the other application we talked about was self-weight.

When you had like a rope swing or power lines.

what we find is, in the condition where we

have a uniform load, the cable forms a, a, parabolic

shape.

Now, under self weight you get a shape which is called a catenary curve.

And I'll let you study about catenary curves on your own.

I'd like to proceed with a span.

Where we have a, a, a uniform load and

again the rest of the bridge, we're going to consider massless.

And I'm going to use symmetry to just look at the

right half of the bridge, and we're going to want to find

the tension in the cable.

and to do that, we're going to first of all look at

the load itself and, I can tell from my earlier course

Introduction to Engineering Mechanics, what this load is, what the resultant

force is because Q sub x or, is, is a constant.

It's Q sub zero and it's expressed in units per, per units of force per length.

And so for

half the bridge the resultant force will

be, that Q sub 0 times L over 2.

And it will act since it's uniform and constant right at the center,

or at L over 4 from the right hand side. And

so, this x at

L over 4 from point B.

And so, once we know F sub R We can now sum forces in the y

direction to find the vertical component of the

tension in the cable up here at point B.

So if I sum forces in the vertical direction, set it equal to 0, I find

that T sub V equals Q sub 0, L over 2. Okay and so that takes care

of the vertical component of the tension. Let's continue, on now.