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

AI Thread Summary
The discussion focuses on analyzing the forces acting on point B of a crane supporting a 20.0-kN load. Participants emphasize the importance of drawing a free-body diagram to identify the three forces: tension in cables AB and BC, and the compressive force in the boom. By applying equilibrium equations for both horizontal and vertical components, users can derive the unknown forces. One participant successfully calculated the vertical and horizontal components, concluding that the boom's compressive force is approximately 36,251 N. The method of breaking down forces and using trigonometry to solve for unknowns is confirmed as correct.
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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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