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

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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
 
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on Phys.org
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.
 
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.
 
So in this case, what is the coordinate?
 
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.
 
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?
 
So z=-300sin30 x=300cos30 and y=500.
 
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.