Archimedes' Principle — Question 1

Question 1

Two apparent weights reveal both volume and density

A rigid object weighs $12.0\,\mathrm{N}$ in air, with air buoyancy neglected. Fully immersed and suspended at rest in water of density $1000\,\mathrm{kg/m^3}$, its support tension is $7.0\,\mathrm{N}$. Use $g=10\,\mathrm{m/s^2}$ and neglect surface tension.

1. Infer the object volume and mean density.

2. Find its support tension if fully immersed in oil of density $800\,\mathrm{kg/m^3}$. Explain why its true weight has not changed.

Question 1 - Solution

Two apparent weights reveal both volume and density

(a) Derivation and calculation

For the immersed object, $T+B-W=0$:$$\boxed{B=W-T=12.0-7.0=5.0\,\mathrm N}.$$$$\boxed{V=B/(\rho_wg)=5.0/(1000\cdot10)=5.00\times10^{-4}\,\mathrm{m^3}}.$$$$m=W/g=1.20\,\mathrm{kg},\quad\boxed{\rho_b=m/V=2400\,\mathrm{kg/m^3}}.$$

(b) Result and interpretation

$$B_o=\rho_ogV=(800)(10)(5.00\times10^{-4})=4.00\,\mathrm N.$$$$\boxed{T_o=W-B_o=12.0-4.00=8.00\,\mathrm N}.$$The true weight $mg$ remains 12.0 N; only the fraction supported by the fluid changes.

Physical check

The less dense oil produces smaller buoyancy and a larger support tension.