What's the optimum angle to pull a wide object over a horizontal plane I have a real life question (not homework so I'm struggling to find an answer online) that I'm hoping someone can assist me with out of curiosity.
So basically I have a wide object (pole in this case) that I want to pull a certain distance towards myself over the ground (so we can ignore any vertical-ness with the angles).
What's the optimum places to attach ropes and pull the object to myself and why?
Some illustrations of what I'm wondering (top down):
Scenario 1
Attaching the ropes in the middle of the object, in line with where I'm standing:

Scenario 2
2 Ropes attached at each point of the object

Scenario 3
1 Rope attached to each point, and pulled with another rope in the middle of the first rope:

Gut feeling
My gut's telling me that scenario 2 or 3 would be best, and most likely the most stable while pulling, but I'm wonder what the math behind this says
Edit 1
Definition of the optimal condition: At which angle or attachment point would the least amount of force be required to move the object
Thanks for pointing this out @basics
 A: Let us suppose that you need a force $F_\text{pull}$ of magnitude $F$ to counteract the friction between the pole and the floor, so that your object can move with a constant velocity. I assume that the rope is ideal, i.e. that tension propagates down it without being reduced.
In the figure drew the force that the ropes apply to the pole in green, friction in red, the ropes' tension in orange, and the net force that you need to apply to the rope in purple. The labels report the magnitude of these forces.

Configuration 1.
You apply a force $F$ to the rope, which pulls with identical force $T=F$ the pole.
Configuration 2.
The net force on the pole - i.e. the magnitude of the vector sum of the two green vectors - is $2 T \sin\alpha$ ($\alpha$ is the angle drawn in green) in the "horizontal" direction; the "vertical" components cancel out. This is the same net force (purple) that you have to apply to the rope, because again the vertical components of tension (orange) cancel out. This means that you pull again with a force $F$, but in this case the tension on each rope is equal to $T = F/(2 \sin\alpha)$, which can be larger or smaller than $F$, depending on the angle. Nevertheless, the work you do is the same.
Configuration 3.
The net force on the pole is again $2 T \sin\alpha$, and the math is identical to config. 2. The only difference is that the tension of the red rope is $F$, while the one on the blue ropes is $F/(2 \sin\alpha)$. You pull again with a force $F$, and you do the same amount of work.
Now let's complicate the problem a bit. Assume that the friction is not uniform along the pole. Then, if you pull in the middle, you can counteract the total friction, but you will unavoidably have some torque acting on the pole, which will then rotate.

In formulas, while $F_\text{pull}$ will perfectly balance $F_1 + F_2$, the total torque will be nonzero and equal to $\frac{L}{2}(F_2-F_1)$ in the "counterclockwise" direction ($L$ is the pole length).
The other two configurations are better from this point of view. Qualitatively, by "tilting" the force that you apply to the rope, you can tune the two tensions so that you counter-balance any torque. I drew an example for config. 2:

Analogous math applies to config. 3. However, in this last case, the red rope itself has to be tilted: I think that, in practice, it is much easier to work with config. 2, especially if you use both hands to pull one rope each...
Hope this helps!
A: To overcome the effect of gravity and friction it is best to attach a rope below the center of gravity of the object.
