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Engineering and Tech / Fundamentals of Engineering Thermodynamics 7 / Chapter 1 / Problem 13P

Fundamentals of Engineering Thermodynamics | 7th Edition | ISBN: 9780470495902 | Authors: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

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Working with Force and MassAt a certain elevation, the

Chapter 1, Problem 13P

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QUESTION:

At a certain elevation, the pilot of a balloon has a mass of \(120 \mathrm{lb}\) and a weight of \(119 \mathrm{lbf}\). What is the local acceleration of gravity, in \(\mathrm{ft} / \mathrm{s}^2\), at that elevation? If the balloon drifts to another elevation where \(g=32.05 \mathrm{ft} / \mathrm{s}^2\), what is her weight, in lbf, and mass, in lb?

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QUESTION:

At a certain elevation, the pilot of a balloon has a mass of \(120 \mathrm{lb}\) and a weight of \(119 \mathrm{lbf}\). What is the local acceleration of gravity, in \(\mathrm{ft} / \mathrm{s}^2\), at that elevation? If the balloon drifts to another elevation where \(g=32.05 \mathrm{ft} / \mathrm{s}^2\), what is her weight, in lbf, and mass, in lb?

ANSWER:

Step 1 of 3

We have to find the local acceleration at a certain elevation where the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf.

The weight $(W)$ of the pilot is given by

$W=m\times{}g$

where, $m=$ mass of the pilot = 120 lb

$g=$ local acceleration of gravity in

$W=$ weight of the pilot = 119 lbf

Conversion of 119 lbf to lb $\cdot{}$

1 lbf = 32.1740 lb $\cdot{}$

So,

$119$ lbf = $119\ \times{}$ 1 lbf

= $119\ \times{}$ 32.1740 lb $\cdot{}$

119 lbf = 3828.706 lb $\cdot{}$

Solving for $g,$

Substituting 3828.706 lb $\cdot{}$ for $W$ and 120 lb for $m$

Therefore, the local acceleration at that elevation of the balloon is $31.9059$ .

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