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Lifting boom

A crate with a mass of 189.5 kg is suspended from the end of a uniform boom with a mass of 84.5 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall. Calculate the tension in the cable.

Added Sun, 05 Jul '15

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m? = mass of the crate = 189.5 kg

m? = mass of the boom = 84.5 kg

g = acceleration by gravity = 9.81 m/s^2

L = length of the boom

? = angle of the boom to horizontal

? = angle of the cable to horizontal

From the image you can see that

tan(?) = 3/9

so

? = arctan(3/9) = 18.434 degrees

From the image you can see that

tan(?) = 4/9

so

? = arctan(4/9) =23.96 degrees

As the system is at equilibrium, you know that the clockwise torque (caused by the weight of the boom and crate) must be equal in magnitude to the counterclockwise torque (caused by the tension in the cable)

So:

m?*g*cos(?)*L/2 + m?*g*cos(?)*L = T*sin(? + ?)*L

(m?/2 + m?)*g*cos(?) = T*sin(? + ?)

T = (m?/2 + m?)*g*cos(?) / sin(? + ?)

T = [(189.5 kg)/2 + (84.5 kg)]*(9.81 m/s^2)*cos(18.434) / sin((18.434) + (23.962))

T = 2474.17 N < - - - - - - - answer

Added Sun, 05 Jul '15

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