Tensions caused by a force in a system of 3 static pillars I was trying to analyze a system like this, which is in static equilibrium.  I have used Newton's first law for forward and rotational motion, and ended up with a peculiar result, where T1 = 0. I was wondering, does it make physical sense, for the second and third support column to be experiencing all the tension, while the first one feels none?
 A: Assuming the rod has no weight, then $T_1$ = 0 is one of many possible answers.  $T_2$ = 0 would also work.
A: 
Tensions caused by a force in a system of 3 static pillars

The system is statically indeterminate due to a redundant tension force.
Assuming $F_y$ is known, there are two equations for static equilibrium, $\sum \vec F=0$, and $\sum \vec M=0$, and three unknowns, $T_1$, $T_2$, and $T_3$. You need to use equations for deformations in addition to the equations for static equilibrium.

I was wondering, does it make physical sense, for the second and third
support column to be experiencing all the tension, while the first one
feels none?

It wouldn't make sense, at least to me, that the tension would be zero for any real structure. But one way it would theoretically make sense for the tension in pillar 1 to be much much less than pillars 2 and 3 is if pillar 1 was much much less "stiff" than pillars 2 and 3.
For example, let pillar 1 be made of rubber with $E=0.01$ GPa and pillars 2 and 3 made of steel with $E=200$ GPa, and the lengths and cross sectional areas of all pillars be the same. Since the stiffness of pillars 2 and 3 is so much greater than pillar 1, and since pillars 2 and 3 are symmetric about the load (and neglecting the weight of the bar), then the deformations of pillars 2 and 3 will be the limiting factor and be equal. Pillar 1 being so flexible will essentially "go along for the ride" and be the same.
The axial deformation, $\delta$, as a function of  length $L$, cross sectional area $A$, Modulus of Elasticity (Young's Modulus) $E$ and tension $T$ is
$$\delta=\frac{TL}{EA}$$
Then assuming the deformation of pillar 1 is the same as pillar 2 (and 3) due to its relatively low $E$ and therefore relatively low resistance to axial stress,
$$\frac{T_{1}L}{E_{1}A}=\frac{T_{2}L}{E_{2}A}$$
$$T_{1}=\frac{T_{2}E_1}{E_2}$$
Finally, if $E_{1}<<E_2$, then $T_{1}<<T_2$.
For the rubber and steel pillars example,
$$T_{1}=T_2\frac{0.01}{200}=0.00005 T_2$$
Hope this helps.
A: There are many solutions for a uniform rod, depending on how much weight e.g. $T_3$ supports and the ratio of the $F_y$ ($=F$) to the weight of the rod.
If $$F=kW$$
Upwards forces equal downward forces, so
$$T_1+T_2+T_3 = (k+1)W$$
taking moments (sum of torques = 0) around each point in turn, from left to right, where an arrow on the diagram meets the rod, gives a set of equations, not all independent.
$$2T_2+4T_3=2W+3F$$
$$2T_1+F = 2T_3$$
$$3T_1+T_2 = W+T_3$$
$$4T_1+2T_2=2W+F$$
All that can be found is $T_1$ and $T_2$ in terms of $T_3$, $k$ and $W$.
$$T_1 = T_3 -\frac{kW}{2}\tag1$$
$$T_2 = (\frac{3k}{2}+1)W-2T_3\tag2$$
So $T_1=0$ makes physical sense if $T_3 = \frac{kW}{2}$ from 1) and $T_2 = \frac{kW}{2} +W$ from 2)
That is $T_2$ and $T_3$ are supporting half the force $F_y$ each and $T_2$ is also supporting the weight of the rod.
