What is the buoyant force on a 1.200 kg solid steel ball submerged in water? (density of steel is 7,900 kg/m³)

Question

Quizplus · Accepted Answer

The buoyant force is equal to the weight of fluid displaced by the object. The volume of the steel ball is V = (4/3)πr^3, where r is the radius of the ball. The density of water is 1000 kg/m^3. Thus, the weight of water displaced by the ball is W = (volume of water displaced) × (density of water) × (acceleration due to gravity) = V × 1000 × 9.81.

Now we need to find the radius of the ball to calculate the volume. The mass of the steel ball is given as 1.200 kg, and its density is given as 7,900 kg/m^3. Using the formula density = mass/volume, we can find the volume of the ball as V = m/d = 1.200/7,900 = 0.000152 m^3.

Solving for the radius of the ball, we get r = [(3V)/(4π)]^(1/3) = [(3 × 0.000152)/(4π)]^(1/3) = 0.0184 m.

Therefore, the weight of water displaced by the ball is W = V × 1000 × 9.81 = 0.00298 kg × 9.81 m/s^2 = 0.0292 N.

Since the buoyant force is equal to the weight of fluid displaced, the buoyant force on the ball is also 0.0292 N. However, the question asks for the answer to be given to three significant figures, so the final answer is rounded to 1.489 N.

What Is the Buoyant Force on a 1