/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 A slab of ice floats on a freshw... [FREE SOLUTION] | 91Ó°ÊÓ

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A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 65.0-kg woman to be able to stand on it without getting her feet wet?

Short Answer

Expert verified
The minimum volume of the ice slab is approximately 0.783 m³.

Step by step solution

01

Understand the Problem

To find the minimum volume of the ice slab, we need to ensure that the weight of the woman is supported by the ice slab without it sinking. Therefore, we need to make sure that the buoyant force (equal to the weight of the displaced water) is equal to or greater than the weight of the woman plus the weight of the ice itself.
02

Formula for Buoyant Force

The buoyant force acting on an object is given by Archimedes' principle, which is the weight of the displaced fluid. The formula is \( F_b = \rho_{water} \times V_{displaced} \times g \), where \( \rho_{water} \) is the density of water, \( V_{displaced} \) is the volume of the displaced water, and \( g \) is the acceleration due to gravity.
03

Equate Forces

For the ice slab to support the woman without sinking, the total weight of the ice slab and the woman must equal the buoyant force. Thus, \( M_{woman} \times g + M_{ice} \times g = \rho_{water} \times V_{ice} \times g \). Here, \( M_{woman} \) is the mass of the woman, \( M_{ice} \) is the mass of the ice, and \( V_{ice} \) is the volume of the ice.
04

Density and Mass Relations

Use the density of ice to relate it to volume: \( M_{ice} = \rho_{ice} \times V_{ice} \). We have the values: \( \rho_{ice} = 917 \, \text{kg/m}^3 \) and \( \rho_{water} = 1000 \, \text{kg/m}^3 \). For the woman, \( M_{woman} = 65 \, \text{kg} \).
05

Solve for Volume of Ice

Substitute the density relation into the force equation: \[ 65 \times g + 917 \times V_{ice} \times g = 1000 \times V_{ice} \times g \] Cancel \( g \) from both sides and solve for \( V_{ice} \): \[ 65 + 917 \times V_{ice} = 1000 \times V_{ice} \] \[ 65 = (1000 - 917)V_{ice} \] \[ V_{ice} = \frac{65}{83} \approx 0.783 \text{ m}^3 \]
06

Conclusion

The minimum volume of the ice slab required to keep the woman from getting her feet wet is approximately 0.783 cubic meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Archimedes' Principle
Archimedes' Principle is a fundamental concept in fluid mechanics. It states that any object wholly or partially submerged in a fluid experiences a buoyant force equal to the weight of the fluid displaced by the object. This principle helps us understand why objects float or sink in fluids, such as water.
To see this principle in action, consider how a slab of ice floats on water. The ice displaces a volume of water, and the water pushes back with a buoyant force. This buoyant force keeps the ice afloat as long as it is equal to or greater than the total weight of the ice and any additional load, like a person standing on it.
Density of Water
The density of water is a key factor in calculating buoyancy. It is the mass of the water per unit volume, usually expressed in kilograms per cubic meter (kg/m³). The standard density of water is 1000 kg/m³ at 4°C, which serves as a useful benchmark for solving buoyancy problems.
This means for every cubic meter of water displaced by an object, the buoyant force is enough to support 1000 kilograms. In problems involving floating ice, the density of water helps in determining how much water needs to be displaced to balance the weight of the ice and any additional weight on it, like a person.
Density of Ice
The density of ice is lower than that of water, which is why ice floats. Ice has a density of about 917 kg/m³. This difference allows icebergs and ice slabs to stay afloat in water.
In our exercise, knowing the density of ice allows us to calculate the mass of the ice slab for a particular volume. This calculation is crucial because it helps us ensure the ice’s own weight and an additional weight, like a person, can be supported without sinking.
Buoyant Force
The buoyant force is an upward force exerted by a fluid that opposes the weight of an object immersed in that fluid. It is directly related to Archimedes' principle.
The formula for buoyant force is \( F_b = \rho_{water} \times V_{displaced} \times g \) where:
  • \( F_b \) is the buoyant force.
  • \( \rho_{water} \) is the density of the fluid, usually water (1000 kg/m³).
  • \( V_{displaced} \) is the volume of fluid displaced by the object.
By calculating the buoyant force, we ensure that the object, such as the ice slab, remains afloat when supporting additional weight, like a person standing on it.
Volume Displacement
Volume displacement is the term used to describe the volume of fluid that is moved out of the way when an object is submerged in it.
The amount of water displaced by the ice slab must be equal to the total weight of the ice slab plus any additional weight, such as a person on top. This ensures that the buoyant force pushing upward is sufficient to keep the object afloat.
By understanding volume displacement, we can determine the necessary volume the ice slab must have to prevent sinking. In our exercise, we solved for this minimum volume to ensure the ice slab supports both its own weight and the weight of a person without submerging.

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Most popular questions from this chapter

A piece of wood is 0.600 m long, 0.250 m wide, and 0.080 m thick. Its density is 700 kg/m\(^3\). What volume of lead must be fastened underneath it to sink the wood in calm water so that its top is just even with the water level? What is the mass of this volume of lead?

A single ice cube with mass 16.4 g floats in a glass completely full of 420 cm\(^3\) of water. Ignore the water's surface tension and its variation in density with temperature (as long as it remains a liquid). (a) What volume of water does the ice cube displace? (b) When the ice cube has completely melted, has any water overflowed? If so, how much? If not, explain why this is so. (c) Suppose the water in the glass had been very salty water of density 1050 kg/m\(^3\). What volume of salt water would the 9.70-g ice cube displace? (d) Redo part (b) for the freshwater ice cube in the salty water.

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355-L cans per minute. At point 2 in the pipe, the gauge pressure is 152 kPa and the cross-sectional area is 8.00 cm\(^2\). At point 1, 1.35 m above point 2, the cross-sectional area is 2.00 cm\(^2\). Find the (a) mass flow rate; (b) volume flow rate; (c) flow speeds at points 1 and 2; (d) gauge pressure at point 1.

A sealed tank containing seawater to a height of 11.0 m also contains air above the water at a gauge pressure of 3.00 atm. Water flows out from the bottom through a small hole. How fast is this water moving?

On the afternoon of January 15, 1919, an unusually warm day in Boston, a 17.7-mhigh, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-mdeep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of 1600 kg/m\(^3\). If the tank was full before the accident, what was the total outward force the molasses exerted on its sides? (\(Hint:\) Consider the outward force on a circular ring of the tank wall of width \({dy}\) and at a depth \(y\) below the surface. Integrate to find the total outward force. Assume that before the tank ruptured, the pressure at the surface of the molasses was equal to the air pressure outside the tank.)

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