Solve Equilibrium 3D Problem: Find Reaction at O and Cable Tensions

[sourlemon]

Homework Statement

Find the reaction in the socket and joint at O and the tension in each cable.
http://img108.imageshack.us/img108/7585/hwstaticod1.jpg

Homework Equations

\sum{F_{x}} = 0
\sum{F_{y}} = 0
\sum{M} = 0
r x F = M

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= \frac{-1}{\sqrt{6}}i + \frac{1}{\sqrt{6}}j - \frac{2}{\sqrt{6}}k

My question is, how can I look at the graph and get i, j, k? I understand where the \sqrt{6} comes from, but not the number on top.

 

[salman213]

that picture is so small i can't see it..

 

[sourlemon]

heheh sorry. I forgot to check preview the post before posting. It's fix :D

 

[salman213]

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)
...----------
...root6

 

[sourlemon]

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

 

[salman213]

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

 

[sourlemon]

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

 

[FredGarvin]

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

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.

 

[sourlemon]

Oohhhh, I forgot about that. Thank you FredGarvin :D

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

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

\sqrt{.75^{2} + 2^{2}} = 2.14

 

[salman213]

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...

 

[sourlemon]

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