Support cable question?

=> will use "L" instead of lower case "L", as given in diagram, for boom length. T = Tension in support cable.

Torques {about the boom's pivot}:

ΣCW Torques = 2400(sin 21°)L + 1425(sin 21°)L/2 = 860L + 511(L/2) = 860L = 255L = 1115L

CCW Torque = T(3L/4) = 0.75TL

Equilibrium of Torques:

1115L = 0.75TL

T = 1115/0.75 = 1487 N ANS (a)

ΣFx = T(cos 21°) = 1487(cos 21°) = 1388 = 1.39 kN ANS (b)

ΣFy = 1425 + 2400 - T(sin 21°) = 3825 - 1487(cos 21°) = 2437 = 2.44 kN ANS (b)

To determine the tension in the cord, let do a torque problem, with the pivot point at hinge.

Counter clockwise torque = T * ¾ * L

For the boom, clockwise torque = 1425 * L/2 * cos 69

For the object, clockwise torque = 2400 * L * cos 69

Total clockwise torque = 1425 * L/2 * cos 69 + 2400 * L * cos 69

T * ¾ * L = 1425 * L/2 * cos 69 + 2400 * L * cos 69

Divide both sides by ¾ * L

T = 1900 * cos 69 + 3200 * cos 69 = 5100 * cos 69

This is approximately 1827.68 N. The tension has a vertical and horizontal component. To determine angle θ, subtract 69˚ from 90˚.θ = 21˚

Vertical = 5100 * cos 69 * sin 21

This is approximately 655 N. This is an upward force.

Horizontal = 5100 * cos 69 * cos 21

This is approximately 1706.3 N.

To determine the vertical component of reaction force, subtract this number from the total weight.

Total weight = 1425 + 2400 = 3825 N

Vertical component = 3825 – 5100 * cos 69 * sin 21 N

This is approximately 317 N.

There two horizontal forces. Since nothing is moving, these two forces are equal.

Horizontal component = 5100 * cos 69 * cos 21 N

You can round to kN.