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

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Homework Help Overview

This discussion revolves around a statics problem involving 3D rigid body equilibrium, specifically focusing on why only the k component of tension is considered when taking moments about the x-axis.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the reasoning behind considering only the k component of tension when calculating moments about the x-axis. There are attempts to clarify the roles of different components of the tension vector and how they relate to the moment calculations.

Discussion Status

The discussion is ongoing, with participants providing insights into the vector components and their contributions to the moment about the x-axis. Some participants express confusion and seek further clarification on the vector set and its implications for the problem.

Contextual Notes

Participants note that the i component of tension is parallel and thus not considered, while the j component is said to cross the x-axis, leading to the conclusion that only the k component contributes to the moment about the x-axis. There is an acknowledgment of the need to understand why certain components are zero in the context of the problem.

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Homework Statement


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

Homework Equations


Sum of Forces = 0
Sum of Moments = 0
{Its in equilibrium}


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 ? ? ? :cry: :cry: :cry: :cry:)
 

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

T=Txi+Tyj+Tzk
 
rock.freak667 said:
Your diagram looks as if the tension vector is just T=-Tj. Is that or is it of the form

T=Txi+Tyj+Tzk

no, its
T=Txi+Tyj+Tzk
 
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)

[tex]\Sigma Mx = 0[/tex] (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 don't get how the part underlined and written in bold comes.
 
sgvaibhav said:
[tex]\Sigma Mx = 0[/tex] (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 don't get how the part underlined and written in bold comes.

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.
 
rock.freak667 said:
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.
ummm I am still kind of confused.
can u show me the vector set / vector form / matrix.
that might clear my confusion.
 
sgvaibhav said:
ummm I am still kind of confused.
can u show me the vector set / vector form / matrix.
that might clear my confusion.

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.

So your answer basically just found what would give i components.
 
[tex]\vec M=\vec r\times \vec T[/tex]
[tex]=(yT_z-zT_y)\vec i+(zT_X-xT_z)\vec j+(xT_y-yT_x)\vec k[/tex]
Maybe Ty and x-axis are the intersection.
 
i think i got it some other way.
I will try explaining what i understood (lol :P)
tell me if I am 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
sgvaibhav said:
i think i got it some other way.
I will try explaining what i understood (lol :P)
tell me if I am 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 ? ? ?

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
Phrak said:
You may consider it, but it is zero, which may, or may not come in useful.

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

THANKS EVERYONE :smile:o:):approve:
 

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