1.52cos(31)+Fbsin(Θb) = (0.305)(9.8)
Fbsin(Θb) = 1.69
Now for horizontal direction:
Fbcos(Θb) = FLsin(ΘL)
cos(Θb) = FLsin(ΘL)/Fb
cos(Θb) = (1.52 x sin31)/1.69
cosΘb=0.464
Θb = 62.35
I thought to find Fb I would just plug the Θb value into one of the trig functions, but apparently both of my...