physics question(statics)

There are 3 forces involved: The weight, the tension of the string along OA and the tension of string along OB.
Let the tension OA be T1 and the tension OB be T2.

Resolve vertically
T1 sin 40 deg + T2 sin 20 deg = weight
T1 sin 40 deg + T2 sin 20 deg = 50 kg x 9.8 m /s^2
T1 sin 40 deg + T2 sin 20 deg = 50 kg x 9.8 m /s^2
T1 sin 40 deg + T2 sin 20 deg = 50 kg x 9.8 m /s^2
0.6428 T1 + 0.3420T2 = 490 N ------------------ (1)

Resolve horizontally
T1 cos 40 deg = T2 cos 20 deg
0.766 T1 = 0.9397 T2 ------------------------------- (2)

Two simultaneous equations (1) and (2) with two unknowns T1 and T2
From equation (2)
T1 = 0.9397 T2/0.766
T1 = 1.227 T2 ---------------------- (3)

Substitute (3) into (1)
0.6428 (1.227 T2) + 0.3420T2 = 490 N
0.7887T2 + 0.3420T2 = 490 N
1.1289T2 = 490 N
T2 = 490 N/1.1289
T2 = 434 N

T1 = 1.227 T2
T1 = 1.227 (434 N)
T1 = 533 N

The two tensions are 434 N and 533 N


2012-10-26 11:03:21 補充：
The unit of tension is Newton (N)