Hi there!
We know that since the object is at equilibrium, the net force must be equal to 0.
In the x-direction:
∑F = 0
Thus:
0 = 30 - OAsin(30) ⇒ Horizontal component of OA
Solve:
OAsin(30) = 30
OA (1/2) = 30
OA = 60 N
To find the weight, we must look at the y-direction.
The only forces in the y-direction are the tension's y-component and the weight, so:
0 = OAcos(30) - W
W = OAcos(30)
Plug in values:
W = 60 · √3/2
W = 30√3 N
The correct answer is B) 30√3, 60
