Question

(a) What is the minimum area of a block of ice \(0.305 \mathrm{~m}\) thick floating on water that will hold up an automobile of mass \(1120 \mathrm{~kg} ?(b)\) Does it matter where the car is placed on the block of ice? The density of ice is \(917 \mathrm{~kg} / \mathrm{m}^{3}\).

Short answer

The minimum area of a block of ice that will hold up an automobile of mass \(1120 \mathrm{~kg}\) is \( 36.7 \mathrm{~m^{2}}\). The placement of the car on the ice block doesn't matter as long as the weight distribution doesn't cause the ice to crack.

Step-by-step solution

Calculating the volume of ice required

Firstly, we need to calculate the volume of ice required to hold up the automobile. We know from Archimedes' principle that the upthrust force equals the weight of fluid displaced by the submerged part of the object. Hence, we equate the weight of the car (which is its mass multiplied by gravitational acceleration, i.e. \(1120 \mathrm{~kg} \times 9.8 \mathrm{~m/s^2}\)) to the weight of the water displaced (which is the density of the water multiplied by the volume of ice submerged and the gravitational acceleration i.e. \(1000 \mathrm{~kg/m^{3}} \times V \times 9.8 \mathrm{~m/s^{2}}\)). So, \( 1120 \mathrm{~kg} \times 9.8 \mathrm{~m/s^{2}} = 1000 \mathrm{~kg/m^{3}} \times V \times 9.8 \mathrm{~m/s^{2}}\). Simplifying, we find the volume of required ice, \(V\), to be \(11.2 \mathrm{~m^{3}}.\)

Calculating the area of the ice

Knowing the volume, we can find the required area of the ice block, \(A\), using the known thickness of the ice, \(0.305 \mathrm{~m}\). Since \( V = A \times \text{thickness} \), we find that \( A = V / \text{thickness} = 11.2 \mathrm{~m^{3}} / 0.305 \mathrm{~m} = 36.7 \mathrm{~m^{2}}.\)

Discussing the car placement

Given the the ice block is uniform and the water underneath it is also uniform, the placement of the car on the ice block doesn't matter. As long as the total weight does not exceed that of the water displaced by the volume of submerged ice block, the ice block will float, regardless of where the car is situated. However, distributing the weight evenly would reduce the risk of the ice cracking.

Key concepts

Buoyancy

Buoyancy is a force that acts in the upward direction on objects immersed in a fluid. It is crucial for understanding why objects float or sink. When an object is placed in a fluid, it displaces a volume of fluid equivalent to its submerged part. According to Archimedes' Principle, the buoyant force on the object is equal to the weight of the displaced fluid.
This principle is key when determining whether an object will float or sink.

• If the buoyant force is greater than the object's weight, the object will rise and float.
• If the buoyant force is equal to the object's weight, the object will stay suspended in the fluid.
• If the buoyant force is less, the object will sink.

For the ice block holding the car, the buoyant force needs to equal the car's weight to achieve flotation.

Density

Density is a measure of mass per unit volume of a substance and is crucial in buoyancy calculations. Higher density means an object is likely to sink if placed in a fluid of lower density. Ice has a density of 917 kg/m³, which is less than water's density (1000 kg/m³). This difference in density explains why ice floats on water. In the exercise, understanding these densities helps in determining the volume of ice needed to support the car. Since the ice has a lower density, it must displace a significant amount of water to counterbalance the car's weight.

Force Balancing

Balancing forces is essential in keeping the car afloat on the ice. It involves ensuring that the downward gravitational force (weight of the car and ice block) is equal to the upward buoyant force. Archimedes' Principle helps us derive the equation:\[ \text{Weight of the car} = \text{Density of water} \times \text{Volume of ice submerged} \times \text{g} \]By solving this equation, we determine the volume of ice needed. More volume means more water displacement and thus a greater buoyant force. When the forces are perfectly balanced, the ice block will float without sinking.

Ice Block Flotation

Ice block flotation is the concept of how ice remains buoyant on water while supporting an additional load, like a car. The crucial part is ensuring the ice displaces enough water to match the weight of the load it carries. To find the minimum area of an ice block needed:

• Determine the volume required to support the weight.
• Use the block's thickness to calculate the surface area.

In the exercise, a 0.305 m thick ice block needs a total volume of 11.2 m³ to support a car weighing 1120 kg. For stability, although placement doesn't affect buoyancy, distributing weight evenly reduces the risk of cracking, especially in practical scenarios.