4 Replies Latest reply on Aug 5, 2009 8:59 PM by Derek Bishop

    Free Body vs Reaction Force

    Mahir Abrahim

      I think know what free body and reaction forces are in theory, but I'd like a little corroboration.  Can someone give me a breakdown of how they apply to the List Result Force tool?  I want to be able to make sense of the numbers without making any false assumptions about what I'm looking at.  For example, what would be the best way to extract the force imposed on a particular section of a model, free body or reaction?

        • Re: Free Body vs Reaction Force
          Mauricio Martinez-Saez

          9080 wrote:

           

          I think know what free body and reaction forces are in theory, but I'd like a little corroboration.  Can someone give me a breakdown of how they apply to the List Result Force tool?  I want to be able to make sense of the numbers without making any false assumptions about what I'm looking at.  For example, what would be the best way to extract the force imposed on a particular section of a model, free body or reaction?

           

          Mahir,

           

          I will try to explain to you the difference between free body and reaction forces...

           

          Free body is the generic term used to describe an object which can be considered as moving as a single unit. The object doesn't have to be "free" in the usual sense of the word. The crucial concept is that you can think of it as a single unit to the extent that it either does or does not move, as the case may be. Often the term is associated with a free body force diagram.  Therefore, Free Body Forces are those that compose the system of forces acting over the body and keeping it steady (equilibrium) or in movement.

           

          Regarding "reaction forces", in classical mechanics, Newton's third law states that forces occur in pairs, one called the Action and the other the Reaction. Both forces are equal in magnitude and opposite in direction. The distinction between action and reaction is purely arbitrary: any one of the two forces can be considered an action, in which case the other (corresponding) force automatically becomes its associated reaction.

           

          Newton's third law of motion is frequently stated in a simplistic but incomplete or incorrect manner through statements such as "Action and reaction are equal and opposite" or "To every action there is an equal and opposite reaction"
            
          These statements fail to make it clear that the action and reaction apply to different bodies.  Also, it is not because two forces happen to be equal in magnitude and opposite in direction that they automatically form an action-reaction pair in the sense of Newton's Third Law.

           

          Action and reaction are often confused with the issue of equilibrium. For example, consider the following statement: A book standing on a table is at rest because its weight, a force pulling it downwards, is balanced by the equal and opposite reaction of the table, a force pushing it upwards.  This statement is misleading since it suggests that the force exerted by the table on the book is the reaction associated with the book's weight. This is not the case, since the two forces are different in nature and are both applied to the book; one cannot be the reaction to the other, since they must apply to different bodies. In fact the force exerted by the table can be seen as the reaction to the contact force exerted by the book on the table, which in turn is equal to the book's weight.

           

          As mentioned, in classical mechanics, Newton's third law states that forces occur in pairs, one called the Action and the other the Reaction. Both forces are equal in magnitude and opposite in direction. The distinction between action and reaction is purely arbitrary: any one of the two forces can be considered an action, in which case the other (corresponding) force automatically becomes its associated reaction.

           

          The ''Reaction'' is one of the least understood of the basic physical concepts, perhaps because it is often incorrectly described, or because Newton's laws of motion may appear counter-intuitive. Here is a modern statement (in words only) of the Third Law of motion:

           

          "If a force acts upon a body, then an equal and opposite force must act upon ANOTHER BODY".

           

          An example is the recoil of a firearm, in which the force propelling the bullet is exerted equally back onto the gun and cause the recoil of the gun. Since the two objects do not have the same mass, the resulting acceleration on each object will be different.

           

          It is essential to understand that the "reaction" applies to another body that the one on which the "action" applies. For instance, in the context of gravitation, when object "A" attracts object "B" (action), then object "B" simultaneously attracts object "A" (with the same intensity and an opposite direction).

           

          Another important point is that the physical nature of the reaction force is identical to that of the action itself: if the action is due to gravity, the reaction is also due to gravity. Hence, claim that an action results in a reaction of a different type (gravitational, electromagnetic, friction, spring, etc.) is incorrect.

           

          Consider a mass hanging at the end of a (non-stretchable) steel cable attached to the ceiling of your room. The mass is pulled towards the Earth (action) by its weight. The corresponding reaction is the gravitational force that mass exerts on the planet. This has nothing to do with the steel cable, in fact, the reaction exists even in the absence of the cable, while the mass will be falling or after it is resting at the floor. On the other hand, if the tension in the cable is pulling the mass upwards and preventing it from falling, then the mass is simultaneously pulling on the cable, with equal intensity and opposite direction. If this simple system is observed to be at rest (in particular not accelerated) with respect to the ceiling, Newton's first law implies that no net force is applied to the mass. Since we have just seen that two distinct forces do apply to the mass (the gravitational pull from the Earth and the tension from the cable), we conclude that these two forces are themselves equal and opposite, i.e., that they compensate each other. However, these latter two forces are not the action and the reaction of each other, these are forces acting on a FREE BODY.

           

          Hope this will be of help to you.

            • Re: Free Body vs Reaction Force
              Mahir Abrahim
              Thanks, Mauricio.  I believe I better understand in a general sense the difference between reaction & free body forces.  So my guess is when you select a surface and have SW Simulation give you a report of the forces on that surface, it's treating every element along that surface as a separate body.  The force vector given is then just the sum of the individual element forces.  If reaction force is chosen, it's the reaction on those elements due to forces from either neighboring elements or external loads/constraints.  If free body is selected, then it's the sum of the external forces on those particular elements.  Still a bit fuzzy, but I think I get it.  So if I want to separate how much load is acting on a particular section, I should stick to free body, not reaction.  Thanks, again.
                • Re: Free Body vs Reaction Force
                  Mauricio Martinez-Saez

                  9080 wrote:

                   

                  Thanks, Mauricio.  I believe I better understand in a general sense the difference between reaction & free body forces.  So my guess is when you select a surface and have SW Simulation give you a report of the forces on that surface, it's treating every element along that surface as a separate body.  The force vector given is then just the sum of the individual element forces.  If reaction force is chosen, it's the reaction on those elements due to forces from either neighboring elements or external loads/constraints.  If free body is selected, then it's the sum of the external forces on those particular elements.  Still a bit fuzzy, but I think I get it.  So if I want to separate how much load is acting on a particular section, I should stick to free body, not reaction.  Thanks, again.

                   

                  One easy way to undestand what SW Simulation provide on every case, is the following:

                   

                  Model a simple assembly, for example a column restrained at the base, with a cantilever beam at the top, conected to the column with flange plates (you can apply bolts or make that connection like "welded"), then apply forces to the end of the beam (one analysis for each force on each axis direction),  make this model in a way that will be easy for you to perform manual calculations of the resulting force diagram, run the analisys and take a look of the values given by the application for free body forces and for reactions (look where the system will give you moments at the base).

                   

                  Next, model a single solid (a regular cube) and apply equal forces to each face and on opposite directions so the cube should be "static" in "equlibrium" in space (do not apply gravity force), run the study and see what happen.  Next, apply gravity force (leave the other forces) and also place a restrain (like floor surface) at the base of the cube, run again the study and see the results.

              • Re: Free Body vs Reaction Force
                Derek Bishop

                I've had a look at this before also. Frankly I think it is another good example of poor programming. The help on this issue is also confusing.