Question

A 3.0 kg object with a volume of 750 cm3 is hanging from a scale. The object is then lowered so it is submerged completely in an unknown fluid. The scale now reads 20.0 N.
A) What is the magnitude of the buoyant force acting on the object?
B) What is the density of the fluid?

Answer 1

A) Let’s assume the upwards to be positive direction and the apparent weight of the object in fluid directed downward, the force of gravity that acts on the object (or weight) directed downward and the buoyant force directed upward:

$$-W_{App} = F_{B} - W,$$ $$F_{B} = W - W_{App},$$

here, $W_{App} = 20 N$ is the apparent weight of the object in fluid, $W = mg$ is the weight of the object in air, $m = 3.0 kg$ is the mass of the object, $g$ is the acceleration due to gravity and $F_{B}$ is the buoyant force.

Then, we get:

$$F_{B} = mg - W_{App} = 3.0 kg \cdot 10 \dfrac{m}{s^2} - 20 N = 10 N.$$

B) The buoyant force acting on the object is equal to the weight of the displaced fluid:

$$F_{B} = \rho_{f}V_{f}g.$$

Since the volume of displaced fluid by the object, $V_{f}$, is equal to the volume of the object, $V_{obj}$, we can write:

$$F_{B} = \rho_{f}V_{obj}g.$$

From this formula we can calculate the density of the fluid:

$$p_{f} = \dfrac{F_{B}}{V_{obj}g} = \dfrac{10 N}{750 cm^3 \cdot \dfrac{1 m^3}{10^6 cm^3} \cdot 10 \dfrac{m}{s^2}} = 1333 \dfrac{kg}{m^3}.$$

Answer:

A) $F_{B} = 10 N.$

B) $p_{f} = 1333 \dfrac{kg}{m^3}.$