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Fluids Assignment 4 – Hydrostatic Forces on Curved Surfaces Description: Companion to Lecture 4 Focused on resolving forces on curved submerged surfaces Helps in mastering vertical and horizontal force components
Typology: Quizzes
1 / 4
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H
= γh
H
= ( 9. 81 kN/m
3
)( 3 m )( 2 m)( 1 m)
H
= 58. 860 kN
V
= γV
V
= ( 9. 81 kN/m
3
)( 1 m)(A
three quarter circle
rectangle
V
= ( 9. 81 kN/m
3
)( 1 m) (
3π
2 m
2
V
= 170. 937 kN
FROM FIG. P4.2, THE 1.20 m DIAMETER CYLINDER, 1.2 m LONG IS ACTED UPON BY WATER ON
THE REACTION AT B IF THE CYLINDER WEIGHTS 19.62 kN.
For Water
H 1
= γh
A = ( 9. 81 kN/m
3
)( 1. 2 m )( 1. 2 m)( 1. 2 m)
H 1
= 16. 952 kN
V 1
= γV = ( 9. 81 kN/m
3
)( 1. 20 m)(A
semicircle
V 1
= ( 9. 81 kN/m
3
)( 1. 20 m) (
π( 0. 60 m)
2
V 1
= 6. 657 kN
For Oil
H 2
= γh
A = ( 9. 81 kN/m
3
)( 0. 80 )( 0. 6 m )( 1. 2 m)( 1. 2 m)
H 2
= 6. 781 kN
V
2
= γV = ( 9. 81 kN/m
3
)( 0. 80 )( 1. 20 m)(A
semicircle
V 2
= ( 9. 81 kN/m
3
)( 0. 80 )( 1. 20 m) (
π
2
V 2
= 5. 326 kN
x
H
1
H
2
B
x
B x
= 16. 952 kN − 6. 781 kN
B x
= 10. 171 kN
y
V 1
V 2
B y
B y
= 19. 62 kN − 6. 657 kN − 5. 326 kN
B y
= 7. 638 kN
oil
= γV
oil
oil
= ( 9. 81 kN/m
3
cylinder
cone
oil
= ( 9. 81 kN/m
3
)( 0. 8 ) (π( 3 tan(15°) m)
2
( 5 m) −
π
3 tan
m
2
( 3 m)
oil
= 63. 726 kN
air
air
air
= ( 20 kPa)(π
3 tan
m
2
air
= 40. 600 kN
y
oil
air
F = 63. 726 kN − 40 .600kN
F = 23. 126 kN
