Equilibrium of a Rigid Body, a boom of a crane

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SUMMARY

The discussion focuses on the equilibrium of a crane boom weighing 2600 N, attached to a frictionless pivot. The center of gravity is located 35.0% of the boom's length from the pivot. The tension in the guy wire is calculated to be 3412 N, and the vertical force component at the lower end of the boom is 7600 N. The horizontal force component remains unresolved, prompting the need for a free body diagram and static equilibrium equations to solve for unknown forces.

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  • Understanding of static equilibrium principles
  • Knowledge of free body diagrams
  • Familiarity with force components in physics
  • Basic grasp of crane mechanics and tension calculations
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  • Study static equilibrium equations in detail
  • Learn how to construct and analyze free body diagrams
  • Research the calculation of horizontal force components in rigid body mechanics
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Students studying physics, particularly those focusing on mechanics and static equilibrium, as well as engineers involved in crane design and load analysis.

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Homework Statement



The boom in the figure below weighs 2600 N and is attached to a frictionless pivot at its lower end. It is not uniform; the distance of its center of gravity from the pivot is 35.0 % of its length.

Homework Equations


Find the tension in the guy wire.
Find the vertical component of the force exerted on the boom at its lower end.
Find the horizontal component of the force exerted on the boom at its lower end.


The Attempt at a Solution



I have already found the correct answers for the tension in the guy wire=3412N, the vertical component of the force on the boom=7600N, but I'm so confused and can't read my notes and what I did to find the Horizontal component. I wrote down "horizontal=guywire plus jasdlkfjas;dlj" so I just don't know what the jaskdfljs is! please advise!
 

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Draw a free body diagram of the boom. Label and insert forces, known and unknown. Write equations of static equilibrium. Solve for unknown forces.
 

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