1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    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: Nov 10, 2007
  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.

  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.

    Last edited: Nov 11, 2007
  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


    User Avatar
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Static: Equilibrium 3D
  1. Statics (Equilibrium) (Replies: 1)