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Mekanika fluida

Mekanika fluida. Disusun oleh :. R.Wisnu M Nur 20110110013 Teknik Sipil. Universitas Muhammadiyah Yogyakarta. Soal no. 1 :. 1. Batu di udara mempunyai berat 506 N, sedang beratnya di dalam air adalah 313 N. Hitung volume dan rapat relatif batu itu. Penyelesaian :.

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Mekanika fluida

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  1. Mekanikafluida Disusunoleh : R.Wisnu M Nur 20110110013 TeknikSipil UniversitasMuhammadiyah Yogyakarta

  2. Soal no. 1 : 1. Batudiudaramempunyaiberat506 N, sedangberatnya didalam air adalah313 N. Hitung volume danrapat relatifbatuitu. Penyelesaian : Gaya apung (FB) samadenganperbedaanantaraberatbatudiudaradandidalam air FB = W diudara– W di air = 506 – 313 = 193 N Hukum Archimedes, Gaya apung (FB) samadenganberat air yang dipindahkanbatu.Sedangkanberat air yang dipindahkanbatusamadenganperkalianantara volume air yang dipindahkan (V) danberatjenis air.

  3. FB = γ . V = ρ.g.V = 1000 x 9,81 x V =9810 V Dari keduanilai FB diatas FB₁ = FB₂ 193 = 9810.V V = 193 / 9810 V = 0,0196 m³ Volume air = volume batu , sehingga volume batu = 0,0196 m³ Beratbatudiudara = beratjenisbatu x volume batu Sehingga : W diudara = γ. V = ρ . g.V 506 = ρx 9,81 x 0,0196 ρ = 506 / 9,81 x 0,0196 = 2635,41 kg / m³ => S = ρ / ρ air = 2,63

  4. Soal no 2 : 2.Silinder berdiameter 45 cm dan rapat relatif 0,9. Apabilasilindermengapungdidalamair dengansumbunyavertikal, tentukantinggimaksimumsilinder.

  5. Penyelesaian : S = γbenda / γ air = 0,9 => γbenda = 0,9 x 1000 = 900 kg/m³ Beratbenda : Fg = (π / 4) D² . H . γbenda = (π / 4) .0,45² x H x 900 = 143,13 H_kgf Gaya apung : Fb = (π / 4 ) D² . d . γair = (π / 4 ) 0,45² x d x 1000 = 159,04 d_kgf

  6. Dalamkeadaanmengapung : Fg = Fb 143,13H = 159,04d d = 0,9H Jarakpusatapungterhadapdasar : OB = d / 2 = 0,9 / 2 =0,45 H Jarakpusatberatterhadapdasar : OG = 1 / 2 = 0,5 H Jarakantarapusatberatdenganpusatapung : BG = OG – OB = 0,5 – 0,45 = 0,05 H

  7. Moment Inersiatampangpelampung yang terpotongmuka air : I₀ = (π / 64) D⁴ = (π / 64) x 0,45⁴ = 2,012 x 10⁻³ m⁴ Volume air yang dipindahkan : V = (π / 4) D². d = (π / 4) o,45² x 0,9H = 0,143 H_m³ Tinggimetasentrum : BM = I₀ / V = 2,01 x 10⁻³ / 0,143 = 0,014 / H Benda akanstabilapabilaBM > BG 0,014 / H > 0,05 H => H < 0,53 m Jaditinggisilindermaksimumadalah0,53m.

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