# Is it possible to create weighing machine with water indicator [closed]

I want to create a weighing machine that works with water.

After researching on Google I got this. $1$ $\text{cm}^3$ of water weights $1$ g.

So, with that equation I'm planned to create this below type of weighing machine. I'm explained weighing machine working process on image. Is my machine logic correct?

NOTE: I know it's very basic physics. But, I have no idea on physics.

## closed as off-topic by user191954, Kyle Kanos, Sebastian Riese, peterh, sammy gerbilSep 26 '18 at 19:18

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• Your weighing machine is based on Pascal's principle. Ref: hyperphysics.phy-astr.gsu.edu/hbase/pasc.html – K_inverse Sep 10 '18 at 7:15
• @K_inverse thanks for this great reference. So, my logic of machine working process is 100% currect. Right? – Chandra Nakka Sep 10 '18 at 7:22
• I am not expert in experiments, but I guess it is ok :). Some remarks are (1) Precise calibrations, including the water level when there is "zero weight". (2) The setup should not be tilted (from calibration to actual measurements). (3) Avoid water evaporation. – K_inverse Sep 10 '18 at 7:53
• The basic principal is ok, but your device is a combined weighing machine / thermometer, because the exact volume of the water depends on its temperature. IIRC, 1g of water occupies 1mL at 14.5°C. – PM 2Ring Sep 10 '18 at 8:20
• Correct, and it would work reliably unless you have a leak or evaporative loss. For perspective: your left piston converts weight into pressure. The rest is just a manometer - a pressure gauge. – orion Sep 13 '18 at 11:32

Consider the following system at equilibrium: Since the two points $A$ and $B$ are at the same altitude in the same fluid in a connected vessel, the pressure at both points is the same: $$P_A=P_B.$$ The pressure at point $A$ is the weight above it divided by the cross sectional area $A_1$ (ignoring the atmospheric pressure since it acts uniformly everywhere): $$P_A=\dfrac{W}{A_1}.$$ At point $B$, there is the hydrostatic pressure: $$P_B=\rho gh.$$
Let $\Delta h_1$ and $\Delta h_2$ represent the change in elevation of water level in the two vessels starting from the initial position when there was no weight placed on the balance.
Let's define $z$ to be equal to $\Delta h_2$, and let's try to find the relation between the weight $W$ and $z$: \begin{align*} P_A&=P_B\\ \dfrac{W}{A_1}&=\rho gh. \end{align*} Now $h=\Delta h_1+\Delta h_2$, and since water is incompressible, the volume decrease of water in vessel 1 is equal to the volume increase in vessel 2, thus $A_1\Delta h_1=A_2\Delta h_2$ which gives $\Delta h_1=\dfrac{A_2}{A_1}\Delta h_2$.
Therefore: \begin{align*} \dfrac{W}{A_1}&=\rho g(\Delta h_2+\dfrac{A_2}{A_1}\Delta h_2)\\ \dfrac{W}{A_1}&=\rho g\Delta h_2(1+\dfrac{A_2}{A_1})\\ \dfrac{W}{A_1}&=\rho gz(1+\dfrac{A_2}{A_1})\\ W&=\rho gz(A_1+A_2). \end{align*} And thus we've got the weight expressed in terms of $z$.