Can anyone explain me how do I calculate torque? 
*

*For my gate automation project, I estimated the amount of force required (in order to estimate the torque) to pull my gate with a spring balance.

*The display on the spring balance read $2 \text{ kg}$. The formula for torque is $T=F \, R \,\sin(\theta)$ 

*I intend to keep the torque in $\text{kg-cm}$.

*My question is, do I need to divide the spring balance's reading i.e $2 \text{ kg}$ by acceleration due to gravity ($9.8 \text{ kg-m/s^2}$)?
This is the first time ever I've posted a question to stack exchange. If anything is inconceivable or inconvenient, my apologies...
 A: The formula for (torque = force × distance) only works with consistent units. If you want the result to be $\text{kg-cm}$ and the force is already measured in $\text{kg}$ there is an implicit conversion from mass to weight by multiplying by gravity on both sides of the equation. As a result, the conversion can be ignored.
$$ (\tau\;\text{[kg-cm]}) = (F\;\text{[kg]}) \times (R\sin\theta\;\text{[cm]} )$$

The SI units for the above formula would be
$$ (\tau\;\text{[N-m]}) = (F\;\text{[N]}) \times (R\sin\theta\;\text{[m]} )$$
and if you divide both sides by gravity $g = 9.80665\;\text{[m/s^2]}$ you have
$$ (\tau\;\text{[kg-m]}) = (F\;\text{[kg]}) \times (R\sin\theta\;\text{[m]} )$$
and finally converting between $\text{[m]}$ and $\text{[cm]}$ on both sides
$$ (100\;\text{[cm/m]}) (\tau\;\text{[kg-m]}) = (F\;\text{[kg]}) \times \left( (100\;\text{[cm/m]}) (R\sin\theta\;\text{[m]} ) \right) $$
$$ (\tau\;\text{[kg-cm]}) = (F\;\text{[kg]}) \times (R\sin\theta\;\text{[cm]} )$$
A: Assuming you mean a physical gate, like that shown below, here's what you do:

The torque is the force applied along the rotation pseudovector, multiplied by the distance at that point to the center of rotation. So let's take our gate and mark a circle of rotation, the direction for measurement:

The red circle is rotation, so we should be measuring our gate along that circle. In practice, this means the spring balance should be held perpendicular to the plane of the gate, which I tried (and that's a very generous use of the word "try") to mark with the blue arrow.
Now while technically you can measure from anywhere along the right edge of the gate, you'll get the best results if you measure along that right edge at the point closest to the vertical center of mass of the gate, approximated with the yellow dot below. That leaves only the math, for which we'll need the purple line.

So take the distance from the point of force measurement at the yellow dot to the center of rotation, along the purple line below, which gives us our distance for the torque calculation.
Finally, because we know torque is $\tau = \textrm{r} \times \textrm{F}$, multiply the measured force along the red axis in the second picture by the distance along the purple line in the third picture, and that will give you the appropriate torque. Note that because we're measuring entirely perpendicular to gravity (which always pulls down), we can disregard gravity here. On the other hand, if we were using a gate that rotated vertically instead of horizontally, gravity would have an impact. Happy physicsing!
Attributions:
Our beautiful gate is courtesy of Bingley Fencing and Timber. I have no idea who they are, they're just the best gate picture I could find in 30 seconds on Google. I hope they don't mind their gate being used for physics.
