Chapter 5, Problem 27P

(a)

To determine

The units and dimensions of the constant $b$ in the retarding force $b{\upsilon }^n$ if $\left(a\right)n=1$ .

Answer to Problem 27P

Dimension of $b=M{T}^{-1}$

The unit of $b\text{ is kg}{\text{s}}^{\text{-1}}$

Explanation of Solution

Given Information:

  $n=1$

Formula Used:

The drag force equation is $F_d=b{v}^n$

Calculation:

The drag force equation is $F_d=b{v}^n$

Dimension of force $=\left[ML{T}^{-}{}^2\right]$

The dimension of speed $=\left[L{T}^{-1}\right]$

Solving the drag force equation for $b$ with $n=1$

  $b=\frac{F_d}{v}$

Substitute the dimensions of $F_d$ and $v$ and simplify to obtain the dimension of $b$ as

  $\begin{array}{l}\left[b\right]=\frac{\left[ML{T}^{-2}\right]}{\left[L{T}^{-1}\right]}\\ \left[b\right]=M{T}^{-1}\end{array}$

So, the units ofb is ${\text{kg s}}^{\text{-1}}$ .

(b)

To determine

Using dimensional analysis, determine the units and dimensions of the constant $b$ in the retarding force $b{\upsilon }^n$ if $n=2$ .

Answer to Problem 27P

Dimension of $b=M{L}^{-1}$

The units of $b\text{ is kg}{\text{.m}}^{\text{-1}}$

Explanation of Solution

Given Information:

  $n=2$

Calculation:

The drag force equation is $F_d=b{v}^n$

Dimension of force $=\left[ML{T}^{-2}\right]$

The dimension of speed $=\left[L{T}^{-1}\right]$

Solving the drag force equation for $b$ with $n=2$ ,

  $b=\frac{F_d}{v^2}$

Substituting the dimensions of $F_d$ and $v$ and simplify to obtain the dimension of $b$ as

  $\begin{array}{l}\left[b\right]=\frac{\left[ML{T}^{-2}\right]}{{\left[L{T}^{-1}\right]}^2}\\ =M{L}^{-1}\end{array}$

So, the units of $b\text{ are }kg{\text{ m}}^{-1}$ .

(c)

To determine

To Show: Air resistance of a falling object with a circular cross section should be approximately $\frac{1}{2}\rho \pi r^2{\upsilon }^2,\text{ where }\rho =1.20{\text{kg/m}}^{\text{3}}$ , the density of air, is dimensionally consistent.

Explanation of Solution

Given Information:

Air resistance of a falling object with a circular cross section should be approximately $\frac{1}{2}\rho \pi r^2{\upsilon }^2,\text{ where }\rho =1.20{\text{kg/m}}^{\text{3}}$ , the density of air.

Calculation:

According to the expression of drag force,

  $F_d=\frac{1}{2}\rho \pi r^2{\upsilon }^2$

So the dimensions of drag force is

  $\begin{array}{l}\left[F_d\right]=\left[\frac{1}{2}\rho \pi r^2{\upsilon }^2\right]\\ or=\left[M{L}^{-3}\right]{\left[L\right]}^2{\left[L{T}^{-1}\right]}^2\\ or=ML{T}^{-2}\end{array}$

The drag force is given as

  $F_d=b{v}^2$

The dimension of drag force is

  $\begin{array}{c}\left[F_d\right]=\left[b{v}^2\right]\\ =\left[M{L}^{-1}\right]{\left[L{T}^{-1}\right]}^2\\ =ML{T}^{-2}\end{array}$

Conclusion:

Hence, the expression for drag force is dimensionallyconsistent .

(d)

To determine

To Find: The terminal speed of the skydiver.

Answer to Problem 27P

  $56.9\text{m/s}$

Explanation of Solution

Given Information:

Mass of skydiver = $56\text{kg}$

Disk of radius = $0.30\text{m}$

Density of air near the surface of Earth is $120{\text{kg/m}}^{\text{3}}$

Formula Used:

The terminal speed is given by relation

  $v_t=\sqrt{\frac{2mg}{\rho \pi r^2}}$

Calculation:

Mass $m=56\text{kg}$

Acceleration due to gravity $g=9.81{\text{m/s}}^{\text{2}}$

Density $\rho =1.2{\text{kg/m}}^{\text{3}}$

Radius $r=0.3\text{m}$

The terminal speed is given by relation

  $\begin{array}{c}{v}_t=\sqrt{\frac{2mg}{\rho \pi r^2}}\\ =\sqrt{\frac{\text{2}\left(\text{56}\text{kg}\right){\left(\text{9}\text{.81m/s}\right)}^{\text{2}}}{\text{π}\left(\text{1}\text{.2}{\text{kg/m}}^{\text{3}}\right){\left(\text{0}\text{.3m}\right)}^{\text{2}}}}\\ =56.9\text{m/s}\end{array}$

(e)

To determine

To Calculate: The terminal velocity at the given height.

Answer to Problem 27P

The terminal velocity is $86.\text{9}\text{m/s}$

Explanation of Solution

Given Information:

Height = $8\text{km}$

Density of air near the surface of Earth is $0.514{\text{kg/m}}^{\text{3}}$

Formula Used:

The terminal speed is given by relation

  $v_t=\sqrt{\frac{2mg}{\rho \pi r^2}}$

Calculation:

Mass, $m=56\text{kg}$

Acceleration due to gravity, $g=9.81{\text{m/s}}^{\text{2}}$

Density, $\rho =0.514{\text{kg/m}}^{\text{3}}$

Radius $r=0.3\text{m}$

Substitute the values:

  $\begin{array}{c}{v}_t=\sqrt{\frac{2mg}{\rho \pi r^2}}\\ =\sqrt{\frac{\text{2}\left(\text{56}\text{kg}\right){\left(\text{9}\text{.81m/s}\right)}^{\text{2}}}{\text{π}\left(\text{0}\text{.514}{\text{kg/m}}^{\text{3}}\right){\left(\text{0}\text{.3}\text{m}\right)}^{\text{2}}}}\\ =272.12\text{m/s}\end{array}$

Conclusion:

Thus, the terminal speed at the given height is $272.12\text{m/s}$