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Need some Math help from math guru!

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Old Aug 1, 2010 | 12:21 PM
  #1  
nMeOnE's Avatar
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From: Charlotte
Default Need some Math help from math guru!

Ok, i got a problem that i need to solve using a 6x6 matrix.

basically a *[6x1]=[6x6][6x1]

my equation are

A = 30 degree

.3(Pcos(A)) - 100(9.81)(0.1) = 0
Raz(.8) - 100(9.81)(.5) + P(.4) - .3(Psin(A)) = 0
Raz - 100(9.81) + Rbz-P = 0
(.3)Pcos(30) - Ray(.8) = 0
Rax - P = 0
Ray - Rby = 0

for the *[6x1] =

P
Rby
Rbz
Ray
Raz
Rax

Can anyone write out this matrix equation for me?

I can solve the problem easy without a matrix but i have to put it in the matrix form as given and solve it for a range of degree from 0-30
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Old Aug 1, 2010 | 12:43 PM
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I started to work on this, but your notation is whacko: in the equations you have P, Rza, Rzb, Ryz, Rxa, Rya, and Ryb,

whereas in the vector you have: P, Rby, Rbz, Ray, Raz, and Rax.

Are Rby and Ryb the same variable? How about Ray and Rya? Where's Rxy in your vector?

This is an easy problem if I know the variables. It's impossible if I don't.
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Old Aug 1, 2010 | 12:45 PM
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From: Charlotte
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hahah sorry about that but yes it is the same, must had been a typo trying to type it so fast. post modified to correct equations.
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Old Aug 1, 2010 | 03:31 PM
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If I read this correctly, your system can be written as:

0.3*cos(A)*P + 0*Rby + 0*Rbz + 0*Ray + 0*Raz + 0*Rax = 100*9.81*0.1
(0.4 - 0.3*sin(A))*P + 0*Rby + 0*Rbz + 0*Ray + 0.8*Raz + 0*Rax = 100*9.81*0.5
-1*P + 0*Rby + 1*Rbz + 0*Ray + 1*Raz + 0*Rax = 100*9.81
0.3*cos(A)*P + 0*Rby + 0*Rbz + -0.8*Ray + 0*Raz + 0*Rax = 0
-1*P + 0*Rby + 0*Rbz + 0*Ray + 0*Raz + 1*Rax = 0
0*P + -1*Rby + 0*Rbz + 1*Ray + 0*Raz + 0*Rax = 0

If we let x be the vector:

[P]
[Rby]
[Rbz]
[Ray]
[Raz]
[Rax],

A be the 6 x 6 coefficient matrix:

[0.3*cos(A) 0 0 0 0 0]
[(0.4 - 0.3*sin(A) 0 0 0 0.8 0]
[-1 0 1 0 1 0]
[0.3*cos(A) 0 0 -0.8 0 0]
[-1 0 0 0 0 1]
[0 -1 0 1 0 0], and

b be the vector:

[100*9.81*0.1]
[100*9.81*0.5]
[100*9.81]
[0]
[0]
[0],

then the system can be written as:

Ax = b.

The solution is:

x = (A^(-1))b

(i.e., A-inverse times b).

I leave it to you to compute A-inverse.
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Old Aug 1, 2010 | 05:02 PM
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I was going to post that, but magician beat me to it.
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Old Aug 1, 2010 | 05:17 PM
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Originally Posted by nMeOnE,Aug 1 2010, 12:21 PM
Ok, i got a problem that i need to solve using a 6x6 matrix.
When you get a solution, please post it.

In what context does this problem arise?
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Old Aug 1, 2010 | 05:58 PM
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Originally Posted by Incubus,Aug 1 2010, 05:02 PM
I was going to post that, but magician beat me to it.
lies!!!!
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Old Aug 1, 2010 | 07:00 PM
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thats why i'm a production operator and not an engineer, i never was good at math...and when they put letters in it, that f**ked my world up
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Old Aug 2, 2010 | 12:56 PM
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You are good magician.
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Old Aug 2, 2010 | 01:11 PM
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Originally Posted by Vik2000,Aug 2 2010, 12:56 PM
You are good magician.
Thanks!
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