Sekalian sama jalannya yaa seutas kawat yang memiliki diameter 3,5 mm, dan panjang 4 m digantung vertikal dan ujung bebasnya diberi beban 46,2 kg. jika g = 10 m/s² dan kawat tsb mulur 3,2 mm, tentukan tegangan, regangan, dan modulus elastisitas kawat!
FadlyHermaja
D = 3,5 mm = 3,5*10⁻³ m Luasnya (A) = π*D² / 4 A = π*(3,5*10⁻³)² / 4 = 9,621*10⁻⁶ m² L = 4 m m = 46,2 kg g = 10 m/s² F = m*g = 46,2*10 = 462 N ΔL = 3,2 mm = 3,2*10⁻³ m Tegangan (σ) = F/A σ = 462/(9,621*10⁻⁶) ≈ 4,8*10⁷ N/m² regangan (e) = ΔL / L e = 3,2*10⁻³ / 4 = 8*10⁻⁴ Modulus elastisitas (E) = σ / e E = 4,8*10⁷ / 8*10⁻⁴ = 6*10¹⁰ N/m²
Luasnya (A) = π*D² / 4
A = π*(3,5*10⁻³)² / 4 = 9,621*10⁻⁶ m²
L = 4 m
m = 46,2 kg
g = 10 m/s²
F = m*g = 46,2*10 = 462 N
ΔL = 3,2 mm = 3,2*10⁻³ m
Tegangan (σ) = F/A
σ = 462/(9,621*10⁻⁶) ≈ 4,8*10⁷ N/m²
regangan (e) = ΔL / L
e = 3,2*10⁻³ / 4 = 8*10⁻⁴
Modulus elastisitas (E) = σ / e
E = 4,8*10⁷ / 8*10⁻⁴ = 6*10¹⁰ N/m²