ME3834 ‐ Fluid Mechanics Problem 3.78 Given: Sphere of =1 ft 3 floating as shown. Find: (a) Specific weight of sphere: . (b) New equilibrium position if 20‐pound weight is removed: d. Solution: Balance forces on sphere, 0 Basic equation: ∑
Computing equation: buoyancy
=
0 Then ∑
∑
0 + Thus 1
20
1
1
2
62.4
51.2
To find new equilibrium position, evaluate force from water on sphere. Basic equations: 1 ME3834 ‐ Fluid Mechanics Problem 3.78 Assumptions: (1) Static liquid. (2)Incompressible. , because it acts (3) Neglect
e everywhere. cos
; ;
2
Then 1
Now =
2
1
1
2
1
2
1
1
6
1
2
1
1
1
2
;
2
2
=
, cos
At 1
, ,so 1
1
3
1
2
1
3
But =
1
1
2
, so 3 4
3
2
buoyancy
Because T
0,
,
1
1
2
=
By iteration using a spreadsheet, 1
1
1
6
;
Thus at equilibrium For 1
3
and SG =
0.547 when =0.620 ft 2 .
.
0.547, 1.46 1
3
ME3834 ‐ Fluid Mechanics Problem 3.78 Thus d
1.46R
1.46
0.62
0.906 The spreadsheets results and plot are shown below. 3