Finding the distance of a sagging mass on rope using tension

[chenying]

Homework Statement

An object of mass m = 3.5 kg hangs from the middle of a (massless) rope of length L =17 m, as the drawing illustrates. It is being pulled with tension T =55 N. Calculate the distance D by which the rope sags in the middle

Homework Equations

F$$_{net}$$ = ma

D = length of sagging rope

L/2 = 8.5 meters

The Attempt at a Solution

So, because the rope is being pulled with a tension of 55 N to keep the object at a certain distance, the tension throughout the string with the mass is 55 N.

I made theta my angle with the horizontal, which is the value I'm trying to find.

The mass is balanced out evenly in the middle, so the two y components of the rope (2Tsin$$\Theta$$) would have to equal the weight of the mass, mg.

I have all the variables except for theta, so solving for theta, I get a value of 18.1688 degrees.

I have the length of the rope, 8.5 meters, and the angle, 18.1688 degrees, so I can use tan $$\Theta$$ = D/8.5

solving for D, I got 2.78954 meters, which was not right.

So, anyone has an idea what I did wrong or how my approach was incorrect?

 

[PhanthomJay]

Without a picture, I think you are using tan theta when you should be using sin theta.

 

[chenying]

Without a picture, I think you are using tan theta when you should be using sin theta.

Sorry for not having a picture, but I set my theta to be the angle at the top right. Imagine that there was an upside down obtuse triangle, with the mass in the middle, and the angle is the one that is horizontal to the triangle.

Sorry again that I don't have a picture, but can anyone help? I don't know what it is I did wrong.

 

[PhanthomJay]

Try $$sin \theta = D/8.5$$ and see if that works. I think the 8.5 m length of rope is measured along the diagonal. D is the vertical 'height' (sag) of your upside down obtuse triangle.