Calculating Moments Using the Method of Joints: A Tutorial

In summary, the author suggests that the moment about B, where Cy and Cx don't act on the boom, can cause the member to turn about B. However, because point C isn't part of the boom, the forces at C won't be included in the equilibrium equation for the boom.
  • #36
billy_joule said:
I'm not sure if there's a hard and fast rule.
Seeing the quickest and easiest approach will come with practice.
This particular problem could be solved much faster with the method of joints.
I am not concerned about the speed of solving the question...
So, if we can't solve the question by moment about B = Cy(0.8) +Ay(0.8)
+ Cx(0.6) = 0
We have to choose to consider moment about a point in 1 force member only?
 
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  • #37
billy_joule said:
I'm not sure if there's a hard and fast rule.
Seeing the quickest and easiest approach will come with practice.
This particular problem could be solved much faster with the method of joints.
Can you show how do you solve the question by method of joints?
 
  • #38
chetzread said:
Can you show how do you solve the question by method of joints?
Best to wait until after you've learned the method in class.
 

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