Hi!
Thanks for the comment. I have attemped the following:
The net force on pivot B is:
0 = F1 + F2 + F3.
Rod AB.
∑Fx = 0: Ax – T cos 60 = 0
∑Fy = 0: Ay + sin 60 – 75 = 0
∑Mx = 0: T sin 60  75 = 0
∑Mx = 0:
75 = T sin 60
T = 75 / sin 60
T = 86,6 N
∑Fx = 0:
Ax = T cos 60
Ax = 86,6 cos 60
Ax = 43,3 N
∑Fy = 0:
Ay = 0
Rod CB.
∑Fx = 0: Cx  T cos 60 = 0
∑Fy = 0: Cy  sin 60 – 75 = 0
∑My = 0: T sin 60  75 = 0
∑My = 0:
75 = T sin 60
T = 75 / sin 60
T = 86,6 N
Cx = T cos 60
Cx = 86,6 cos 60
Cx = 43,3 N
Cy  sin 60 – 75 = 0
Cy = sin 60 + 75
Cy = 75,9 N.
Rod AB is pivoted at points A and B and since it is horizontal, it can only exert force in the horizontal direction. So the vertical force exerted by rod AB on pivot A is zero.
Rod CB is pivoted at points B and C and since it is at a 60 deg angle to the horizontal and the vertical, it exert a horizontal force of 75 N, and a vertical force on pivot C of 75 kN.
