Static equilibrium?

Question

1. A uniform board is suspended from the ceiling and horizontally by two wires, A and B. The board is 2m long and has a mass of 25kg. What is the tension in each of the two wires when a sign of 35kg is hung on the board a distance of 0.5m from one end?

see picture
http://i23.photobucket.com/albums/b354/Sunshinegirl93534/untitled-4.jpg

Answer 1

The board by itself adds
(12.5 kg)*(9.81 m/s^2) = 122.625 N
to each of the wires, because its mass is equally distributed.

The 35 kg sign is 1/4 of the way from what I assume is wire A (on the left) to wire B (on the right). Therefore, its weight is spread so 3/4 of the weight goes into wire A, and 1/4 into wire B.

Wire A gets
(35 kg)*(3/4)*(9.81 m/s^2) = 257.5125 N

Wire B gets
(35 kg)*(1/4)*(9.81 m/s^2) = 88.8375 N

So, in total, wire A has
257.5125 N + 122.625 N = 380.1 N of tension

and wire B has
88.8375 N + 122.625 N = 211.5 N of tension

Answer 2

Static equilibrium requires:
- Sum of all forces = 0
- Sum of all torques = 0

The forces are:
- T-l = tension on left-hand wire, at x = 0
- T-r = tension on right-hand wire, at x = 2
W-b = 25g, at x = 1 (where the center of mass is)
W-s = 35g, at x = 0.5

a) Because all forces = 0,
T-l + T-r = (25 + 35) g = 60 g

so T-r = 60g - T-l

b) 0 = 0 *T-l - (0.5)W-s - (1)W-b + 2*T-r
= 0 - 12.5 g - 35 g + 2T-r

So:
2 T-r = (12.5 + 35)g = 47.5g

T-r = 47.5g/2 = 23.75g = 237.5 N (because g = 10 m/s^2, about)

T-l = 60g - T-r = 600 - 237.5 = 362.5 N