This is my first question here so I hope I get warmly welcomed by answers that would resolve this perplexion :-)

This is a statics problem from Hibbeler's Mechanics for Engineers: Statics (13th Edition)... I have access to its solution from the solution manual, but there is something I don't understand here:


  1. How is AC here a two-force member (as the ISM says) when it has that 2-kN load acting on it at pin A?

  2. When AC is a two-force member, Ax = Cx, and Ay = Cy, then the FBD of member AC would have only the reaction forces at the two pins A and C, how is it that the 2-kN force is not considered here?

I know both questions seem quite similar, but this is for the purpose of clarification as to where the confusion lies.

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    $\begingroup$ I wouldn't get hung up on someone else's artificial classification of the problem. I hadn't hear the term "two force member" before, but what does that matter anyway? I don't know what the problem is asking for, but it is easily possible to solve for the forces on the pins, force on member A-C, force on the bottom support, etc. If you're stuck solving this problem, ask about that specifically. $\endgroup$ – Olin Lathrop Dec 13 '15 at 13:22
  • $\begingroup$ Well, a two-force member is any structure that has only two forces acting on it. In equilibrium, these forces must be equal, opposite, and collinear... (You could've looked that up)...Anyway, my question is: how would you draw the FBD of member AC? And why would you draw it that way? My question is not on the whole problem but just to clear some misconception that I might have here :) $\endgroup$ – user101197 Dec 13 '15 at 13:35
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    $\begingroup$ It should be obvious that A-C only has two forces on it (other than gravity, since weights weren't given I'm assuming you're supposed to ignore this). It is in compression, so the pin at each end is pushing on the member in the direction of the other pin. Since A-C isn't accellerating, those two forces are equal and opposite. What's the problem? $\endgroup$ – Olin Lathrop Dec 13 '15 at 13:38
  • $\begingroup$ Can I consider that 2-kN load to be acting on pin A, and hence have a component acting on AC? This way AC wouldn't have only two forces acting on it...If no, then why not? $\endgroup$ – user101197 Dec 13 '15 at 13:41
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    $\begingroup$ If that's what you want to know, then again, ask about that. Your question as it currently stands is all hung up about whether A-C is a "two force" member or not. Write the question properly about what the forces are at pin A or on member A-C, if that's what you really want to know. You also seem to be misreading the diagram, or at least jumping to a incorrect conclusion. $\endgroup$ – Olin Lathrop Dec 13 '15 at 14:49

Here you have Four equal forces on link AB. There is no any other external force on any of the other links. All other members will bear the forces produced as a result of these forces. I don't know what makes you think that the 2kN force on AB is acting on AC. It is clearly visible that the 2kN force is on AB not on pin AC.


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