The boom weighs 2600N. The boom is attached with a frictionless pivot at its lower end. It is not uniform; the distance of its center of gravity from the pivot is 35% of its length. a) Find the tension in the guy wire and the horizontal and vertical components of the force exerted on the boom at its lower end. b) Does the line of action of the force exerted on the boom a tits lower end lie along the boom?

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Elberte · Accepted Answer

Taking torques about the pivot point A TLsin⁡60=5000Lcos⁡60+2600(0.35L)cos⁡60 or T=3.14 kN The vertical force exerted on the boon by the pivot Fv=5000 N+2600 N=7.06 N the Horizontal force Fb=T=3.14 kN No, tan⁡(FvFh) is not equal to zero

The boom weighs 2600N. The boom is attached with a frictionless pivot at its lower end. It is not uniform; the distance of its center of gravity from

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