91Ó°ÊÓ

A foam plastic \(\left(\rho_{p}=0.58 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is to be used as a life preserver. What volume of plastic must be used if it is to keep 20 percent (by volume) of an 80 -kg man above water in a lake? The average density of the man is \(1.04 \mathrm{~g} / \mathrm{cm}^{3}\). Keep in mind that a density of \(1 \mathrm{~g} / \mathrm{cm}^{3}\) equals \(1000 \mathrm{~kg} / \mathrm{m}^{3}\). At equilibrium $$\begin{array}{l} F_{B} \text { on man }+F B \text { on plastic }=\text { Weight of man }+\text { Weight of plastic } \\ \quad\left(\rho_{w}\right)\left(0.80 V_{m}\right) g+\rho_{w} V_{p} g=\rho_{m} V_{m} g+\rho_{p} V_{p} g \end{array}$$ or $$\left(\rho_{w}-\rho_{p}\right) V_{p}=\left(\rho_{m}-0.80 \rho_{w}\right) V_{m}$$ where subscripts \(m, w\), and \(p\) refer to man, water, and plastic, respectively. But \(\rho_{m} V_{m}=80 \mathrm{~kg}\) and so \(V_{m}=(80 / 1040) \mathrm{m}^{3}\). Substitution gives $$\left[(1000-580) \mathrm{kg} / \mathrm{m}^{3}\right] V_{p}=\left[(1040-800) \mathrm{kg} / \mathrm{m}^{3}\right]\left[(80 / 1040) \mathrm{m}^{3}\right]$$ from which \(V_{p}=0.044 \mathrm{~m}^{3}\).

Step by step solution

01

Understand the Problem

We are given the density of foam plastic and need to find the volume required to keep 20% of an 80 kg man above water. The man has an average density, and we know the density of water.

02

Apply Buoyant Force Equation

Use the provided buoyancy balance equation: \[ (\rho_{w} - \rho_{p}) V_{p} = (\rho_{m} - 0.80 \rho_{w}) V_{m} \] where \( \rho_{w} \) is the density of water, \( \rho_{p} \) is the density of plastic, \( \rho_{m} \) is the density of man, and \( V_{p} \) and \( V_{m} \) are the volumes of plastic and man, respectively.

03

Calculate Man's Volume \( V_m \)

The man's density is \( 1.04 \text{ g/cm}^3 \) or \( 1040 \text{ kg/m}^3 \). Given the man's weight is 80 kg, calculate his volume: \[ V_{m} = \frac{80 \text{ kg}}{1040 \text{ kg/m}^3} = 0.07692 \text{ m}^3 \]

04

Substitute Values in Equation

Substitute the known values into the buoyancy equation. Given \( \rho_{w} = 1000 \text{ kg/m}^3 \) and \( \rho_{p} = 580 \text{ kg/m}^3 \), substitute these along with \( V_m \) calculated in Step 3: \[ (1000 - 580) V_{p} = (1040 - 0.80 \times 1000) \times 0.07692 \]

05

Solve for \( V_p \)

Simplify the equation: \[ 420 V_p = 240 \times 0.07692 \] Now multiply and solve for \( V_p \): \[ 420 V_p = 18.46 \] \[ V_p = \frac{18.46}{420} = 0.04395 \text{ m}^3 \]

06

Conclusion

The volume of plastic required to keep 20% of an 80 kg man above water is approximately \( 0.044 \text{ m}^3 \).

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Key Concepts

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

Density Calculations

Density is a measure of how compact matter is within a particular volume. It tells us how heavy a material is relative to its size. Mathematically, density (\( \rho \)) is expressed as the mass of the object divided by its volume:\( \rho = \frac{m}{V} \). This means that if you know the mass (in kilograms) and the volume (in cubic meters) of a substance, you can calculate its density in kg/m³. In the original exercise, we calculated the density of a man and a piece of foam plastic. For the man, given that his weight is 80 kg and his density is 1040 kg/m³, we learned that his volume can be derived by rearranging the density equation to \( V = \frac{m}{\rho} \). Thus, \( V_m = \frac{80 \text{ kg}}{1040 \text{ kg/m}^3} \approx 0.07692 \text{ m}^3 \).Understanding density helps predict how substances will interact when submerged in a fluid. This interaction is fundamental in buoyancy – whether an object will sink or float depends on their density relative to the fluid.

Volume Displacement

Volume displacement is an important concept when considering how objects behave in a fluid, such as water. When an object is submerged, it pushes a certain volume of fluid out of the way or "displaces" it. The volume of fluid displaced is equivalent to the submerged portion of the object's volume. In the exercise, we're concerned with keeping 20% of a man out of the water, which implies 80% of his volume is submerged, thus displacing an equivalent volume of water. This principle is crucial for calculating the buoyant force acting on the man. The displaced water exerts an upward force – equal to the weight of the fluid displaced – which we call the buoyant force. When considering the foam plastic alongside the man, the combined displaced volume needs to support the weight of both the man and the plastic to ensure buoyancy. This requires carefully calculating the volume of foam needed to achieve the necessary displacement.

Archimedes' Principle

Archimedes' principle is a fundamental principle in fluid mechanics which states that any object submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object. This principle is crucial in the exercise because it helps us understand that the buoyant force keeping the man afloat needs to account for not only his body weight but also part of the foam plastic. The buoyancy balance equation \( (\rho_{w} - \rho_{p}) V_{p} = (\rho_{m} - 0.80 \rho_{w}) V_{m} \) connects these ideas.This equation means the buoyant force acting on the plastic and the man comes from the water displaced by both their submerged volumes. By rearranging and substituting known values into this equation, we determined the necessary volume of foam plastic to keep 20% of the man’s volume above water. This application shows Archimedes' principle at work in the real world, crucial for designing floating devices.