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Expert-verified Found in: Page 444 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # Review: The true weight of an object can be measured in a vacuum, where buoyant forces are absent. A measurement in air, however, is disturbed by buoyant forces. An object of volume V is weighed in air on an equal-arm balance with the use of counterweights of density ${\mathbit{\rho }}$ . Representing the density of air as ${{\mathbit{\rho }}}_{\mathbf{a}\mathbf{i}\mathbf{r}}$ and the balance reading as ${{\mathbit{F}}}_{{\mathbf{g}}}{\mathbf{\text{'}}}$ , show that the true weight ${{\mathbit{F}}}_{{\mathbf{g}}}$ is${{\mathbit{F}}}_{{\mathbf{g}}}{\mathbf{=}}{{\mathbit{F}}}_{{\mathbf{g}}}{\mathbf{\text{'}}}{\mathbf{+}}\left(v-\frac{{F}_{g}\text{'}}{\rho g}\right){\mathbit{\rho }}{}_{\mathbf{a}\mathbf{i}\mathbf{r}}{\mathbit{g}}$

Hence it is proved that:

${{F}}_{{\mathrm{g}}}{=}{{F}}_{{\mathrm{g}}}{\text{'}}{+}\left(\mathrm{v}-\frac{{\mathrm{F}}_{\mathrm{g}}\text{'}}{\mathrm{\rho g}}\right){\rho }{}_{{\mathrm{air}}}{g}$

See the step by step solution

## Identification of the given data

The given data can be listed below as,

⦁ The volume of object is, V.

⦁ The density of object is, $\rho$ .

⦁ The density of air is, ${\rho }_{air}$ .

⦁ The balance reading weight is, $F{\text{'}}_{g}$ .

⦁ The true weight is, ${F}_{g}$ .

## Significance of the buoyant force

When an object is partially or fully submerged in a fluid, the fluid exerts on the object an upward force called the buoyant force. According to Archimedes’s principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object:

${\mathbit{B}}{\mathbf{=}}{{\mathbit{\rho }}}_{\mathbf{f}\mathbf{l}\mathbf{u}\mathbf{i}\mathbf{d}}{\mathbit{g}}{{\mathbit{V}}}_{\mathbf{f}\mathbf{l}\mathbf{u}\mathbf{i}\mathbf{d}}$

## Determination of the true weight  Fg

When both weight and the body are in balance, the force relation by using a above concept of step (2) is given by:

${F}_{g}-B={F}_{g}^{\text{'}}-B\text{'}\phantom{\rule{0ex}{0ex}}{F}_{g}-{\rho }_{air}gv={F}_{g}^{\text{'}}-{\rho }_{air}gv\text{'}$

The value $v\text{'}$ is expressed as,

$v\text{'}=\frac{{m}^{\text{'}}}{\rho }$

And

${m}^{\text{'}}=\frac{{F}_{g}^{\text{'}}}{g}$

Substitute all the value in the above equation,

${F}_{g}-{\rho }_{air}gv={F}_{g}^{\text{'}}-{\rho }_{air}g\frac{m\text{'}}{\rho }\phantom{\rule{0ex}{0ex}}{F}_{g}-{\rho }_{air}gv={F}_{g}^{\text{'}}-{\rho }_{air}g\frac{{F}_{g}\text{'}}{\rho g}\phantom{\rule{0ex}{0ex}}{F}_{g}={F}_{g}^{\text{'}}-{\rho }_{air}g\frac{{F}_{g}^{\text{'}}}{\rho g}+{\rho }_{air}gv\phantom{\rule{0ex}{0ex}}{F}_{g}={F}_{g}^{\text{'}}+{\rho }_{air}g\left(-\frac{{F}_{g}\text{'}}{\rho g}+v\right)\phantom{\rule{0ex}{0ex}}{F}_{g}={F}_{g}^{\text{'}}+{\rho }_{air}g\left(v-\frac{{F}_{g}\text{'}}{\rho g}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence the true weight ${F}_{g}$ is, ${F}_{g}={F}_{g}^{\text{'}}+{\rho }_{air}g\left(v-\frac{{F}_{g}\text{'}}{\rho g}\right)$ ### Want to see more solutions like these? 