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 θ = 29° with both wall and beam. ?

##a)T=420 . cos 29=367.34 N##
##b)##
##”horizontal component of net force :”98,716 . sin 2 theta=83,716 N##
##c)##
##”vertical component of net force :”98,716 .cos 2 theta=52,312N##

##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=222,566 N##
##”net force on point hinge :” 321,282-222,566=98,716 N##
##”horizontal component of net force :”98,716 . sin 2 theta=83,716 N##
##”vertical component of net force :”98,716 .cos 2 theta=52,312N##