# Why is only the k component considered? (Details inside)

1. Jun 15, 2010

### sgvaibhav

1. The problem statement, all variables and given/known data
This question is for statics - 3D rigid body equilibrium
When taking moments about the x axis, why is only the k component considered?
I have attached a picture of my situation

2. Relevant equations
Sum of Forces = 0
Sum of Moments = 0
{Its in equilibrium}

3. The attempt at a solution
Actually, moments are taken here about the x-axis, since there is no moment reaction about x-axis
Moment is taken about x-axis to determine the tension.
But only the K component is considered of the tension (WHY WHY WHY ? ? ? )

#### Attached Files:

• ###### Freebodydiag2.PNG
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2. Jun 15, 2010

### rock.freak667

Your diagram looks as if the tension vector is just T=-Tj. Is that or is it of the form

T=Txi+Tyj+Tzk

3. Jun 15, 2010

### sgvaibhav

no, its
T=Txi+Tyj+Tzk

4. Jun 15, 2010

### sgvaibhav

this is the exact working.

Tbc= (0.3/0.7Ti) + (-0.6/0.7Tj) + (0.2/0.7Tk)
MAx = 0 (Since there is no reaction)

$$\Sigma Mx = 0$$ (Taking moments about x axis)
(300 x 0.6) - Tbc (0.2 / 0.7) x 0.6 = 0
We obtain tension from this, but i dont get how the part underlined and written in bold comes.

5. Jun 15, 2010

### rock.freak667

If you take moments about the point A, you will have the terms MAyj and MAzk (note how there is no MAxi)

If the sum of moments is zero. Then the i component is zero. So you are crossing the moment arms and the forces to give you i

they are crossing 0.6j with Tbck since that will give you an i

j x k = i

Similarly, if you wanted to you could write out the entire vector set and just equate each component to zero and you will see it.

6. Jun 15, 2010

### sgvaibhav

ummm im still kind of confused.
can u show me the vector set / vector form / matrix.
that might clear my confusion.

7. Jun 15, 2010

### rock.freak667

If you sum the moments about the point it you will get this

r x F + MAyj + MAzk = 0

Crossing r andF might give you something like this (A-TBC)i+(D+TBC)j+(E+TBC)k (A,D,E are arbitrary constants). If put everything together you will get:

(A-TBC)i)i +(D+TBCMAy)j + (E+TBC+MAzk) = 0

If you equate each component to zero, the i component can be used to get TBC which can then be used to get MAy and MAz.

8. Jun 15, 2010

### mo_0820

$$\vec M=\vec r\times \vec T$$
$$=(yT_z-zT_y)\vec i+(zT_X-xT_z)\vec j+(xT_y-yT_x)\vec k$$
Maybe Ty and x-axis are the intersection.

9. Jun 15, 2010

### sgvaibhav

i think i got it some other way.
I will try explaining what i understood (lol :P)
tell me if im wrong or not...
(this is the exact question : https://www.physicsforums.com/attachment.php?attachmentid=26455&d=1276534148)
ok so we are taking moments about x-axis
when considering the moments tension about x-axis.

i component tension for moment will not be considered --- since its parallel
j component of tension would not be considered -- since it would cross the x axis
and the k component FOR THE WIN :D. it successfull exists :P

so we only consider the k component for the moment of tension (Tbc) about x axis.
am i correct ? ? ?

10. Jun 15, 2010

### Phrak

You may consider it, but it is zero, which may, or may not come in useful. But in any case, you've found out correctly why it is zero.

11. Jun 15, 2010

### sgvaibhav

yeah, but in begining stages, it is better to understand why its zero.
later on, it will just take a second to understand it....

THANKS EVERYONE