Question

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°) –  F g

Answer 1

B: Fnety = (FT)(sin 32°) – Fg

Step-by-step explanation:

Answer 2

Fnety = (FT)(sin 32°) – Fg Or the answer B, I checked it.