Chapter 5, Problem 51E

To determine

The number of $65\text{kg}$ people that the raft can support.

Answer to Problem 51E

The raft can support $24$ people with mass $65\text{kg}$ each.

Explanation of Solution

Given info: The length, width and the thickness of the raft is $3\text{m}$, $2\text{m}$ and $30\text{cm}$ respectively and the weight of each person is $65\text{kg}$ the density of water is $1.00\times {10}^3{\text{kg/m}}^{\text{3}}$ and the density of the balsa wood is $1.3\times {10}^2{\text{kg/m}}^{\text{3}}$.

Write the expression for the volume of a raft.

$V=LWH$

Here,

$V$ is the volume of the raft

$L$ is the length of the raft

$W$ is the width of the raft

$H$ is the thickness of the raft

Substitute $3\text{m}$ for $L$, $2\text{m}$ for $W$ and $30\text{cm}$ for $H$ to find the volume of the raft.

$\begin{array}{c}V=\left(3\text{m}\right)\left(2\text{m}\right)\left(30\text{cm}\right)\\ =\left(3\text{m}\right)\left(2\text{m}\right)\left(30\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =1.8{\text{m}}^{\text{3}}\end{array}$

Write the expression for the maximum upward force the raft can exert.

$F_{\text{up}}=F_b-F_G$

Here,

$F_{\text{up}}$ is the maximum upward force the raft can exert

$F_b$ is the buoyant force

$F_G$ is the gravitational force on the raft (weight of the raft)

The expression (1) can be elaborated as,

$\begin{array}{c}{F}_{\text{up}}=d_{\text{water}}gV-d_{\text{balsa}}gV\\ =\left(d_{\text{water}}-d_{\text{balsa}}\right)gV\end{array}$

Here,

$d_{\text{water}}$ is the density of water

$d_{\text{balsa}}$ is the density of balsa wood from which the raft is made

$g$ is the acceleration due to gravity

$V$ is the volume of the raft

Substitute $1.00\times {10}^3{\text{kg/m}}^{\text{3}}$ for $d_{\text{water}}$, $1.3\times {10}^2{\text{kg/m}}^{\text{3}}$ for $d_{\text{balsa}}$, $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $1.8{\text{m}}^{\text{3}}$ for $V$ to find $F_{\text{up}}$.

$\begin{array}{c}{F}_{\text{up}}=\left(1.00\times {10}^3{\text{kg/m}}^{\text{3}}-1.3\times {10}^2{\text{kg/m}}^{\text{3}}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(1.8{\text{m}}^{\text{3}}\right)\\ =1.53\times {10}^4\text{N}\end{array}$

Write the expression for the gravitational force on each person.

${F^{\prime }}_G=mg$

Here,

${F^{\prime }}_G$ is the gravitational force on each person (weight of each person)

$m$ is the mass of a person

Substitute $65\text{kg}$ for $m$ and $9.8{\text{m/s}}^{\text{2}}$ for $g$ to find ${F^{\prime }}_G$.

$\begin{array}{c}{F^{\prime }}_G=\left(65\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\\ =637\text{N}\end{array}$

Write the expression for the number of persons ($N$) that the raft can support.

$N=\frac{F_{\text{up}}}{{F^{\prime }}_G}$

Substitute $1.53\times {10}^4\text{N}$ for $F_{\text{up}}$ and $637\text{N}$ for ${F^{\prime }}_G$ to find $N$.

$\begin{array}{c}N=\frac{1.53\times {10}^4\text{N}}{637\text{N}}\\ =24\end{array}$

Conclusion:

Therefore, the raft can support $24$ people with mass $65\text{kg}$ each.