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

waigaK 2020-12-01 Answered
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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Expert Answer

Elberte
Answered 2020-12-02 Author has 95 answers

image
Taking torques about the pivot point A
TLsin60=5000Lcos60+2600(0.35L)cos60
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

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