Finding Gauge Pressure in this Manometer

[Steve4Physics]

In addition to Post #30 ...

$P_{3} = \left( 800 \ \frac{kg}{m^3} + 9.8 \ \frac{m}{s^2} + 0.15 \ m \right) + P_{4}$

So, the pressure difference between point 3 and point 4 is:

$P_{3} - P_{4} = \left( 800 \ \frac{kg}{m^3} + 9.8 \ \frac{m}{s^2} + 0.15 \ m \right) = 1,176 \ Pa$

Is this correct?

The final value is correct but there are some errors in what you have typed. You have typed additions ($+$) where you should have multiplications ($\times$).

 

[Lnewqban]

I don't know. Please explain.

If you would imaginarily remove the pipe and tank and liquid above cross-section 4, how the remaining liquid columns would naturally move to find a new balance?
How much effort would be needed to avoid that movement?