Example 2.A Width is 1 m. Find: F and the position that the fluid pressure exerted. ΣF=0=Fy-Fy - W FH -2 m Fv tw Fx F 4m y Fy=pA = 9810*4*2*1 = 78.5 kN π 2 W = WV = y = y₁² 4 = 9810*0.25 **4*1 = 30.8 kN F₁ = 30.8+78.5 y F₁ = 109.3 KN *1 ΣF=0=FH-Fx Fx = PA X = 5*9810*2*1 Fx = 98.1kN
Example 2.A Width is 1 m. Find: F and the position that the fluid pressure exerted. ΣF=0=Fy-Fy - W FH -2 m Fv tw Fx F 4m y Fy=pA = 9810*4*2*1 = 78.5 kN π 2 W = WV = y = y₁² 4 = 9810*0.25 **4*1 = 30.8 kN F₁ = 30.8+78.5 y F₁ = 109.3 KN *1 ΣF=0=FH-Fx Fx = PA X = 5*9810*2*1 Fx = 98.1kN
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Please explain these slides step by step so I can understand
![Y cp
FH
W
LIX
Fy
F
Example 2.A
4 m
x cp
x cp
Ycp = y +
xw
I
JA
XcpFy = Fy *1+W* xw
ср у
=
4r
3π
5+0.067 5.067m
=
= 0.957 m
0.849 m
78.5*1+30.8*0.849
109.3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b3f294e-d3c8-4475-8e8e-f2b525027412%2Fd440959e-7f2e-42e7-83b7-8eb32dd49118%2Fmzuujzh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Y cp
FH
W
LIX
Fy
F
Example 2.A
4 m
x cp
x cp
Ycp = y +
xw
I
JA
XcpFy = Fy *1+W* xw
ср у
=
4r
3π
5+0.067 5.067m
=
= 0.957 m
0.849 m
78.5*1+30.8*0.849
109.3
![Example 2.A
Width is 1 m. Find: F and the position that the fluid pressure exerted.
ΣF=0=Fy-Fy – W
FH
Fx
W
F
4m
y
Fy = PA
= 9810*4*2*1
= 78.5 kN
T
W = y = y ²¹ r² *1
4
= 9810*0.25*π *4*1
= 30.8 kN
Fy = 30.8+78.5
F₁, = 109.3 kN
ΣF=0= FH - Fx
X
Fx = pA
= 5*9810*2*1
F₂ = 98.1kN
X](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b3f294e-d3c8-4475-8e8e-f2b525027412%2Fd440959e-7f2e-42e7-83b7-8eb32dd49118%2F5mzqt2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example 2.A
Width is 1 m. Find: F and the position that the fluid pressure exerted.
ΣF=0=Fy-Fy – W
FH
Fx
W
F
4m
y
Fy = PA
= 9810*4*2*1
= 78.5 kN
T
W = y = y ²¹ r² *1
4
= 9810*0.25*π *4*1
= 30.8 kN
Fy = 30.8+78.5
F₁, = 109.3 kN
ΣF=0= FH - Fx
X
Fx = pA
= 5*9810*2*1
F₂ = 98.1kN
X
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