Explanation Given Information: Let diameter of the balloon be d 0 Weight of the treasure = 7000lbf = 31.12 kN (1 lbf = 0.004448 kN) Dimensions of the box − 2 ft wide, 5 ft length, 1.5 ft height Weight of the empty balloon = 250 lbf = 1.112 kN Pressure in the balloon = 3 atm = 303.925 kN/m 2 For this problem, the volume of the balloon is equal to the volume of the fluid displaced. V s = 4 π 24 × d o 3 Area of a cross-section of balloon = A L i f t f o r c e o n t h e b a l l o o n = P × A L i f t f o r c e o n t h e b a l l o o n = ( W e i g h t o f t h e b o x + W e i g h t o f t h e e m p t y b a l l o o n - B u o y a n c y f o r c e o n t h e b o x ) × 1

8th Edition

ISBN: 9780073398273

Author: Frank M. White

Publisher: McGraw-Hill Education

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Properties of Fluids

The term fluid represents a type of substance that can easily change its shape under any external pressure/force and can flow. Fluids can acquire any shape in which it is stored and can resist normal stresses but cannot resist shear stress in rest condit…

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Chapter 2, Problem 2.118P

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To determine the diameter of the balloon for 20% extra Lift force than the required.

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Chapter 2, Problem 2.117P

Chapter 2, Problem 2.119P

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