With force fixed, a larger checked area lowers normal stress. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Keep the force fixed before changing area

This contrast holds the force vector steady. Only the checked area marker is larger than the base case.

F=12 NA=6 m2F=12\ \text{N}\qquad A=6\ \text{m}^{2}
Fixed-force area contrastOnly the area marker changes in this case.forceareaforce=12 Narea=6 m^2stress=2 Pa

Larger checked area lowers stress

The force column is unchanged. As area grows, each exact stress entry gets smaller.

FAσ12 N3 m24 Pa12 N6 m22 Pa12 N12 m21 Pa\begin{array}{c|c|c}F&A&\sigma\\12\ \text{N}&3\ \text{m}^{2}&4\ \text{Pa}\\12\ \text{N}&6\ \text{m}^{2}&2\ \text{Pa}\\12\ \text{N}&12\ \text{m}^{2}&1\ \text{Pa}\\\end{array}

Doubling area halves the stress when force stays fixed

The force is unchanged, so the larger checked area lowers the stress.

σ=12 N6 m2=2 Pa\sigma=\frac{12\ \text{N}}{6\ \text{m}^{2}}=2\ \text{Pa}
Double-area stress ledgerOnly the cross-sectional area changes in this contrast case.forceareaforce=12 Narea=6 m^2stress=2 Pa