I don't know either of the sags at point P1 or P2, but I do know in this case that

the maximum tension allowable at any segment of the cable is 1400 pounds.

First thing I'd like you to do on your own

is to determine the reactions at point A and B.

And so you can go through and do that, and once you've completed

that come on back, and I'll show you what you should have done together.

'Kay so you should have drawn, whoops, a free body diagram of the entire structure.

And if you did that, some moments about A, you could

find By, some forces on the y direction, you find Ay.

And if there is some forces in the x direction, you can't solve for Ax

or Bx explicitly again, but you can solve

for their relationships, which is that they're equal.

And so having done that.

those are the results.

We want to find Bx and Ax, and so what I'll do is, I'll go ahead and use

the fact that the x components are constant throughout the cable, and

so we can therefore find the section with the largest tension.

And the one with the largest tension is

the one that will have the largest y component.

Because if you take the x

component which is the same throughout and you use

Pythagorean's theorem to find the overall tension in the segment.

The one with the y component that's the largest

will be the segment with the largest overall tension.

And so in this case, that will be

section BP2, because it has the, the highest slope.

And so in doing that, we find that TB2 then is1400

pounds in tension.

So we've solve for one of the ta, cable segments.

and then let's do a joint cut at B to find B sub x.

And so here is the joint cut a joint B, I know what it, B sub y is, I

don't know what B sub x is. I do however now know that the TB2 is 1400

pounds, I just don't know the angle at which it'd, which it's at.

And so to do that, here is my joint cut again, what I'm going

to do is I'm going to apply the equations of equilibrium to that joint cut.

First of all some forces in the y direction, set it equal to zero.

I'll choose up as positive. And so I'll get 853.3

minus the 1400 pound tension in B2.

But just it's y component which is going to be the sign of

theta B2 equals 0. so I'll find sine of

theta B2 equals 0.6095, or theta

[COUGH]

excuse me, B2 is equalled to 37.56 degrees.

Now knowing that I can sum forces in the x direction.

Set it equal to 0, and I get the x component of the 1400 pound

force which is going to be negative according with my sign convention.

So this is going to be minus 1400 times the cosine

of theta B2, which was 37.56 degrees,

plus BX equals 0. And so therefore,

BX equals 1109.9 pounds to

the right, which means that Ax is

going to be equal to 1100.9,

1109.9 pounds to the left.

there's the joint cut at B. Bx is now found, By

was found, I know that TB2 was 1400

pounds. And so I have this distance is 6 feet.

I don't know what the distance y2 is, but I can use the geometric

relationship using the angle here. So I have got opposite

over adjacent will be the tension function.

So the opposite side is y2

over 6 is equal to the tangent

of 37.56 degrees.

Or y2 ends up equaling

4.613 feet, which

is the sag at point P2.