A foam plastic (ρp=0.58 g / cm^3) is to be used as a life...

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 $V_{p}=0.044 \mathrm{~m}^{3}$.

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 $V_{p}=0.044 \mathrm{~m}^{3}$.

Key Concepts

Mechanical Equilibrium in Fluids

Mechanical equilibrium in a fluid occurs when the net forces acting on an object are zero, meaning the downward gravitational force is balanced by the upward buoyant force. Establishing this equilibrium is vital for ensuring that floating objects remain stable and for designing systems that maintain a predetermined level of submersion.

Density and Volume Relationship

Density, defined as mass per unit volume, is a key parameter when analyzing buoyancy and flotation. Differences in density between an object and the surrounding fluid determine the object's tendency to sink or float. This concept allows one to calculate the mass or volume of a substance given its density, which is crucial when balancing forces in a flotation problem.

Buoyant Force

The buoyant force is the upward force exerted by a fluid on an immersed or partially immersed object. It is directly proportional to the volume of fluid displaced and the density of the fluid, and it counters the weight of the object. Calculations involving buoyant force are essential in designing flotation devices.

Archimedes' Principle

Archimedes' Principle states that any object submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid that it displaces. This principle is fundamental in understanding why objects float or sink and is used to determine the necessary volume of displaced fluid needed to support a given weight.

Transcript

00:01 In this problem of fluids at rest we have given a foam plastic having the density is equal to 0 .58 grams per centimeter cube is to be used as a life preserver.

00:17 What volume of plastic must be used if it is to keep 20 % volume of an 80 kilograms man above water in a leak and we have given the density of the average density of man is equals to this is 1 .04 grams per centimeter cube so here we would apply the equilibrium equation so this would be fb on man that is a bant force on man plus bind force on plastic so this is plastic is equals to weight of man plus weight of plastic so here weight of man plus weight of plastic now here weight this is the bind force on man so this would be in water this would be density of the water multiplied with this would be 80 % man is submerged and 20 % is out of the water so this would be 0 .80 and volume of this man and this is multiplied with g plus force on plastic so this would be density of water and volume of this plastic multiplied with g is equal to here weight of man so weight of man is given by m multiplied with g and m is given by density of the man multiplied with volume of the man and this would be multiplied with g so here this would be density of man multiplied with volume of man multiplied with g plus this would be density of the plastic volume of the plastic multiplied with g so from here g term is cancelled in everywhere and we can write this term as density of water minus density of plastic multiplied with volume of plastic is equal to density of man minus 0 .8 times of of density of water is multiplied with volume of man and now density multiplied with volume is equal to m so this would be is equal to 80 kilograms so this is 80 kilograms and from here we can calculate vm so vm is equal to 80 divide with this would be 1 .40 kilogram per meter cube so this would be 1040 and this would be in meter cube now we can put these values so here density of water is 1000 minus density of plastic so this would be 580 kilogram per meter cube so this would be 580 and vp so here this would be again 1 040 minus this would be here simply 800 so here 800 and this would be vm is equal to 80 divided with 1040.

03:56 Now, vp, so this would be in meter cube...

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