Center of Gravity, Non-Uniform Bar

1. Aug 28, 2004

JasonRox

This is a question I need to do, I'm just going to put an example on what the question is like. In case a teacher of mine comes on here, I know I'm not "cheating".

Here is the image:

l\
lz.\................/l
l....\............./xl
l......\______/...l
l.......l____q_l...l
l.....................l

*Ignore dots.

z=56.2 degrees
x=90-56.2 degrees
q=center of gravity
lenght of bar is 4.3m

Find the distance from the right side of the bar, where the center of gravity is located.

This is all the information they give you. Simliar style, but not the same.

I started with using the Force of the bar as 10N. Easy number.

Then I find the force on the ropes, using trigonometry. I also find the horizontal forces as well.

The part I'm stuck on is how much force is actually used by each pole to keep the bar up.

The total vertical force must equal 10N, or else it won't stay up, or go up.

Do I use the horizontal forces to determine them?

I was thinking of using the difference between the two horizontal forces, and use the ratio to determine how far it is from the center.

Is this correct?

Last edited: Aug 28, 2004
2. Aug 29, 2004

jepe

Yes, you can use the horozontal forces to solve the problem.
The clue is that, since the bar is in equilibrium, the total horizontal force = zero.
Hence, the horoizontal components of the ropes are equal and in opposite directions.
Work it out from here

3. Aug 29, 2004

JasonRox

I found it now.

I think I did anyways.

Last edited: Aug 29, 2004