How Do You Calculate Cable Length and Vector Forces in Statics?

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Homework Help Overview

The discussion revolves around calculating the length of a cable and expressing a force as a Cartesian vector in a statics problem involving a window held open by the cable AB. Participants are examining the coordinates of points A and B in relation to the geometry of the setup.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to clarify the coordinates of points A and B, with some questioning the accuracy of the z-coordinate for point A. There is a focus on the geometry of the setup, including the inclination of the plate and its effect on the coordinates.

Discussion Status

The discussion is active, with participants providing insights and corrections regarding the coordinates. Some have offered guidance on interpreting the diagram and the implications of the plate's angle. Multiple interpretations of the coordinates are being explored, particularly for point A.

Contextual Notes

There are constraints related to the geometry of the problem, including the inclination of the plate and the specific lengths associated with the coordinates. The original poster's diagram has been modified to clarify the position of point A, indicating ongoing adjustments in understanding the setup.

sevag00
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Homework Statement



The window is held open by the cable AB. Determine the length of the cable and express the 30N acting at A along the cable as a Cartesian vector.
http://img802.imageshack.us/img802/9599/ru5y.jpg

Homework Equations



vector(F) = FuAB

The Attempt at a Solution



Here's my solution
http://img833.imageshack.us/img833/6996/n7b0.png
 
Last edited by a moderator:
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Check the x and z coordinates of A.

ehild
 
Check the coordinates of B.
 
Yeah the B coordinates should be B(0,150,250) but i don't know what's wrong with the A coordinates especially the z one.
 
The plate is not horizontal, but inclined at 30°angle downward. 300 mm is the length of the edge.

ehild
 
300mm is the length of the edge which is the x coordinate, just like 500 which is the y coordinate.
 
sevag00 said:
300mm is the length of the edge which is the x coordinate, just like 500 which is the y coordinate.

Nope. Sketch the side view:

attachment.php?attachmentid=62900&stc=1&d=13817679824.gif


EDIT: Added point A to the diagram.
 

Attachments

  • Fig1.gif
    Fig1.gif
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Last edited:
But isn't the y of point A at the end of the window and not at 0?
 
sevag00 said:
But isn't the y of point A at the end of the window and not at 0?

Yes. The sketch is looking along the y-axis (the y-axis is coming straight out of the page), so all y-coordinates will be seen as flat against the x-z plane. Point A is at the end of the heavy line, not at the origin.
 
  • #10
So in this case, what is the coordinate?
 
  • #11
sevag00 said:
So the y coordinate will be 300, right?

How do you get that? I thought we were talking about the x and z coordinates for point A.
 
  • #12
sevag00 said:
So in this case, what is the coordinate?

I modified my diagram to show the location of point A (I thought it was obvious, but it never hurts to clarify). What x and z coordinates would you assign to point A?
 
  • #13
So z=-300sin30 x=300cos30 and y=500.
 
  • #14
gneill said:
How do you get that? I thought we were talking about the x and z coordinates for point A.

No. That was stupid. I edited my post.
 
  • #15
sevag00 said:
So z=-300sin30 x=300cos30 and y=500.

Yes, that looks better.
 
  • #16
Thanks for the help.
 

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