Answer to Question #88013 in Mechanics | Relativity for gc

Question

An ebony log with volume 17.2 ft3 is submerged in water. What is the buoyant force on it (in lb)? (Enter the magnitude.)

Answer 1

The buoyant force acting on the ebony log is equal to the weight of the displaced water:

$$F_B = \rho_{water}V_{water}g.$$

Since the volume of displaced water by the ebony log, $V_{water}$, is equal to the volume of the ebony log, $V_{log}$, we can write:

$$F_B = \rho_{water}V_{log}g,$$

here, $F_B$ is the buoyant force, $\rho_{water}$ is the density of the water, $V_{log}$ is the volume of the ebony log, $g$ is the acceleration due to gravity.

Then, we get:

$$F_B = 1000 \dfrac{kg}{m^3} \cdot 17.2 ft^3 \cdot \dfrac{0.02832 m^3}{1 ft^3} \cdot 9.8 \dfrac{m}{s^2} = 4774 N \cdot \dfrac{0.22481 lbf}{1 N} = 1073 lbf.$$

Answer:

$F_B = 1073 lbf.$