Mekanika Fluida Tugas 3.docx

GAYA PADA BIDANG MEKANIKA FLUIDA

GILANG RIZKIAWAN HADI PUTRA

PROGRAM STUDI TEKNIK SIPIL FAKULTAS TEKNIK UNIVERSITAS RIAU KEPULAUAN 2018/2019

( 17.07.0.025 )

16 OKTOBER 2018

Tugas :

Jawaban : Diketahui : 𝜃 = 45°

𝐴 = 0,6 𝑥 0,6 = 0,36 𝑚2

𝑔 = 9,81 𝑚⁄𝑠 2

ℎ = 12 𝑚

𝜌𝑎𝑖𝑟 = 1000 𝑘𝑔⁄𝑚3

𝑑 = 0,3 𝑚

Ditanya : 𝑃?

𝐹?

𝑦 ?

𝑦𝑝 ?

Penjelasan : 𝐹 =𝑃 ×𝐴

𝑦=

𝐹 = 𝜌×𝑔×ℎ×𝐴

𝑦 =

𝐹 = 1000 × 9,81 × (12 + 0,3. 𝑠𝑖𝑛 45° ) × (0,62 ) 𝑦 =

ℎ sin 45° 12 + 0,3 𝑠𝑖𝑛 45° 𝑠𝑖𝑛 45° 12 + 0,212132034 0,707106781

12,212132034

𝐹 = 9810 × 12,212 × 0,36

𝑦 =

𝐹 = 43128,36549 𝑁 → 43,128 𝐾𝑁

𝑦 = 17,27056275 𝑚

0,707106781

1⁄ × 0,6 × 0,63 12 𝑦𝑝 = 17,27056275 + 2 0,6 × 17,27056275 𝑦𝑝 = 17,27056275 +

0,0108 6, 217402591

𝑦𝑝 = 17,27056275 + 1,737059784 × 10−3 𝑦𝑝 = 17,27229981 𝑚

𝐴 × 𝑃 = ( 𝑦𝑝 − 𝑦 +

1 2

𝐴)𝐹

0,6 𝑃 = ( 17,27229981 − 17,27056275 + 0,3 ) × 43,128 0,6 𝑃 = 0,30173706 × 43,128 0,6 𝑃 = 13,01331592 𝑃=

13,01331592 0,6

𝑃 = 21,68885987 𝐾𝑁