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)
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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