Consider a long (length = 15 m) uniform wooden beam (mass = 60 kg) attached horizontally...

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

Expert Solution

draw your FBD diagram. 60 kg acts on mid span of the beam.

Consider point A where the cable is connected
and 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)+M@B=15Rb
[?Forces Vertical=0] Ra +Rb=60+40

3 unknowns, 3 eqxn, solving simultaneously

Reaction @ A=41.23 Kg (404.5663KN) Tension on the cable
Reaction @ B=58.77 Kg (576.5337 KN)
Moment @ B=263.15 Kg-m (2581.5015 N-m)

genius_generous answered 2 years ago

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