Tension of 2 Ropes Find the Angle

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Tension of 2 Ropes are known angle and mass unknown help please

Homework Statement


A block mass M hangs in equilibrium. A horizontal rope attached to the block has a tension of 30N. Another rope attached to the cieling with a tension of 90N, and makes an angle with the ceiling. What is the angle


Homework Equations


I was thinking the inverse tangent
tan = y/x being y= 90 and x = 30

The Attempt at a Solution


inverse tan = 30/90 = 0.3333 18 degrees

i'm probably doing something wrong cause i think gravity here is important i just don't see how to put all this together
Any suggestions or corrections would be helpful
 
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Draw a picture of the situation.

The rope making an angle with the ceiling MUST have a certain tension in the horizontal direction. What is that tension and why must it have that magnitude?

Think on that.
 
http://img254.imageshack.us/img254/2228/tensionandanglesjy0.png
The rope with tension 90N makes an angle with the ceiling

I kind of thought of drawing in this triangle to help me out
http://img508.imageshack.us/img508/8627/trianglerc6.png

R = sqrrt{(30)^2 + (90)^2}
sorry made a mistake in the triangle C is 95 N confused with another problem
using this to help me find the angles
what do u guys think?
 
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So what is the horizontal component of the angled rope? Remember, the object is in equilibrium. What does that mean about all of the forces in the problem?


Mattowander said:
Draw a picture of the situation.

The rope making an angle with the ceiling MUST have a certain tension in the horizontal direction. What is that tension and why must it have that magnitude?

Think on that.
 
So for M to be in equilibrium it must be acted on by no forces? the net force is zero i think so if it is zero then what?
 
Yes, the net force must be zero. That means the Net force in the x direction must be zero. So what is the horizontal component of the angled rope?
 
I hit the wall with that one, sorry I'm
still trying to figure out
 
for M in equilibrium each component of the net force must be zero so
[tex]\Sigma[/tex]Fx = 0 and [tex]\Sigma[/tex]Fy = 0
 
That's correct. So if the sum of the x forces is zero, then what would be T Sin Theta?
 
Then it would be the inverse of Sine theta = 30/90 = 0.33 which is 19 degrees