Force needed to push a syringe plunger: does one add force associated with downstream back-pressure to frictional plunger force? I am trying to figure out how much force $F$ is needed to push a syringe plunger.
The plunger needs to overcome the friction force $F_1$ and (a much smaller) inertia force $F_2=ma$, giving the total $F$ needed: $F=F_1+F_2$. Now there is a constriction downstream creating a back-pressure $dP$ (or force $F_3=dP \times A$). This is where I am getting confused.
Does the back pressure mean that the total force needed to push the plunger is $F=F_1+F_2+F_3$ or is it just $F=F_1+F_2$ provided $F_1+F_2 > F_3$?

 A: The force you have to exert on the plunger is $F_1 - F_2 + F_3$. If you didn't include the back pressure term it would be just as easy to squirt treacle as it would be to squirt water, which obviously isn't the case.
A: Yes, you absolutely need to count the back pressure.  Otherwise the force required would be essentially independent of flow rate or the properties of the fluid.  For a given fluid, you need to assess the pressure needed inside the syringe to make the desired flow. The force on the plunger needs to overcome that.  In many cases that will be the dominant contribution.
A: You need to count the back pressure.  I know that, because I've felt the difference, in a syringe filled with oil versus one with alcohol.  You also need to understand that you are moving the fluid, which has its own mass, and does not have the same center-of-mass acceleration as the piston does (because some of fluid mass is in the narrow channel shown). 
There may be some justification for ignoring the air behind the plunger, as 'light weight', but it, too,  has mass, and maybe
even its own contribution to velocity-dependent (drag) resistance.  If one was really accurate, there would be additional forces related to surface tension and the relative wetting of air and fluid against your syringe material, and a drop of fluid forming at the tip of the syringe, in air, would also have a surface tension force associated with it.  
It would be nice to calculate all (eight?) terms, but it's more usual to stop somewhere short of that.  :>
