Download Fluids Assignment 3 – Hydrostatic Forces on Plane Surfaces and more Quizzes Fluid Mechanics in PDF only on Docsity!
A TANK 5.4 m DEEP AND 2 m WIDE IS LAYERED WITH 2.4 m OF OIL SP. GR. (0.8) , 1.8 m OF WATER
AND 1.2 m OF MERCURY. (A) COMPUTE THE HYDROSTATIC FORCE (kN) ON THE UPPER LAYER
OF THE OIL ONLY. (B) COMPUTE THE HYDROSTATIC FORCE OF THE 4.2 m CONTAINING OIL AND
WATER
SOLUTION:
F
oil
= P
cg
A
P
cg
= ( 9. 81 kN/m
3
)( 0. 8 )( 1. 2 m)
P
cg
= 9. 418 kPa
F
oil
= ( 9. 418 kPa)( 2. 4 m)( 2 m)
F
oil
= 45. 204 kN
F
water
= P
cg
A
P
cg
= ( 9. 81 kN/m
3
)( 0. 8 )( 2. 4 m) + ( 9. 81 kN/m
3
)( 0. 9 m)
P
cg
= 27. 664 kPa
F
water
= ( 27. 664 kPa)( 1. 8 m)( 2 m)
F
water
= 99. 591 kN
F
oil+water
= 45. 204 kN + 99. 591 kN
F
oil+water
= 144. 796 kN
IN THE FIG. P3.2, STOPPER B WILL BREAK IF THE FORCE ON IT REACHES 40 kN. FIND THE
CRITICAL WATER DEPTH. THE LENGTH OF THE GATE PERPENDICULAR TO THE SKETCH IS 1.5 m
SOLUTION:
M
HINGE
F
- 5 + e
F = yh
A
F = ( 9. 81 kN/m
3
)(h
)( 1 m)( 1. 5 m)
F = 14. 715 h
e =
I
g
y̅A
; y̅ = h
e =
( 1. 5 m)( 1 m)
3
h
( 1. 5 m)( 1 m)
12 h
- 715 h
12 h
h
= 5. 270 m
h = h
h = 5. 770 m
A TRIANGULAR THROUGH AS SHOWN IN FIG. P3.4 CARRIES WATER AT A DEPTH OF 5 m. THE
TROUGH IS HINGED AT A AND IS HELD TOGETHER BY CABLE BC AT THE TOP. THE ANGLE BAC IS
EQUAL TO 110°. (A) COMPUTE THE TOTAL PRESSURE ON SIDE AB PER METER LENGTH. (B)
COMPUTE THE TENSION AT THE CABLE IF IT HAS A SPACING OF 1 m. PERPENDICULAR TO THE
PAPER.
SOLUTION:
F = yh
A
F = ( 9. 81 kN/m
3
)( 2. 5 m) (
sin
m) ( 1 m)
F = 213. 790 kN
e =
I
g
y̅A
e =
( 1 m) (
sin
m)
3
sin(35°)
m
) ( 1 m) (
sin
m)
e = 1. 453 m
∑ M
HINGE
T( 2 m + 5 m) = 213. 790 kN (
sin
m
− e)
T = 88. 746 kN
A CIRCULAR GATE 1.5 m IN DIAMETER IS INCLINED 30 DEGREES WITH THE VERTICAL. IF
CENTER OF THE GATE IS 10 m BELOW THE SEA WATER SURFACE. (A) CALCULATE THE TOTAL
HYDROSTATIC FORCE ON THE GATE. (B) CALCULATE THE DISTANCE OF THE TOTAL FORCE FROM
THE BOTTOM OF THE GATE MEASURED ALONG THE GATE. (C) IF THE GATE IS HINGED AT TOP,
FIND THE NORMAL FORCE APPLIED AT THE BOTTOM REQUIRED TO OPEN THE GATE.
SOLUTION:
REQUIRED:
P
A
− P
B
SOLUTION:
F = ( 9. 81 kN/m
3
)( 1. 03 )( 10 m) (
π( 1. 5 )
2
F = 178. 558 kN
e =
I
g
y̅A
; cos
y̅
e =
π
- 75 m
4
cos(30°)
π( 1. 5 )
2
e = 0. 012 m
d = 1. 5 m − 0. 75 m − e
d = 1. 5 m − 0. 75 m − 0. 012 m
d = 0. 738 m
M
HINGE
P( 1. 5 ) = F( 0. 75 m + e )
P( 1. 5 ) = ( 178. 558 kN)( 0. 75 m + 0. 012 m )
P = 90. 729 kN
