Question

A pole is sticking out from the right side of a brick wall and perpendicular to it. There is a downward vector at the end of the pole labeled F Subscript g Baseline. The second vector points in a direction 32 degrees north of east and labeled f Subscript T Baseline.
Consider a sign suspended on a boom that is supported by a cable, as shown. What is the proper equation to use for finding the net force in the y direction?

Fnety = (FT)(sin 32°) + Fg
Fnety = (FT)(sin 32°) – Fg
Fnety = (FT)(cos 32°) + Fg
Fnety = (FT)(cos 32°) – Fg

Answer 1

Fnet,y = (FT)(sin 32°) − Fg

Reason: the vertical component of the tension is FT sin 32° upward while the weight Fg acts downward, so net vertical force = upward component − weight. (For equilibrium set this equal to 0, so FT sin32° = Fg.)