Numerical 1: Extension of a Steel Wire

Given: Length L = 2 m, Cross-sectional area A = 1 $mm^{2}$ = $1 \times 10^{-6}$$m^{2}, Applied force F = 100 N, Young's modulus Y = \2 \times 10^{11}$ Pa.

Find: Extension $\Delta L$.

Formula: $\Delta L$ = FL / (AY)

Step-by-step:

Step 1: Identify values with units. F = 100 N, L = 2 m, A = $1 \times 10^{-6}$$m^{2}, Y = \2 \times 10^{11} Pa = \2 \times 10^{11}$N/$m^{2}$

Step 2: Substitute. $\Delta L$ = (100 N × 2 m) / ($1 \times 10^{-6}$$m^{2} × \2 \times 10^{11}$N/$m^{2}$)

Step 3: Numerator. 100 × 2 = 200 N·m

Step 4: Denominator. $1 \times 10^{-6} × \2 \times 10^{11} = \2 \times 10^{5}$N·$m^{2}$/$m^{2} = \2 \times 10^{5}$ N

Step 5: $\Delta L$ = 200 N·m / ($2 \times 10^{5} N) = \1 \times 10^{-3}$ m = 1 mm

Dimensional check: [N][m] / ([$m^{2}$][N/$m^{2}$]) = [N·m] / [N] = [m] ✓$