A weight of mass 47 kg is suspended from a point near the right-hand upper end o

Question

A weight of mass 47 kg is suspended from
a point near the right-hand upper end of a
uniform boom of mass 23 kg . This boom is
supported by a cable (with a tension of 513 N)
that runs from the boom to a point on the wall
(the left-hand vertical coordinate at a height
of 10 m) and by a pivot (at the origin of the
coordinate axes) on the same wall.

A weight of mass 47 kg is suspended from a point near the right-hand upper end of a uniform boom of mass 23 kg . This boom is supported by a cable (with a tension of 513 N) that runs from the boom to a point on the wall (the left-hand vertical coordinate at a height of 10 m) and by a pivot (at the origin of the coordinate axes) on the same wall. Calculate the weight of the boom. The acceleration of gravity is 9.8 m/s2. Answer in units of N Your answer must be within +- 3.0%

Explanation / Answer

at the upper end

tan theta = (10-6.5)/6.5

theta = 28.3

net torque = 0

T*cos28.3*y + T*sin28.3*x - (9*9.8*4.5) - (mw*9.8*6.5) = 0

(175*cos28.3*6.5) +(175*sin28.3*6.5) - (9*9.8*4.5) - (mw*9.8*6.5) =   0

1540.81 - 396.9 - 63.7*mw = 0

mw = 17.95 kg     (18 kg nearly)

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