Question

Calculate the buoyant force in air on a kilogram of titanium (whose density is about 4.5 grams per cubic centimeter). compare with the weight mg of this much titanium.

Answer 1

1) The buoyant force acting on an object immersed in a fluid is:
$B=d_f V_d g$
where $d_f$ is the density of the fluid, $V_d$ is the volume of displaced fluid, and $g=9.81~m/s^2$ is the gravitational acceleration.

2) We must calculate the volume of displaced fluid. Since the titanium object is completely immersed in the fluid (air), this volume corresponds to the volume of 1 Kg of titanium, whose density is $d=4.5~g/cm^3 = 4.5\cdot10^3~Kg/m^3$. Using the relationship between density, volume and mass, we find
$V_d= \frac{m}{d}= \frac{1~Kg}{4.5\cdot10^3Kg/m^3}=2.22\cdot10^{-4}~m^3$

3) Now we can recall the formula written at step 1) and calculate the buoyant force. The air density is $d_f = 1~Kg/m^3$, so we have
$B=d_f V_d g=1~Kg/m^3 \cdot 2.22\cdot10^{-4}~m^3 \cdot 9.81~m/s^2=2.22\cdot10^{-3}~N$

4) The weight of 1 Kg of titanium is instead:
$W=mg=1~Kg \cdot 9.81~m/s^2=9.81~N$
So, the buoyant force is negligible compared to the weight.