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Fluids Assignment 6 – Buoyancy, Quizzes of Fluid Mechanics

Fluids Assignment 6 – Buoyancy Description: Linked to Lecture 6 Problems on buoyant force, Archimedes’ principle, and volume displacement Includes cases of both floating and submerged objects

Typology: Quizzes

2021/2022

Available from 06/06/2025

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PROBLEM 1
A HOLLOW CYLINDER 1m IN DIAMETER AND 2 m HIGH WEIGHS 3825 N.
(A) HOW MANY kN OF LEAD WEIGHING 110 kN/m3 MUST BE FASTENED TO THE OUTSIDE
BOTTOM OF THE CYLINDER TO MAKE IT FLOAT WITH 1.5 m SUBMERGED IN WATER?
(B) HOW MANY kN OF LEAD IF IT IS PLACED INSIDE THE CYLINDER?
SOLUTION:
(A) FASTENED TO THE OUTSIDE BOTTOM OF THE CYLINDER
Fy= 0
BFC+BFL= WC+ WL
γfluidVd+ γfluidVd= γCVC+ γLVL
(9.81 kN/m3)(π(1 m)2(1.5 m)
4) + (9.81 kN/m3)(VL)= 3.825 + (110 kN/m3)(VL)
VL= 0.0772 m3
WL= 0.0772 m3(110 kN/m3)
WL= 8.489 kN
(B) PLACED INSIDE THE CYLINDER
Fy= 0
BFC= WC+ WL
γfluidVd= γCVC+ γLVL
(9.81 kN/m3)(π(1 m)2(1.5 m)
4) = 3.825+ WL
WL= 7.732 kN
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Download Fluids Assignment 6 – Buoyancy and more Quizzes Fluid Mechanics in PDF only on Docsity!

A HOLLOW CYLINDER 1m IN DIAMETER AND 2 m HIGH WEIGHS 3825 N.

(A) HOW MANY kN OF LEAD WEIGHING 110 kN/m

3

MUST BE FASTENED TO THE OUTSIDE

BOTTOM OF THE CYLINDER TO MAKE IT FLOAT WITH 1.5 m SUBMERGED IN WATER?

(B) HOW MANY kN OF LEAD IF IT IS PLACED INSIDE THE CYLINDER?

SOLUTION :

(A) FASTENED TO THE OUTSIDE BOTTOM OF THE CYLINDER

∑ F

y

BF

C

+ BF

L

= W

C

+ W

L

γ

fluid

V

d

  • γ

fluid

V

d

= γ

C

V

C

  • γ

L

V

L

( 9. 81 kN/m

3

π( 1 m)

2

( 1. 5 m)

4

) + ( 9. 81 kN/m

3

)(V

L

) = 3. 825 + ( 110 kN/m

3

)(V

L

V

L

= 0. 0772 m

3

W

L

= 0. 0772 m

3

( 110 kN/m

3

W

L

= 8. 489 kN

(B) PLACED INSIDE THE CYLINDER

∑ F

y

BF

C

= W

C

+ W

L

γ

fluid

V

d

= γ

C

V

C

  • γ

L

V

L

( 9. 81 kN/m

3

π( 1 m)

2

( 1. 5 m)

4

) = 3. 825 + W

L

W

L

= 7. 732 kN

A CUBE 2.2 ft ON AN EDGE HAS ITS LOWER ON AN EDGE HAS ITS LOWER HALF OF SG = 1.6 AND

UPPER HALF OF SG = 0.7. IT RESTS IN A TWO-LAYER FLUID, WITH LOWER SG = 1.4 AND UPPER

SG = 0.8. DETERMINE THE HEIGHT H OF THE TOP OF THE CUBE ABOVE THE INTERFACE.

SOLUTION :

∑ F

y

BF

1

+ BF

2

= W

A

+ W

B

BF = γ

fluid

V

d

BF

1

  1. 4 lb/ft

3

2 ft

2

  1. 2 ft − h

BF

2

  1. 4 lb/ft

3

2 ft

2

h

W = γV

W

A

= ( 62. 4 lb/ft

3

)( 1. 6 )( 2 ft)

2

( 1. 1 ft)

W

B

  1. 4 lb/ft

3

2 ft

2

  1. 1 ft

( 62. 4 lb/ft

3

)( 1. 4 )( 2 ft)

2

( 2. 2 ft − h) + ( 62. 4 lb/ft

3

)( 0. 8 )( 2 ft)

2

(h) =

  1. 4 lb/ft

3

2 ft

2

  1. 1 ft
  1. 4 lb/ft

3

2 ft

2

  1. 1 ft
  1. 2 ft − h

h

  1. 1 ft
  1. 1 ft
  1. 08 − 1 .4h + 0 .8h = 1. 76 + 0. 77

h = 0. 917 ft

FROM THE FIGURE BELOW, IT IS SHOWN THAT THE GATE IS 1.0 m WIDE AND IS HINGE AT THE

BOTTOM OF THE GATE. COMPUTE THE FOLLOWING:

(A) THE HYDROSTATIC FORCE IN kN ACTING ON THE GATE

(B) THE LOCATION OF THE CENTER OF PRESSURE OF THE GATE FROM THE HINGE,

(C) THE MINIMUM VOLUME OF CONCRETE (UNIT WEIGHT = 23.6 kN/m

3

) NEEDED TO KEEP THE

GATE CLOSED POSITION.

SOLUTION :

(A) THE HYDROSTATIC FORCE IN kN ACTING ON THE GATE

F = γhA

F =

  1. 81 kN/m

3

( 1 m)( 1 m)( 2 m)

F = 19. 62 kN

(B) THE LOCATION OF THE CENTER OF PRESSURE OF THE GATE FROM THE HINGE

e =

I

g

Ay̅

; I

g

bh

3

e =

( 1 m)( 2 m)

3

( 1 m)( 2 m)( 1 m)

e =

d = 1 − e

d = 0. 667 m

(C) THE MINIMUM VOLUME OF CONCRETE

M

A

T( 3 m) = 19. 62 kN( 0. 667 m)

T = 4. 36 kN

∑ F

y

T + BF = W

T + γ

fluid

V

d

= γ

concrete

V

o

  1. 36 kN + ( 9. 81 kN/m

3

)(V

O

) = ( 23. 6 kN/m

3

)(V

O

V

O

= 0. 316 m

3