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##