Force on every particle? Say there is a horizontal rod, where a force $P$ is applied to the right hand side of the rod to drag it horizontally to the right. If I drew a force diagram, I'd put some vector with a length of $\| \mathbf{P} \|$ coming out of the right end of the rod. In actuality, if the rod is composed of ten particles, then is $\| \mathbf{P} \|$ distributed evenly? So if I felt like creating a more accurate force diagram, then each particle would have a smaller vector going to the right with magnitude $\| \mathbf{P} \|/10$?
Is this true? Is the reason we usually only put 1 vector of length $\| \mathbf{P} \|$ for simplicity? Likewise when we assume the force of gravity $F_g$ on an extended object can be put at the center of gravity of that object as if it was a point mass. But in actuality, there are like a billion really small $F_g$ vectors on each particle?
 A: Consider a train consisting of two cars (for simplicity) both of mass $m$ - it is connected by a long rope to the engine. In the first instance, let's assume there is no friction, and the engine is pulling with force $2F$. Since the two cars are connected, they will accelerate at the same rate $a$, where $a = \frac{2F}{2m}$. If I wanted to represent the entire train as a single object, I would use the single force vector $2F$ and consider the entire train as having mass $2m$ - and all would be well.
If I look closer, I see that the last car must experience a force $F$ (not $2F$) - which tells me that the force on the coupling between the two cars is $F$. That means that the net force on the first car is $2F - F = F$ - the tension in the rope pulling minus the force on the coupling with which it is pulling the second car. With this net force, the first car is accelerating at the same rate $a$ as the second car, and again all is well.
You can make the analysis more complicated by adding friction, changing the mass of the cars, etc. My point is that the net force on each "car" should be such that it moves (accelerates, or maintains velocity) at the same rate as the other "cars" in the "train" (or points in the object). Exactly what this force balance looks like will depend on the details of the setup. For the rod, if it's a frictionless situation then the force per unit length must be constant - but the net force on each segment is only a fraction of the total force on the rod.
If you hang the rod vertically, then each element of mass feels a corresponding force (since $F_g = mg$; but again if you consider the force at the top of the rod, that force has to carry the weight of the entire rod, so the tensile stress there will be greater than lower down in the rod.
A: If we are considering "rigid" objects (ones where the deformation of the material simply isn't large enough to be relevant to our problem), then no, there's no decrease.  You can think of each of those ten elements having the same force from one to the next.  $P$ pulls on #10, #10 pulls on #9, and so on.  Between each element is the same force.  
Also, the location of the force makes no difference for linear translations or accelerations.  You could pull the end of the rod, the middle, or anywhere else.  (It does matter where the force is applied if you are interested in the torque or rotation of the object).  
The location would affect your diagram though.  If instead of pulling on the end of the rod you pulled on the middle, then the wall would feel the same force, but the end of the rod would not.  Only the elements from the pull point to the wall would have your force $P$.
Now gravity is different.  There, you are correct that each individual piece is being pulled slightly.  However, Newton did the math to show that in many cases, we don't have to worry about the complexity.  Instead, the result is the same as if we consider a simple force acting on the object through the center of gravity.  In cases where we cannot generalize, then you might want to actually sum up all of the infinitesimal forces (via calculus perhaps).

I was reading your first question differently.  As in "if you pull on the end, what do the forces between the elements look like".  I think now you were asking "Is pulling on the end with P really just something pulling on each element with a fraction of the force".
I would say that it's not the same, but both would yield similar reactions on a rigid body. Since you can just do a vector summation of the forces, then one force of $P$, or 10 forces each with one-tenth the magnitude are equivalent.  But that second way of seeing it is not "more accurate".
Once you move beyond rigid bodies and care about the internal stress, then where the force is applied matters.  When the force is applied at the end, then each linear element pulls on the next with the same force.  But if the force is applied in the middle, only the elements from there to the wall have the forces between them.  If the force is distributed over all elements (as gravity would do), then the forces between each element are different.
