An intrepid treasure-salvage group has discovered a steel box, containing gold doubloons and other valuables, resting in 80 ft of seawater. They estimate the weight of the box and treasure (in air) at 7000 lbf. Their plan is to attach the box to a sturdy balloon, inflated with air to 3 atm pressure. The empty balloon weighs 250 lbf. The box is 2 ft wide, 5 ft long, and 18 in high. What is the proper diameter of the balloon to ensure an upward lift force on the box that is 20% more than required

the proper diameter is 6.137 ft

Explanation:

first we find the volume of box using the relation, which is;

V = 2 x 5 x 1.5 = 15 ft³

we  find the buoyant force on the box by calculating the weight of water displaced.

FB = V x y

where  y is the specific weight of sea water(62.4 lbf/ft³)

so we Substitute

FB = 15 x 62.4 = 936 lbf  

now we find the upward force required by the balloon

FR = (W - FB) x 120%

= 1.2 (W - FB)

where W is the weight of the box treasure(7000 lb)

so we  Substitute,

FB = 1.2( 7000 - 936 ) = 7276.8 lbf

Because the universal gas constant contains a Rankine in its units, we make use of Rankine for our temperature

so we find the density of air at 3 atm using ideal gas relation,

Pair = p/RT

Here, p is the pressure acting (3 atm), R is the universal gas constant (1716 ft²/S²-R), and T is the temperature (520°R),

Substitute so we substitute

Pair = (3 * 2116.22) / (1516 * 520)

= 0.007114  lbf.ft³

next we find the specific weight of air;

Yair = Pair * g

g is acceleration due to gravity(32.2 ft/s²)

Yair =  

0.007114 * 32.2

= 0.23 Ibf /ft³

Now we find diameter of the balloon by balancing the net force required

FR = (y - yair) * V - Wb

= (y - yair) x (π/6)d³ - Wb  

d is the diameter of the balloon.

so we Substitute, 7276.8 lbf for FR,

62.4 Ibf/ft³ for  y, 0.23 for lbf/ft³ for Yair, 350 lb for Wb

so

7276.8 = (62.4 - 0.23)πd³  - 250

d³ = 231.23 ft³

d = 6.137 ft   

Therefore, the proper diameter is 6.137 ft

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach