Find the tension in each rope "torque"

[poloboy]

1. A uniform plank of length 2m and mass 30kg is supported by three ropes. Find the tension in each rope when 700N person is 0.5m from the left end.

2. diagram
↑ /
| /
| /40 degrees ↓ (rotating clockwise)
←-----person-----------------/---


3. The attempt at a solution

So i began with isolating the X and Y variables
ie: X on left, fcos40
y going up, fsin40
how would i go about breaking down the mass of the person + plank to figure out the tension in each rope?

[Astronuc]

For a statics problem the sum of the moments about any pivot is zero. It's a little difficult to tell from the illustration where the ropes are, but I'm guessing all three are attached at the left end?

Treat the person as a point load from at 0.5 m from the end, or 1.5 m from the pivot. Treat the weight of the plank at its center of mass.

[poloboy]

sorry the diagram didn't show up better

theres a rope pointing left, one pointing up (on the left side) and one rope pointing tot he right at 40 degrees North east

[Astronuc]

And the pivot is on the right?

If there is no tension in the plank then then the x- or horizontal component in the rope at 40° will balance the tension in the horizontal rope.

Try to write the equation for the sum of the moments about the pivot.

[poloboy]

awsome, figured it out.
i set x coord t1cosO - T3 and y coord as T1sinO + T2 - W(person+plank)

set my t1 = 1/4(Wperson +2Mplank)/sinO then i just plugged in T1 to my other forumlas.

thanks!