# Are the so called force balance drawings for motion systems “counterintuitive”?

Lets take the standard mass-spring-damper system with an external force. Please also see the image. As we all know making mistakes with signs in these questions is a common error. There is even an earlier question about confusion on the sign here on stackexchange.

In order not to make mistakes with signs, I always teach my students to start with

F = ma

Then from here, I ask them to think a standard Cartesian coordinate system on top of the picture. For this example, it would thus mean that the external force has a positive sign in this picture, but the forces related to the spring and damper have a negative sign. So now the left-hand side becomes:

$$- F_k - F_v + F_{ext} = ma$$

Now rearranging and substitution will give them:

$$F_{ext} = ma + F_k + F_v$$

$$F_{ext} = m \ddot{x} + f_v \dot{x} + kx$$

Unfortunately, if we put this equation into a force balance picture as shown here: It becomes very difficult for the students to grasp. They understand that the external force is pointing to the right and they also understand that the spring force and friction force are pointing to the left. However, when it comes to the newtonian force to the left they get confused and find it counter intuitive although we have just derived the equation by ourselves as shown above.

I seem not to be able to explain intuitively where this force $$F_{newton}$$ comes from. Could someone help me to explain this more intuitively to the students? I think the reason why I cannot explain it so well is because in the picture $$F_{newton}$$ is not visible explicitly, while all the other forces are all visible: an external force, e.g. pulling by hand, a spring and a damper, but how can this mass out of nothing exert a force and why is it in the other direction.

• Are you sure you should draw $F_{newton}$? From what I understand it is just the net force as in $F_{net}=ma$. You don't need a force balance to explain what is happening: when you sum all the forces you get some net force which causes the acceleration of the block. Introducing this artificial $F_{newton}$ just causes confusion. – AccidentalTaylorExpansion Jan 26 '19 at 22:25
• What do you mean by "newtonian force"? Is it the sum of the three real forces? If so, do not include it in a force diagram. If you know the direction of the net force because, for example, you know the direction of acceleration, draw the arrow somewhere else. – garyp Jan 26 '19 at 22:35
• Presumably $F_{\text{newton}}$ is an attempt to show that the mass has inertia? A better way to do all of this IMO is draw two diagrams. One shows all the forces, the other shows all the changes of momentum (or "mass $\times$ accelerations" if you are being simple-minded). Then write equations to make the two diagrams "equivalent" (It's intuitively obvious what "equivalent" means - so don't confuse your students by trying to make a rigorous definition!) – alephzero Jan 26 '19 at 22:40
• To second the above, I think that the only forces that should be drawn on Free Body Diagrams are forces that you would plug into Newton's 2nd law. But I guess I'm not 100% sure what $F_{newton}$ is, so maybe I could be persuaded otherwise? – Bunji Jan 27 '19 at 0:54
• I see: $F_\mathrm{Newton} = ma$. But $ma$ is not a force. There is a list of reasons for writing Newton's second law as $a=F_\mathrm{net}/m$ instead of $F=ma$, and this confusion adds to the list. – garyp Jan 27 '19 at 2:02

The diagram can be thought of as a statics situation in the accelerated (relative to the ground) frame of reference of the mass with $$F_{\rm newton}$$ as a fictitious force.
Another example is an object moving in a circle undergoing centripetal acceleration under the influence of a force $$F$$ whose direction is towards the centre of circle.
In the frame of reference of the object there is a force $$F$$ but no acceleration of the object.
So that Newton’s laws can be used in that accelerated frame a fictitious force equal in magnitude and opposite in direction is introduced and often labelled as $$ma_{\rm centrifugal}$$ where $$m$$ is the mass of the object. The magnitudes of the centripetal and centrifugal acceleration are the same but they are opposite in direction.