What is the Tension in the 37-Degree Cable Holding a 4000kg Wrecking Ball?

[xmflea]

a large wrecking ball is held in place by two steel cables
http://img378.imageshack.us/img378/3994/yf0543ei1.jpg
if the mass m, of the wrecking ball is 4000kg, what is the tension Tb in the cable that makes an angle 37 degrees with the vertical?

F=T + W = ma
i drew my FBD
i tried Tb=(4000*9.8)/cos37, but it got me nowhere
i also tried Tb= 4000cos37, which also didnt work.

maybe i am reading the qustion wrong, need guidance.

 

[hage567]

Why is the angle 40 degrees in the picture?

 

[xmflea]

yeah i don't know why either, the picture shows 40 degrees but the problem asks 37 degrees. and I am confused. i know how to solve problems like these when the cables are attatched to flat vertical and horizontal walls, but i am thrown off when the cables are attached to an angled crane

 

[hage567]

Try it with 40 degrees instead of 37? The first equation you wrote for Tb should work. Is there more to the question than what you've written?

 

[xmflea]

no , that is all they give, and then the second part of the question asks what is the tension Ta of the horizontal cable when the wrecking ball is 4000kg, but i figured, if i could solve the first part, i could solve the second part.
i also tried 40 degrees instead of 37, but it stil comes up with a wrong answer.

 

[hage567]

What is the number that you get out of your calculation?

 

[xmflea]

plugging in for the first equation i wrote for Tb, i get 51172kg, which the computer says is a wrong answer. i also tried using sin in the equations as well, which still ended up wrong.