   Chapter 9, Problem 86AP

Chapter
Section
Textbook Problem

A helium-filled balloon, whose envelope has a mass of 0.25 kg, is lied to a 2.0-m-long, 0.050-kg string. The. balloon is spherical with a radius of 0.40 m. When released, it lifts a length h of the string and then remains in equilibrium, as in Figure P9.86. Determine the value of h. Hint: Only that part of the string above the floor contributes to the load being supported by the balloon. Figure P9.86

To determine
The value of h .

Explanation
The weight of air displaced by the balloon equals the total weight of the balloon plus the string that lies above the ground. The weight of the string in the air is (h/L)ms . The force of buoyancy at equilibrium condition is (ρairVballoon)g=mballoong+(h/L)msg and now this expression is rearranged for the height of string. The volume of the balloon is Vballoon=4πr3/3 .

Given info: Mass of the balloon is 0.25kg , mass of the string is 0.050kg , radius of the balloon is 0.40m , density of air is 1.29kg/m3 , and density of helium is 0.179kg .

The formula for the value of h is,

h=[(ρairρhelium)(4πr3/3)mballoonms]

• ρair is density of air.
• ρhelium is density of helium

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