Answer to Question #236473 in Mechanics | Relativity for Favice

Question

An object of volume 4x10^4m^3 and density 6000kg/m^3 is immersed inside a fluid of density 5000kg/m^3. The force exerted by the fluid on the object is 0.7N. Calculate the density of the fluid

Answer 1

According to this principle, the buoyant force on an object equals the weight of the fluid it displaces. In equation form, Archimedes’ principle is

F_B = w_fl - m_objectg

where F_B is the buoyant force and w_fl is the weight of the fluid displaced by the object.

$F_B =0.7\,N \\F_B=w_{fl}-m_{object}\cdot g=\rho_{fl}\cdot g\cdot V_{object}-V_{object}\cdot \rho_{object} \cdot g \\ \text{ } \\ \implies \rho_{fl}=\dfrac{F_B+V_{object}\cdot \rho_{object} \cdot g}{g\cdot V_{object}}=\dfrac{F_B}{g\cdot V_{object}}+\rho_{object} \\ \text{ } \\ \implies \rho_{fl}=\dfrac{0.7\,N}{(9.81\,{m/s^2})(4\times10^{-4}\,m^3)}+6000\, \dfrac{kg}{m^3} \\ \text{ } \\ \implies \rho_{fl}=6178.39\dfrac{kg}{m^3}$

In conclusion, the density of the fluid is about 6178.39 kg/m³.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.