Pascal's Principle — Question 3

Question 3

Piston weights and unequal surroundings enter the force balance

A pair of ideal pistons at equal elevations encloses one connected liquid. The input piston has area $A_1=4.0\,\mathrm{cm^2}$, weight $W_1=6.0\,\mathrm{N}$ and exterior pressure $100\,\mathrm{kPa}$. The output piston has area $A_2=40\,\mathrm{cm^2}$, weight $W_2=40\,\mathrm{N}$ and exterior pressure $90\,\mathrm{kPa}$. A downward applied input force of $30\,\mathrm{N}$ is maintained.

1. Find the common absolute liquid pressure and the additional load supported by the output piston.

2. How much of the supported load would be lost if output exterior pressure rose to $100\,\mathrm{kPa}$? Explain why simply multiplying $30\,\mathrm{N}$ by the area ratio is incomplete.

Question 3 - Solution

Piston weights and unequal surroundings enter the force balance

(a) Derivation and calculation

Input equilibrium requires$$pA_1=p_{e1}A_1+W_1+F_1.$$$$\boxed{p=100000+\frac{6.0+30}{.000400}=190000\,\mathrm{Pa}}.$$At the output,$$pA_2=p_{e2}A_2+W_2+L\Rightarrow\boxed{L=(190000-90000)(.00400)-40=360\,\mathrm N}.$$

(b) Result and interpretation

$$\Delta L=-A_2\Delta p_{e2}=-(.00400)(10000)=\boxed{-40\,\mathrm N}.$$$$\boxed{L^{\prime}=320\,\mathrm N}.$$The input weight adds pressure, the output weight consumes supporting force, and unequal external pressures do not cancel.

Physical check

The simple applied-force area ratio omits three actual terms in the piston balances.