Question

A nail weighs 0.2254 N in air and 0.1245 N when fully submerged in oil of density 800 kg/m^3. What is the density of the nail?

Answer 1

Answer on Question #58497, Physics / Mechanics | Relativity

A nail weighs 0.2254 N in air and 0.1245 N when fully submerged in oil of density 800 kg/m^3.

What is the density of the nail?

Find: $\rho_{\mathrm{nail}} - ?$

Given:

$F_{\mathrm{air}} = 0,2254 \mathrm{~N}$

$F_{\mathrm{nail}} = 0,1254 \mathrm{~N}$

$\rho_{\mathrm{oil}} = 800 \mathrm{~kg} / \mathrm{m}^{3}$

Solution:

Consider the forces which acting on a nail.

Of the figure $\Rightarrow F_{\mathrm{oil}} = F_{\mathrm{air}} - F_{\mathrm{A}}$ (1),

where $F_{\mathrm{A}}$ – Archimedes force.

Archimedes force: $F_{\mathrm{A}} = \rho_{\mathrm{oil}} V_{\mathrm{nail}} g$ (2)

Weight of nail in the air: $F_{\mathrm{nail}} = \rho_{\mathrm{nail}} V_{\mathrm{nail}} g$ (3)

Of (3) $\Rightarrow V_{\mathrm{nail}} = \frac{F_{\mathrm{nail}}}{g \rho_{\mathrm{nail}}}$ (4)

(4) in (2): $F_{\mathrm{A}} = \frac{\rho_{\mathrm{oil}}}{\rho_{\mathrm{nail}}} F_{\mathrm{nail}}$ (5)

(5) in (1): $F_{\mathrm{oil}} = F_{\mathrm{air}} - \frac{\rho_{\mathrm{oil}}}{\rho_{\mathrm{nail}}} F_{\mathrm{nail}}$ (6)

Of (6) $\Rightarrow \frac{\rho_{\mathrm{oil}}}{\rho_{\mathrm{nail}}} F_{\mathrm{nail}} = F_{\mathrm{air}} - F_{\mathrm{oil}}$ (7)

Of (7) $\Rightarrow \rho_{\mathrm{nail}} = \frac{F_{\mathrm{nail}}}{F_{\mathrm{air}} - F_{\mathrm{oil}}} \rho_{\mathrm{oil}}$ (8)

Of (8) $\Rightarrow \rho_{\mathrm{nail}} = 1787,12 \mathrm{~kg} / \mathrm{m}^{3}$.

Answer:

1787,12 kg/m³

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