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I was discussing the first part of this question with a friend, and he told me that while taking moments about $A$ and $B$, we can ignore the reaction force at C. Indeed this is correct, because ignoring the reaction force at C gives the right answer. However, what I don't understand is that why can we do this?

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  • $\begingroup$ I saw classical mechanics as I tag. Isn't this included in that? $\endgroup$ – user140161 Feb 9 '15 at 19:46
  • $\begingroup$ Also, I included that this applies while taking moments about A and B. How does the line of action of the reaction at C pass through the A or B? $\endgroup$ – user140161 Feb 9 '15 at 19:48
  • $\begingroup$ Then why would we take the reaction for at B while calculating moments about A, or the reaction force at A, while taking moments about B? $\endgroup$ – user140161 Feb 9 '15 at 20:02
  • $\begingroup$ You should know when you are talking about a system and when you are talking about an isolated object. $\endgroup$ – Troy Woo Feb 9 '15 at 20:11
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The reaction at C is within the system. It's internal. Therefore, the free body diagram of the whole system is the whole body itself and the forces acting on the body from the outside. The reactions at A and B are external to the whole body. The force of gravity is also external to the whole body.

Now, if a force were applied externally to point C, then you would have to consider that in your computation.

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