I know there is a downward force at the center of the wire of 539 newtons. Also, the upward components of applied force at either end is 269.5 newtons. However, I do not know where to go from there.

I did, but I don't think I can recreate it in a post very accuratelly...

Tag, Chris...you’re “it.”

Here's a free body diagram:



You've already pointed out that the system is in equilibrium, which means there is no acceleration, so therefore no net force.

So add up the forces in the x-direction to get one equation, and up the forces in the y-direction to get another equation. Set both equal to zero.

You're going to need to do some trigonometry and use the lengths given to determine the angle with the horizontal (shown as the dotted line).

Now let's see the equations you come up with.

ShawnBradley wrote:

So, would it be 2*T*sin(theta)=539 where theta is the angle T makes with the horizontal?

In the y-direction, yes.



so,



or,



Now all you have to do is figue out .

Something that is pretty neat is if you look at the above equation, as gets smaller and smaller, the tension gets bigger and bigger. This is why you can never pull a string completely straight (=0) if there is any mass on it. Try it. Go get a long piece of string, a small mass, and a buddy. Put the mass in the middle of the string and see if you and your buddy can pull the string taunt.

So, the answer is 539/(2*3.5/23)=1771, right?

I would argue that the wire could not sag without stretching—that the weight of the tightrope walker would cause the wire to increase in length. Such would yield a little more than a half-meter increase in the wire’s total length; instead of ShawnBradley‘s 1771 N, the result would be 1791 N.

Chris wrote:

What's 20 N among friends?

Among friends (or strangers): about 1%.