Question

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 r. representing the density of air as ${\mathit{\rho }}_{\mathbf{a}\mathbf{i}\mathbf{r}}$and the balance reading as ${\mathit{F}}_{\mathbf{g}}\mathbf{\text{'}}$, show that the true weight ${\mathit{F}}_{\mathbf{g}}$is

${\mathit{F}}_{\mathbf{g}}\mathbf{=}{\mathit{F}}_{\mathbf{g}}\mathbf{\text{'}}\mathbf{+}\left(v-\frac{F_g\text{'}}{\rho g}\right)\mathit{\rho }{}_{\mathbf{a}\mathbf{i}\mathbf{r}}g$

Short answer

Hence it is proved that:

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

Step-by-step solution

Step 1:

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:

$B={\rho }_{fluid}g{V}_{fluid}$

Step 2:

When both weight and the body are in balance, the force relation by using a above concept of step (1) is given by:

$$\begin{gathered} F_y-B=F_y\text{'}-B\text{'} \\ {F}_y-{\rho }_{air}gv=F_y\text{'}-{\rho }_{air}gv\text{'} \end{gathered}$$

$$\begin{gathered} F_y-{\rho }_{air}gv=F_y\text{'}-{\rho }_{air}g\frac{m\text{'}}{\rho } \\ {F}_y-{\rho }_{air}gv=F_y\text{'}-{\rho }_{air}g\frac{F_g\text{'}}{\rho g} \\ {F}_y=F_y\text{'}-{\rho }_{air}g\frac{F_g\text{'}}{\rho g}+{\rho }_{air}gv \\ {F}_y=F_y\text{'}+{\rho }_{air}g\left(-\frac{F_g\text{'}}{\rho g}+v\right) \\ {F}_y=F_y\text{'}+{\rho }_{air}g\left(v-\frac{F_g\text{'}}{\rho g}\right) \end{gathered}$$