Confusion regarding Equilibrium Analysis Procedures

I have trouble understanding the bolded sentence in the following Equilibrium Analysis Procedures:

1. Draw a boundary around the system, so that you can clearly separate the system you are considering from its environment.
2. Draw a free-body diagram showing all external forces that act on the system and their points of application. External forces are those that act through the system boundary that you drew in step 1; these often include gravity, friction, and forces exerted by wires or beams that cross the boundary. Internal forces (those that objects within the system exert on each other) should not appear in the diagram. Sometimes the direction of a force may not be obvious in advance. If you imagine making a cut through the beam or wire where it crosses the boundary, the ends of this cut will pull apart if the force acts outward from the boundary. If you are in doubt, choose the direction arbitrarily, and if you have guessed wrong your solution will result in negative values for the components of that force.

What does it mean by "the ends of this cut will pull apart"? How does "pulling apart" look like? Why would "the ends of this cut will pull apart"?

What does it mean by "the ends of this cut will pull apart"? How does "pulling apart" look like? Why would "the ends of this cut will pull apart"?

Actually, the statement was "the ends of this cut will pull apart if the force acts outward from the boundary". For a beam or wire that would mean the beam or wire is in tension. I emphasized if because the ends of the cut will not pull apart if the force acts inward from the boundary. For a beam that would mean the beam is in compression. A wire would collapse. Perhaps the best way to visualize this is with an example from statics.

The first figure below shows a simply supported truss where it is asked what the force is in a particular member (member BD). Here I am using what is called the method of sections in which I am isolating the section (which can be thought of as the "system") from the rest of the truss (which becomes the surroundings) by cutting members with the dotted boundary. Before proceeding one determines the vertical reaction at support A by applying the requirements for equilibrium, i.e., the sum of the vertical forces equals zero ($$\sum\vec F_{V}=0$$) and the sum of the moments about H equals zero ($$\sum\vec M_{H}=0$$).

The second figure then isolates section ABC from the rest of the truss by the red boundary creating a free body diagram (FBD) for section ABC.

The external forces acting on section ABC are then $$\vec F_{CD}$$ , $$\vec F_{BD}$$ (the unknown being sought), $$\vec F_{BE}$$, reaction $$R_A$$, and the external vertical 5 KN load.

The internal forces in this section (system) are $$F_{AC}$$, $$F_{AB}$$ and $$F_{CB}$$ and are therefore not shown.

Now here is the key point. In the FBD diagram I have shown all three unknown member forces, including the member in question $$\vec F_{BD}$$ pointing outward, i.e., showing all members in tension. If the forces are actually pointing outward, then if the members are cut they will "pull apart". In this example, the value for $$F_{BD}$$ turns out to be negative, meaning it is in compression and would therefore not "pull apart" if cut. T

This approach is what the author mean by the following statement:

If you are in doubt, choose the direction arbitrarily, and if you have guessed wrong your solution will result in negative values for the components of that force.

Hope this helps.

• Thank you! Just to make sure, so does the word "pull apart" here just mean that when something is in tension, cutting it is as if the tension pulls the rod/wire apart? Jul 7 at 0:59
• @Cheryl Yes, when something pulls apart it means it fails in tension. Jul 7 at 1:24