How do I solve this physics exercise (Equilibrium)?

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The discussion revolves around solving a physics exercise related to equilibrium involving cable tensions. The participant calculates that the 500-pound vertical load is supported entirely by cable CD, with 375 pounds distributed horizontally to the other two cables. Each of these cables, being symmetric, carries 187.5 pounds horizontally. Another participant confirms the calculations, arriving at a slightly different value of 197.64 pounds, suggesting minor discrepancies may be due to rounding. The consensus indicates that the original approach is largely correct, with only minor adjustments needed for precision.
Tapias5000
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Homework Statement
Determine the force required on each cable to sustain the load of 500lb
Relevant Equations
Σfx=0, Σfy=0, Σfz=0
1632608403797.png

I tried to solve it and I got the following, is it correct?
aa.png
 
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Looks ok.

Insofar as 2 of the cables are in the horizontal plane, then the 500 pound vertical load must all be carried by the vertical comp of cable CD. And with that 3-4-5 triangle, 3 fourths of it or 375 pounds , must go horizontally to the other 2 cables, and their being symmetric, that’s half again or 187.5 borizontally to each, or cable tension root 10/3 times that. Saves a few lines. At least on this one.
 
PhanthomJay said:
Looks ok.

Insofar as 2 of the cables are in the horizontal plane, then the 500 pound vertical load must all be carried by the vertical comp of cable CD. And with that 3-4-5 triangle, 3 fourths of it or 375 pounds , must go horizontally to the other 2 cables, and their being symmetric, that’s half again or 187.5 borizontally to each, or cable tension root 10/3 times that. Saves a few lines. At least on this one.
Are my answers correct?
 
Tapias5000 said:
Are my answers correct?
Near enough. I get 197.64...
 
haruspex said:
Near enough. I get 197.64...
ooh... maybe then only decimals are involved in the procedure... well I'll settle for that
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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