Question

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. ?

How to find (a) the tension in the wire and the (b) horizontal and (c) vertical components of the force of the hinge on the beam.

Answer 1

`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`

Explanation:

`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`