Archimedes' Principle — Question 6

Question 6

The scale reading depends on where the support is attached

A beaker of water stands on a scale. A rigid object of mass $0.60\,\mathrm{kg}$ and volume $1.0\times10^{-3}\,\mathrm{m^3}$ is held fully submerged. Water density is $1000\,\mathrm{kg/m^3}$ and $g=10\,\mathrm{m/s^2}$. First it is tethered by a light string to the beaker bottom; then the string is replaced by an external downward support. There is no overflow or other contact.

1. Find the downward holding force on the object and the scale increase in the bottom-tethered arrangement relative to the original beaker and water.

2. Find the scale increase with the external support. Explain why the two scale readings differ although the submerged object has the same buoyancy.

Question 6 - Solution

The scale reading depends on where the support is attached

(a) Derivation and calculation

$$B=\rho_wgV=(1000)(10)(.00100)=10.0\,\mathrm N,\quad W=mg=6.00\,\mathrm N.$$$$B-W-T=0\Rightarrow\boxed{T=4.00\,\mathrm N\ \text{downward}}.$$For the bottom-tethered arrangement,$$\boxed{\Delta N=B-T=10.0-4.00=6.00\,\mathrm N}.$$

(b) Result and interpretation

With an external downward support, its upward reaction bypasses the scale:$$\boxed{\Delta N_{\mathrm{external}}=B=10.0\,\mathrm N}.$$$$\Delta N_{\mathrm{external}}-\Delta N_{\mathrm{tethered}}=4.00\,\mathrm N=T.$$The internal tether subtracts an upward force on the beaker; the external support does not.

Physical check

The beaker-plus-water-plus-object system in the tethered case gains exactly the object weight.