As shown, a uniform beam of length l = 5.90 and 48.2 lb is attached to a wall with a pin at point B. A cable attached at point A supports the beam. The beam supports a distributed weight = 21.0 lb/ft. If the support cable can sustain a maximum tension of 300 lb , what is the maximum value for W1 ? Under this maximum weight, what is FBy , the vertical component of the support's reaction force at point B?
[itex]\Sigma[/itex] F = 0
[itex]\Sigma[/itex] M = 0
The Attempt at a Solution
I began by trying to draw a free-body diagram with everything I know put in place. I then came up with 3 equations: The sum of x-axis forces, the sum of y-axis forces, and the sum of moments at point B.
(1) [itex]\Sigma[/itex]Fx: 300 cos(60o)+ Bx= 0
(2) [itex]\Sigma[/itex]Fy: 300 sin(60o)- 48.2 - 123.9 - W1 - By= 0
(3)( this is my problem equation)[itex]\Sigma[/itex]M: -300cos(60o )(5.9ft) - (127.1)(2.95) - W1(0) - By(0) = 0
The problem I can't quite figure out is that if the unknowns FBy and W1 are applied at B, they create no moments, and that equation becomes useless for solving my system. So, I'm left with an indeterminate system of 2 equations and 3 unknowns. I've attached the Figure referenced in the problem statement and a MS Paint version of my FBD. I just need a nudge in the right direction I think.
I also replaced the distributed load with an equivalent Force of 123.9 lb directed at the center of the beam.
Addendum: I also tried with By oriented upward, AND I tried maybe summing the moments at A instead. I still get an indeterminate answer.
Thank you in advance for your help