Weight Distribution: Solving for Tension in a Hawser System

[lendav_rott]

Homework Statement

In the illustration there is a hawser attached to the ceiling at points A and B. Attached to point C is a mass of 400kg.

Find the tension in the hawsers AC and BC.

Homework Equations

The Attempt at a Solution

I have no idea how to go about this one. I can't find any materials in English about this that are understandable and in my own language things are vague to begin with :/
What is tension? This much I have figured out in my head that it's a force and its unit is 1N or that would make sense.
But how do I find the tension.

I know that the weight of the mass is 3920N and it should be distributed between those 2 holding hawsers in some proportion. Does it mean that the sum of the tensions in the hawsers will have to be mg?

 

[mishek]

Hello,

I hope my attached photo helps.

 

[lendav_rott]

So what I understand is that point B is pulling the weight attached to point C by F1 and point A is pulling it by F2. Considering the mass is still then according to Newton the sums of these forces will have to be 0.
so F1 + F2 + mg = 0 (sum of vectors)

If that's true I understand the idea, but how do I put it on paper with this assignment.

Can I use this?
the sine law or theorem or how it s called in English:
A/sina = B/sinb so in this assignment F2/sin30 = F1/sin50 and using the a/b = c/d => ad = cb rule I get that
F1/F2 = sin50/sin30
sin50/sin30 = 1.532 so F1 = 1.532F2

But something doesn't feel right, isn t F2 supposed to be > F1, I mean, it's closer to being in a vertical position compared to F1 so isn t it supposed to have more tension not less :s

 

[mishek]

hell,

i would use sum on x and sum on y.

In this example F1x=F2x; F1*cos30=F2*cos50...

 

[lendav_rott]

Oh okay, so F1 = F2*cos50/cos30 - so my assumption is correct, the closer the hawser gets to a vertical position, the bigger the tension gets?
Thank you, mishek, I understand how you approached this, cheers.