Question

a plastic ball with a density of 500 kg/m³ is partially submerged in a pool of water. if the volume of the ball is 0.5 m³, what is the buoyant force acting on it? ? n ρ_water = 1,000 kg/m³

Explanation:

Step1: Recall Archimedes' Principle

The buoyant force \( F_b \) is equal to the weight of the displaced fluid, given by \( F_b=
ho_{fluid}V_{displaced}g \). Since the ball is partially submerged and in equilibrium (floating), the weight of the ball equals the buoyant force. First, find the mass of the ball: \( m =
ho_{ball}V_{ball} \), where \(
ho_{ball}=500\ kg/m^3 \), \( V_{ball}=0.5\ m^3 \). So \( m = 500\times0.5 = 250\ kg \). Then weight of the ball \( W = mg \), \( g = 9.8\ m/s^2 \), so \( W = 250\times9.8 = 2450\ N \). Alternatively, using Archimedes' Principle directly: for a floating object, \( F_b = W_{object} \). But also, \( F_b=
ho_{water}V_{displaced}g \). Since the ball floats, \(
ho_{ball}V_{ball}g=
ho_{water}V_{displaced}g \), so \( V_{displaced}=\frac{
ho_{ball}}{
ho_{water}}V_{ball}=\frac{500}{1000}\times0.5 = 0.25\ m^3 \). Then \( F_b=
ho_{water}V_{displaced}g = 1000\times0.25\times9.8 = 2450\ N \).

Step2: Calculate Buoyant Force

Using Archimedes' Principle, \( F_b =
ho_{water}V_{displaced}g \). Since the ball floats, \( V_{displaced} \) is the volume of water displaced, which can also be found by equating the weight of the ball to the buoyant force. Weight of the ball: \( m =
ho_{ball}V_{ball}=500\ kg/m^3\times0.5\ m^3 = 250\ kg \). Weight \( W = mg = 250\ kg\times9.8\ m/s^2 = 2450\ N \). So the buoyant force equals the weight of the ball (since it's floating), so \( F_b = 2450\ N \).

Answer:

2450