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A uniform beam of length 1.0 m and mass 10 kg is attached to a wall...

A uniform beam of length 1.0 m and mass 10 kg is attached to a wall by a cable. The beam is free to pivot at the point where it attaches to the wall. What is the tension in the cable? (THE ANGLE IS 30 degrees)

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Answer #1

Take the moments about the point of wall attachment. The Center-of-Gravity (CG) of the beam is at 0.5m, and the weight of the beam is m*g = 10*9.8 = 98 N. Only the vertical component of the cable tension contributes to the moment calculation, since the horizontal component has a zero radius or "moment arm" relative to the wall point of attachment. Let "T" be the tension in the cable, and then the vertical component of the cable tension is:

F_v = T*Sin(30)

Equating the two moments:

0.5*m*g = 1.0*T*Sin(30)

.5*98/Sin(30) = T = 98N

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