Problem 35 - College Physics

Question

A hollow plastic sphere is held below the surface of a fresh- water lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.650 \(\mathrm{m}^{3}\) and the tension in the cord is 900 \(\mathrm{N} .\) (a) Calculate the buoyant force exerted by the water on the sphere. (b) What is the mass of the sphere? (c) The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

Step-by-Step Solution

Answer

(a) 6376.5 N; (b) 558.3 kg; (c) 0.859 (fraction submerged)

Step 1: Calculate the Buoyant Force

The buoyant force exerted on the sphere can be determined using Archimedes' principle, which states that the buoyant force is equal to the weight of the displaced water. The volume of the sphere is given as \( V = 0.650 \, \text{m}^3 \), and the density of water is \( \rho = 1000 \, \text{kg/m}^3 \). The gravitational acceleration is \( g = 9.81 \, \text{m/s}^2 \). The buoyant force \( F_b \) is calculated by the formula: \[ F_b = \rho \cdot V \cdot g = 1000 \, \text{kg/m}^3 \cdot 0.650 \, \text{m}^3 \cdot 9.81 \, \text{m/s}^2 = 6376.5 \, \text{N}. \]

Step 2: Calculate the Mass of the Sphere

To find the mass of the sphere, we use the fact that the net force acting on the sphere is zero while it is held by the cord. The tension in the cord is \( T = 900 \, \text{N} \), and this tension balances the difference between the buoyant force and the weight of the sphere. Let the mass of the sphere be \( m \). The weight of the sphere is \( W = m \cdot g \). The equation for equilibrium is: \[ F_b = W + T \] \[ 6376.5 = m \cdot 9.81 + 900 \] Solving for \( m \), we have: \[ m \cdot 9.81 = 6376.5 - 900 \] \[ m \cdot 9.81 = 5476.5 \] \[ m = \frac{5476.5}{9.81} \approx 558.3 \, \text{kg}. \]

Step 3: Determine the Submerged Volume Fraction

When the sphere comes to rest at the surface, the buoyant force equals the weight of the sphere. Since the sphere is less dense than water, it will partially float. Let \( V' \) be the submerged volume and \( \rho_{sphere} \) be the sphere's density. The buoyant force is now equal to the sphere's weight (\( W = m \cdot g \)) and is also equal to the weight of the displaced water (\( \rho \cdot V' \cdot g \)). Therefore, \[ m = \rho \cdot V'. \] We know \( m = 558.3 \, \text{kg} \) and \( \rho = 1000 \, \text{kg/m}^3 \), so \( V' = \frac{m}{\rho} = \frac{558.3}{1000} = 0.5583 \, \text{m}^3 \). The fraction of the volume that is submerged is: \[ \frac{V'}{V} = \frac{0.5583}{0.650} \approx 0.859. \]

Key Concepts

Archimedes' Principle

Archimedes' principle is a fundamental concept in fluid mechanics that describes how objects behave when submerged in a fluid. According to this principle, any object, whether fully or partially immersed in a fluid, experiences an upward buoyant force exerted by the fluid. This force is equal to the weight of the fluid that the object displaces.

In practical terms, this means:

• When you submerge a solid object in water, it pushes water out of the way, and the water tries to push back with a force equal to the weight of the water displaced by the object.
• The more water it displaces, the greater the buoyant force.

In our exercise, since the sphere is hollow and immersed in water, it is subjected to this buoyant force, which we calculated using Archimedes' principle. Understanding this principle is crucial for determining whether an object will float or sink.

Density

Density is a measure of how much mass an object has in a given volume. It is commonly expressed in kilograms per cubic meter (kg/m³). In simpler terms, it tells us how "heavy" or "light" a material is for its size.

The formula for density is given by: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]Water, for example, has a density of approximately 1000 kg/m³.

• When an object is denser than the fluid it is in, such as a rock in water, it will sink.
• If it is less dense, like a hollow plastic sphere, it will float.

In the context of the exercise, understanding the sphere's density compared to the water's density helped us determine how much of the sphere would be submerged when it reached equilibrium on the water's surface.

Buoyant Force

The buoyant force is the upward force exerted by a fluid that opposes the weight of an object immersed in the fluid. This force plays a crucial role in determining whether an object will float or sink.

According to Archimedes' principle, the magnitude of the buoyant force is equal to the weight of the fluid that the object displaces. The formula to calculate the buoyant force is:\[ F_b = \rho \cdot V \cdot g \]where:

• \( \rho \) is the fluid density
• \( V \) is the volume of fluid displaced
• \( g \) is the acceleration due to gravity

In our exercise, the sphere experienced a buoyant force of 6376.5 N while submerged, counteracting the gravitational force trying to pull it downward.

Equilibrium

Equilibrium in physics refers to a state where opposing forces are balanced and there is no net change in the motion of an object. This principle is essential in understanding how objects behave when immersed in fluids.

In the exercise, the equilibrium of the sphere is achieved when the following happens:

• The buoyant force from the displaced water equals the weight of the sphere.
• The tension in the cord equals the difference between the buoyant force and the weight of the sphere while it is submerged.

Once the cord breaks, the sphere rises to the surface, and equilibrium is again established when the buoyant force perfectly balances its weight. This equilibrium position allows us to calculate what fraction of the sphere remains submerged, giving insight into its overall density relative to water.