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TUGAS MEKANIKA FLUIDA

TUGAS MEKANIKA FLUIDA. YULI TRIAWAN 20110110005. NOMOR 1. Sebuah balok ponton dengan panjang L= 12m dan lebar B= 6m dan dan benda yang tenggelam dalam air d=1,5m seperti gambar ini : Tentukanlah : a.berat balok b.(d) apabila dalam air laut dengan ρ =1025 kg/m 3. jawaban.

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TUGAS MEKANIKA FLUIDA

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  1. TUGAS MEKANIKA FLUIDA YULI TRIAWAN 20110110005

  2. NOMOR 1 Sebuahbalokpontondenganpanjang L= 12m danlebar B= 6m dandanbenda yang tenggelamdalam air d=1,5m sepertigambarini : Tentukanlah: a.beratbalok b.(d) apabiladalam air lautdenganρ=1025 kg/m3

  3. jawaban A. dalamkeadaanterapung, beratbendaaknsamadenganberat air yang dipindahkanbendaFb makaFb = Fg = ρ₁.g.B.L.d d = 1000x9,81x6x12x1,5 d = 1.059.480N

  4. B dalamkeadaanterapungmakaberatbendasamadenganberat air yang dipindahkanbenda. Fb = Fg = ρ₂.g.B.L.d d = Fg/ρ₂.g.B.L d = 1.059.480 N/1025x9,81x6x12 d = 1,463 m

  5. NOMOR 2 • Kubusdengansisi 25 cm danrapatrelatif 0,9 mengapung di air denganslahsatusisinyasejajarmukaair.hitunglahberatbeban agar kubusdapattenggelam.

  6. jawaban Sisikubus 0,25m Rapatrelatif 0,9x1000= 900 kg/m3 Beratkubus = Fg = V.ρ₂.g = B³. ρ₁.g Misalakantinggikubus yang tenggelamadalah d gayaapungFb= A.d. ρ₁.g = B².d. ρ₁.g Fg = Fb d= (ρ₂/ ρ₁)X B = 0,9 x 0,25 = 0,225 m

  7. Jikadiataskubusdiberibebansebesar W₂ makaberat total kubusadalah: ∆W = W₁ + W₂ = Fg + W₂ 0,25³X900X9,81 + W₂ = 137,953 + W₂ jikabendaterendamseluruhnyamakadalamkeadaantersebut : Fb= V. ρ₁.g = 0,25³X1000X9,81 = 153,281 N

  8. Jikapersamaandisaamakanmaka : ∆W = Fb 137,953 + W₂ = 153,281 N W₂ = 15,328 n

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