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

AI Thread Summary
The discussion focuses on calculating the length of cable AB and expressing the force at point A as a Cartesian vector in a statics problem. Participants clarify the coordinates for points A and B, noting that point A is not at the origin but at the end of the window. The correct coordinates for point A are debated, with emphasis on the x, y, and z values derived from the geometry of the setup. The inclination of the plate at a 30° angle is highlighted as a crucial factor in determining these coordinates. Overall, the thread emphasizes the importance of accurate diagram representation and coordinate assignment in solving statics problems.
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.
 

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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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