The International System of Units — Question 2

Question 2

One force, two contact areas

Two ideal pistons exert uniform normal forces on their contact surfaces. Ignore all other forces. Define pressure by $p=F/A$, where $A$ is contact area. The newton satisfies $1\ \mathrm{N}=1\ \mathrm{kg\,m\,s^{-2}}$.

1. Derive the coherent SI unit of pressure in base units. Its special name is the pascal, symbol $\mathrm{Pa}$.

2. Calculate both pressures and their ratio $p_B/p_A$. Explain physically why equal forces need not give equal pressures.

3. A student proposes $1\ \mathrm{Pa}=1\ \mathrm{N\,m^2}$ and consequently computes $p=FA$. Use both units and the piston comparison to diagnose the error.

Treat the given central values as exact for this unit-structure exercise; no measurement uncertainty is being estimated.

Question 2 - Solution

One force, two contact areas

(a) Follow the quotient in the quantity equation.

$$[p]=\frac{[F]}{[A]}=\frac{\mathrm{kg\,m\,s^{-2}}}{\mathrm{m^2}} =\boxed{\mathrm{kg\,m^{-1}\,s^{-2}}}.$$ Thus $1\ \mathrm{Pa}=1\ \mathrm{N/m^2}$. Coherence means that no additional numerical scale factor is needed when the inputs use coherent SI units.

(b) Compare force per area.

$$p_A=\frac{12.0}{0.030}\ \mathrm{Pa}=400\ \mathrm{Pa},\qquad p_B=\frac{12.0}{0.060}\ \mathrm{Pa}=200\ \mathrm{Pa}.$$ Therefore $\boxed{p_B/p_A=1/2}$. Spreading the same force over twice the area halves the force per unit area.

(c) Two independent diagnostics.

$$[FA]=\mathrm{kg\,m^3\,s^{-2}}\ne\mathrm{Pa},\qquad\frac{FA_2}{FA_1}=2\ne\frac{p_2}{p_1}=\frac12.$$

Multiplying force by area produces $\mathrm{kg\,m^3\,s^{-2}}$, which is not the required pressure unit. The proposed expression would also double its value when area doubles at fixed force, whereas the definition of pressure demands a decrease by a factor of two. Neither renaming $\mathrm{N\,m^2}$ as pascals nor changing the numerical label can repair the quantity equation.