Static: Equilibrium 3D

  • Thread starter sourlemon
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  • #1
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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 [Broken]

Homework Equations


[tex]\sum{F_{x}}[/tex] = 0
[tex]\sum{F_{y}}[/tex] = 0
[tex]\sum{M}[/tex] = 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= [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.
 
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Answers and Replies

  • #2
302
1
that picture is so small i cant see it..
 
  • #3
77
1
heheh sorry. I forgot to check preview the post before posting. It's fix :D
 
  • #4
302
1
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
 
  • #5
77
1
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]
 
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  • #6
302
1
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]
 
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  • #7
77
1
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
 
  • #8
FredGarvin
Science Advisor
5,066
8
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.
 
  • #9
77
1
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
 
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  • #10
302
1
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...
 
  • #11
77
1
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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