Answer:
[tex]f_{net} = 0.1853 lbf[/tex] {upward}
Explanation:
Byoncy force is given as
[tex]fb = \gamma_{water} v_{ball}[/tex]
[tex]fb = \gamma_{water} \frac{4}{3} \pi R^3[/tex]
At 60 degree F [tex]\gamma_{water} = 62.37 lb/ft^3[/tex]
R is 1.5 inch = 1.5/12 ft
so solving for Fs we get
fb = 0.5102 lb
By using ideal gas equation calculate mass of air in sphere
[tex]m = \frac{pV}{R_{air}T}[/tex]
p = 200 atm = 2939.19 psi
[tex]R_{air} = 0.3704 psi ft^3 /lbm -R[/tex]
[tex]m = \frac{2939.19 psi a\times [\frac{4}{3} \pi [\frac{1.5}{12}]^2]}{0.3704 \times (90+459.67}[/tex]
m = 0.1249 lbm
total weight of air and ball os
wa +wb = g(ma+mb)
[tex]= 32.174 (0.1249+0.2)\times \frac{1 lbf}{32.174\ lbm ft/s^2}[/tex]
wa+wb = 0.3249 lbf
net foce on the ball is
[tex]f_{net} = fb - (wa+wb)[/tex]
[tex]f_{net} = 0.5102 - 0.249[/tex]
[tex]f_{net} = 0.1853 lbf[/tex]
