What are the forces acting on point B in the crane diagram?

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SUMMARY

The discussion centers on analyzing the forces acting on point B in a crane diagram supporting a 20.0-kN load. Participants detail the process of drawing a free-body diagram and applying equilibrium equations, specifically Sigma(Fx) = 0 and Sigma(Fy) = 0, to determine the unknown tensions in cables AB and BC, as well as the compressive force in the boom. The final calculations yield a boom compressive force of 36,251 N, confirming the method's correctness through systematic component breakdown and equilibrium analysis.

PREREQUISITES
  • Understanding of free-body diagrams in static equilibrium
  • Knowledge of tension and compression forces in structural analysis
  • Familiarity with trigonometric functions for force component resolution
  • Proficiency in applying equilibrium equations (Sigma(Fx) and Sigma(Fy))
NEXT STEPS
  • Study the principles of static equilibrium in mechanical systems
  • Learn about tension and compression in cables and structural members
  • Explore advanced trigonometric applications in force analysis
  • Investigate software tools for structural analysis, such as AutoCAD or SAP2000
USEFUL FOR

Engineering students, structural analysts, and professionals involved in crane operations or load management will benefit from this discussion, particularly those focused on understanding force dynamics in mechanical systems.

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The Crane in P19 (Attached) supports a 20.0-kN load.. to simplify things take the bales to be represented by two separate bundles: Cable-AB and cable-BC. (a) Draw a free-body diagram of point-B putting in the forces exerted by the boom, cable-AB, and cable -BC. (b) determin the horizontal and the vertical components of each force. (c) Using Sigma(Fx) = 0 and Sigma(Fy) = 0, determine the compressive force on the boom.


Since Fy, and Fx = 0, The load must be in Equilibrium.


I'm kind of stuck on this question, I had skipped over it before but now I still want to figure it out.

Any help would be much appreciated!
 

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Take the suggestions in parts a, b, and c. There are 3 forces acting on B, one is known (tension in BC, pulling away from B), the others are not (tension in AB pulling away from B), and compression in the boom (pushing towards point B). Break the forces into components and apply the equilibrium equations in the x and y directions. You'll get 2 equations with 2 unknowns, solve for T_AB and the boom compressive force. Watch your geometry and trig and plus and minus signs!
 
PhanthomJay said:
Take the suggestions in parts a, b, and c. There are 3 forces acting on B, one is known (tension in BC, pulling away from B), the others are not (tension in AB pulling away from B), and compression in the boom (pushing towards point B). Break the forces into components and apply the equilibrium equations in the x and y directions. You'll get 2 equations with 2 unknowns, solve for T_AB and the boom compressive force. Watch your geometry and trig and plus and minus signs!

Sorry for the late response I was caught up with other work, anyways,
so I drew the free-body diagram as suggested in part a,
For part b,
F_BCy = 20000N down
F_BCx = 0
F_BAy = 20000sin40 = 12855N down
F_BAx = 20000cos40 = 15320N left
Since in equilibrium, the boom must counteract these forces yielding,
F_boomy = 20000 + 12855 = 32855N up
F_boomx = 0 + 15320 = 15320N right

therefore the compression of the boom would be a^2 + b^2 = c^2,
sqrt(32855^2 + 15320^2) = 36251N

is this method correct ?
 

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