00:01
All right, we've got a rope that's 25 meters away, connected to something that's 25 meters away.
00:18
Now, the rope can handle a tension force of 29 kilonutons, but we're only going to design it to handle, to undergo 2 .9 kiloons.
00:51
So it's going to bend, i don't want to draw that there, it's going to bend downward like this, and this is going to be a distance x.
01:07
There's a mass here, and the mass is 72 kilograms.
01:22
Okay.
01:26
Now, what's the distance to meet that safety rate? so we're going to have tension here and tension here.
01:39
Now, if you look at the triangle, this angle is going to be the inverse tangent of x over d.
02:00
And that angle is this angle.
02:05
So also thinking about it, i want to draw it here, if this is theta, then this is this is x and this is d.
02:16
Wait a minute, that's only d over two.
02:22
Okay.
02:24
So, now, some of the forces in the y direction, 2t vertically times the vertical direction.
02:47
That's going to be the sign of theta.
02:55
Has to counter out m .g.
02:58
But the sign of theta is going to be x over the square root of x squared plus d over two squared.
03:10
So 2t times the sign to be x over the square root of x squared plus 2 squared.
03:28
T squared over 4 equals m g all right now we need to figure out x so i'm going to multiply by the square root of x squared uh all that stuff on both sides that's going to give me 2 t x equals m g square root of x squared plus d squared over four um but that really still doesn't help me much.
04:10
So i'm going to subtract 2tx from both sides.
04:15
I'm going to erase it here.
04:18
And then, wait a minute.
04:22
That did help me.
04:23
Let's go back.
04:28
Because now i'm going to square both sides...