Consider a long (length = 15 m

Consider a long (length = 15 m) uniform wooden beam (mass = 60kg) attached horizontally to a wall that can only support avertical load ( The horizontal component of the force of the wallon the beam is identically zero). There is a chandelier (mass = 40kg) hanging at a distance = 4.21 meter from the end of the beamthat is attached to the wall. There is a vertical cable hangingdown from the ceiling that is attached to the other end of the beamthat is in mid air. The entire structure is in static equilibrium.Determine the tension in the vertical cable.

draw your FBD diagram. 60 kg acts on mid span of thebeam.Consider point A where the cable is connectedand point B as the fixed end.Take moments.Counter clockwise positive.Moment @ B = -60(7.5)-40(10.79)+15Ra[?Moment @ A=0] 40(4.21)+60(7.5)[email protected]=15Rb[?Forces Vertical=0] Ra +Rb=60+403 unknowns, 3 eqxn, solving simultaneouslyReaction @ A=41.23 Kg (404.5663KN) Tension on thecableReaction @ B=58.77 Kg (576.5337 KN)Moment @ B=263.15 Kg-m (2581.5015 N-m)

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