In the figure one end of a uniform beam of weight 420 N is hinged to a wall; the other end is supported by a wire that makes angles = 29 with both wall and beam. ?

##a)T=420 . cos 29=367.34 N##
##b)##
##horizontal component of net force :98716 . sin 2 theta=83716 N##
##c)##
##vertical component of net force :98716 .cos 2 theta=52312N##

##P=420 .sin 2 theta##
##R=420 .cos 2 theta##
##L=T .cos theta##
##K=T . sin theta##
##M=T .sin theta##
##N=T .cos theta##
##K .2.l=P. l( torque for point A)##
## R and L have no torque for point A##
##K.2=P##
##2.T.sin theta=420 . sin 2.theta##
##T=210 .(sin 2 theta)/sin theta##
##sin 2. theta=2. sin theta .cos theta##
##T=210. (2.sin theta .cos theta)/sin theta##
##T=420 .cos theta##
##a)T=420 . cos 29=367.34 N##
##L=367.34 . cos 29=321.282 N##
##R=420 . cos 58=222566 N##
##net force on point hinge : 321282-222566=98716 N##
##horizontal component of net force :98716 . sin 2 theta=83716 N##
##vertical component of net force :98716 .cos 2 theta=52312N##