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## Introduction to Fluid Mechanic

by: Jolie Shields

53

0

6

# Introduction to Fluid Mechanic CE 321

Marketplace > Michigan State University > Civil Engineering > CE 321 > Introduction to Fluid Mechanic
Jolie Shields
MSU
GPA 3.85

Roger Wallace

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COURSE
PROF.
Roger Wallace
TYPE
Class Notes
PAGES
6
WORDS
KARMA
25 ?

## Popular in Civil Engineering

This 6 page Class Notes was uploaded by Jolie Shields on Saturday September 19, 2015. The Class Notes belongs to CE 321 at Michigan State University taught by Roger Wallace in Fall. Since its upload, it has received 53 views. For similar materials see /class/207369/ce-321-michigan-state-university in Civil Engineering at Michigan State University.

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Date Created: 09/19/15
Lecture 5 Forces on Submerged Bodies Work Problems 234 Two square gates close two openings in a conduit connected to an open tank of water as shown in Fig P234 When the water depth 11 reaches 5 m it is desired that both gates open at the same time Determine the weight of the Nhomogeneousuhqrizpntal gate and the horizontal force R acting on the vertical gate that is required to keep the gates closed until this depth is reached The weight of the verti cal gate is negligible and both gates are hinged at one end as shown Friction in the hinges is negligible T w 39 quot Hinge Horizontal gate 439 X 439 Vertical gate 4m gtlt 4m Hinge 234 solution 01 H m 5 Aar7a77L4 yak ZMH 0 50 7714i A Allure L39s 77w zom t ortsrw an 77 did 3 Adam fwfnce 777u5 dZpZ j 50 771424 N 70 7300071 3 sz m xiim 341 234 solution For Vrr cql yadg H A MAN 15 7m JD 77 g kiwi 37WJMMX4m Moo x A F 7 laa c FL 5 L75 1 Jaw3944 z M 37 LA 5C 7M9Mx m 1 7M 5r ailErik Z M 0 50 774424 I HmHva 7 7m M H7AV 239 An open tank has a vertical partition and on one side contains gasoline with a density p 700 kgrn3 at a depth of 4 m as shown in Fig 13239 A rectangular gate that is 4 m high and 2 m wide and hinged at one end is located in the partition Water is slowly added to the empty side of the tank At what depth h will the gate start to open Partition F125 2 3 4k A 239 Solution where 7 heFers 0 gasoipne 5 700 ZM Hoxlojlv Mm FEW b 1 A where bu refers 40 waiev 3N h m FE Homo i z xk when 1 is clcFTh 0 waJRY FR Mo meMLZ

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