A nonuniform beam 4.42 long and weighing 1.06 makes an angle of 25 below the hor

Question

A nonuniform beam 4.42 long and weighing 1.06 makes an angle of 25 below the horizontal. It is held in position by a frictionless pivot at its upper-right end and by a cable a distance of 3.04 farther down the beam and perpendicular to it (see Figure 11.31 in the textbook). The center of gravity of the beam is a distance of 2.00 down the beam from the pivot. Lighting equipment exerts a downward force of 4.91 on the lower-left end of the beam.
I have solved for the Tension already but I am stuck on the components of force exerted by the beam on the pivot in the x and y directions.

Thanks for all help and I will rate accordingly!

Explanation / Answer

Applying the conditions of equilibrium to the beam

To calculate the tension by taking the torques about the pivot beam

a ) T(3.04 m ) = ( 1.06 ) ( 2.00 m ) cos 25 + ( 4.91 N)( 4.42 m ) cos 25

                   = 7.19N which is the vertical component

The vertical component of the force exerted on the beam by the pivot is net weight minus the upward component of T

          5.97 N - T cos 25

        = 0.54 N

horizontal component is T sin 25

                                       = 7.19 N sin 25

                                       = 3.03 N

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