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Homework Help: Static: Equilibrium 3D

  1. Nov 10, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the reaction in the socket and joint at O and the tension in each cable.
    http://img108.imageshack.us/img108/7585/hwstaticod1.jpg [Broken]

    2. Relevant equations
    [tex]\sum{F_{x}}[/tex] = 0
    [tex]\sum{F_{y}}[/tex] = 0
    [tex]\sum{M}[/tex] = 0
    r x F = M

    3. The attempt at a solution
    My teacher actually solved the problem for us. But I when I go back and tried to follow what he did, I came across a few problem. So I'll just start with question 1 first, hopefully I'll be able to solve the rest.

    First, my teacher list out all the unit vector.

    Tac= [tex]\frac{-1}{\sqrt{6}}[/tex]i + [tex]\frac{1}{\sqrt{6}}[/tex]j - [tex]\frac{2}{\sqrt{6}}[/tex]k

    My question is, how can I look at the graph and get i, j, k? I understand where the [tex]\sqrt{6}[/tex] comes from, but not the number on top.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 10, 2007 #2
    that picture is so small i cant see it..
  4. Nov 10, 2007 #3
    heheh sorry. I forgot to check preview the post before posting. It's fix :D
  5. Nov 11, 2007 #4
    hmm k well its pres easy just look at the dimensions given. From the picture look at the AXIS first, i repersents x , j represents the y axis, and k represents z.

    SO TAC is going in the negative x and z direction while since its going UP not down its positive y. Now look at the dimensions. At the moment since ur having problems visualizing i think rather then just looking at picking the i j and k components DIRECTLY you should find the (x,y,z) of point c and (x,y,z) of point a.

    so tehrefore

    point c (-1m,+1m,0m)
    point a (0m,0m,+2m)

    now u want Tac so subtract c-a = (-1,1,-2) BUt to find a unit vector the form is

    Tac = T*lambda(ac) = T*vector ac/magnitudeac

    vector ac u know (-1,1,-2)
    magnitude of ac is just the root of the sum of the squares root((-1)^2+(1)^2+(-2)^2) = root6

    Tac = T*(-1,1,-2)
  6. Nov 11, 2007 #5
    Thank you salman!! This way seems much easier. Hehe I got point a, but I still don't see how you got c. From the origin, it seems to me, z = 1, and x = 0.

    http://img115.imageshack.us/img115/6337/hwstaticaj3.jpg [Broken]
    Last edited by a moderator: May 3, 2017
  7. Nov 11, 2007 #6
    ahh i see umm the problem is ur viewing it wrong i think

    that 1 m u circled in red is ALONG THE X axis NOT the z axis. Basically its not from the Y axis OUT OF THE PAGE its HORIZONTALLY along the X axis.

    http://img214.imageshack.us/img214/3694/14739152uh7.jpg [Broken]
    Last edited by a moderator: May 3, 2017
  8. Nov 12, 2007 #7
    Thank you. I see it now. But I have another question. The next unit vector is T(be) = 0i + 0j - 1k.

    But from the picture, I see

    E = 1.5i + 0j + 0k
    B = 1.5i + 0j + 2k

    T(be) = 0i + 0j - 2k
  9. Nov 12, 2007 #8


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    Science Advisor

    You list the vector, not the unit vector. Don't forget to divide by the magnitude. Or you can think of it this way...the magnitude of a unit vector is 1. Since you have only one component, it's value must be = 1.
  10. Nov 12, 2007 #9
    Oohhhh, I forgot about that. Thank you FredGarvin :D

    Heheh I have a new question, again with unit vector.

    I have the weight as [tex]\frac{.75}{2.14}[/tex]i + 0j + [tex]\frac{2}{2.14}[/tex]k, but my teacher has it as 0i -1j + 0k.

    [tex]\sqrt{.75^{2} + 2^{2}}[/tex] = 2.14
    Last edited: Nov 12, 2007
  11. Nov 12, 2007 #10
    when ur looking at the mass what points did u subtract?
    the weight acts DIRECTLY downwards therefore it only has a Y COMPONENT.

    When dealing with weights just multiply the mass by 9.81
    400 kg in this case by 9.81 to get force (f=mg)

    that force in NEWTONS will basically be the y component..

    (0i - (400*9.81)j + 0K)

    it is negative since the weight is acting DOWNWARDS...
  12. Nov 12, 2007 #11
    I didn't subtract it from anything. I got the points from the origin.

    I worked out the problem, it matches with my teacher now :D Thank you.

    Thank you very very much for answering all my questions. You make static seems like a learnable subject ;) I'll be sure to come back for more question. :D
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