Problem 26 - University Physics with Modern Physics

Question

A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 45.0 -kg woman to be able to stand on it without getting her feet wet?

Step-by-Step Solution

Answer

The minimum volume of the ice slab is approximately 0.545 m³.

Step 1: Understand the Problem

We need to determine the minimum volume of an ice slab floating on freshwater that can support a woman weighing 45.0 kg, keeping her feet dry. This involves calculating the buoyant force required to counteract the weight of both the ice slab and the woman.

Step 2: Identify the Principle

To find the required volume, use the principle of buoyancy, where the weight of the displaced water equals the total weight on the slab. This principle is given by Archimedes' principle: the weight of the displaced fluid equals the weight of the object.

Step 3: Determine the Forces

Calculate the forces at play. The weight of the woman is given by her mass times the gravitational acceleration: \( F_{woman} = m_{woman} \times g = 45.0\, \text{kg} \times 9.8\, \text{m/s}^2 \).

Step 4: Weight of the Ice Slab

Let the ice slab have a density \( \rho_{ice} = 917\, \text{kg/m}^3 \) and volume \( V_{ice} \). Its weight is:\( F_{ice} = \rho_{ice} \times V_{ice} \times g \).

Step 5: Weight of the Displaced Water

The weight of the displaced water, which balances the total weight, is:\( F_{displaced\,water} = (\rho_{water} \times V_{ice}) \times g \),with \( \rho_{water} = 1000\, \text{kg/m}^3 \).

Step 6: Set Up the Equation

The total weight must equal the weight of the displaced water for the slab to just barely keep the woman dry:\[ F_{woman} + F_{ice} = F_{displaced\,water} \]

Step 7: Solve for the Volume

Substitute and solve the equation:\[ (45.0\, \text{kg} \times 9.8\, \text{m/s}^2) + (917\, \text{kg/m}^3 \times V_{ice} \times 9.8\, \text{m/s}^2) = 1000\, \text{kg/m}^3 \times V_{ice} \times 9.8\, \text{m/s}^2 \]Simplifying, we find:\[ V_{ice} = \frac{45.0}{1000 - 917} \approx 0.545 \text{m}^3 \]

Step 8: Conclusion

The minimum volume of the ice slab required to support the woman without her feet getting wet is approximately 0.545 cubic meters.

Key Concepts

Buoyancy

When it comes to floating objects, buoyancy is a key concept. Buoyancy is the upward force exerted on an object immersed in a fluid, which, in our case, is the water in the lake. This force is crucial because it determines whether an object will float or sink. The principle that explains this force is Archimedes' principle.
According to Archimedes' principle, the buoyant force on an object in a fluid is equal to the weight of the fluid displaced by that object. This means if the upward buoyant force is equal to or greater than the downward force of gravity on the object, the object will float.
In the context of our problem, the slab of ice and the woman together must displace enough water to balance their combined weights. To achieve this, the volume of water displaced must be sufficient to produce a buoyant force equal to the weight of both the ice and the woman.

Density of Ice

Density plays a significant role in determining buoyancy. Density is defined as mass per unit volume and is often represented by the symbol \( \rho \). For ice, the density \( \rho_{ice} \) is approximately 917 kg/m³. This is crucial for understanding how ice floats.
Since ice is less dense than water (which has a density of about 1000 kg/m³), it is able to float. This difference in density means that ice displaces a lesser volume of water when compared to an equal mass of liquid water.
In our problem, we need to ensure that the ice slab, with the woman standing on it, doesn't sink. We use the density of ice to calculate how much ice is needed to displace enough water to equal the weight of the ice plus the woman. This involves using the equation: \( F_{ice} = \rho_{ice} \times V_{ice} \times g \), where \( V_{ice} \) is the volume of the ice slab.

Weight of Displaced Water

The weight of displaced water is a central factor in applying Archimedes' principle. This weight is essentially what provides the upward buoyant force.
To support a given weight on the water, the amount of water displaced must have a weight equal to that of the object being supported. This means that to keep the woman dry, the ice needs to displace enough water almost equal to or greater than the weight of both the ice and the woman.
In mathematical terms, the weight of the displaced water is given by \( F_{displaced\,water} = (\rho_{water} \times V_{ice}) \times g \). Here \( \rho_{water} = 1000 \text{ kg/m}^3 \) is the density of water, and \( V_{ice} \) is the volume of ice we are calculating. Solving the balance of forces, \( F_{woman} + F_{ice} = F_{displaced\,water} \), allows us to find the minimum volume of the ice needed to achieve this equilibrium.