No? You saying I someone cheated? Humphff!!! ;)

    Normal force and weight of the beam meet at mg, thus:

    Sum of all moments on mg = 0 ; -Na(cos45) x 1 + T(sin24.11) x 1 = 0
    Sum of all forces of X = 0 ; Na - T(sin 45+24.11) = 0

    Hmm...but according to this T = 0 .....
     
  21. tiny-tim

    tiny-tim 26,041
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    where's mg ? :confused:

    anyway, if the normal force goes through mg, how can its moment about that point be anything other than 0 ? :redface:
     
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