# Calculating the water pressure knowing the circuit that tracks it

I'm not very good at physics, but I've got a problem that I really need to solve.

On the second picture you can see a pipe with flowing water. The point of this mechanism is to lit a LED whenever the water pressure becomes too high. We're using a comparator to track the pressure. The piston on the second picture is connected to the linear sliding potentiometer R1.

• $$R2 = 5kΩ$$
• $$R3 = 30kΩ$$
• Length of the potentiometer $$d=50mm$$
• When the potentiometer slider is at its lowest position, the voltage at its ends is $$U=0V$$
• Voltage provided to the circuit is $$U=5V$$ (excuse me if I'm not using the proper terminology, I'm not a native english speaker)
• Water density may be assumed equal $$1000\frac{kg}{m^3}$$
• Gravitational acceleration may be assumed equal $$10\frac{m}{s^2}$$

The problem is to calculate the minimum water pressure that would cause the LED to lit.

As far as I understand, the first thing to do would be to calculate the current at the negative end of the comparator. According to the Ohm's law, the current in that branch would be $$I=\frac{U}{R2 + R3}=\frac{5}{35000}=\frac{1}{7000}A$$. Now the voltage at the negative comparator end is $$U=R2\times I=5000\times\frac{1}{7000}=\frac{5}{7}V$$.

Now comes the part where I have no idea, what to do. My assumption is that if the voltage at the potentiometer's lowest point is 0V and at its highest point it is (probably) 5V, then it must become $$\frac{5}{7}V$$ at $$\frac{1}{5}\times\frac{5}{7}=\frac{1}{7}$$ its length, so $$\frac{1}{7}\times50mm=\frac{50}{7}mm$$. Even if these ideas are correct, I've got no idea, what to do next. I'd appreciate any help