Chapter 10, Problem 91P

(a)

To determine

The maximum stress on a single strand and also the angle between the web and the horizontal.

Answer to Problem 91P

The angle between the web and the horizontal is $42.2°$.

Explanation of Solution

The Young’s modulus of the spider silk is $4.0\text{GPa}$ and the maximum stress is $1.4\text{ GPa}$.

Write the expression for maximum stress

$\frac{F}{A}=Y\frac{\Delta L}{L}$                                                                                       (I)

Here, $\frac{F}{A}$ is the maximum stress, $Y$ is the Young’s modulus, $\Delta L$ is the change in length of the strand and $L$ is the original length of the strand.

Rearranging the above expression

$\frac{\Delta L}{L}=\frac{F/A}{Y}$                                                                                    (II)

Substitute $4.0\text{GPa}$ for $Y$ and $1.4\text{ GPa}$ for $F/A$  in (II) to find $\frac{\Delta L}{L}$

$\begin{array}{c}\frac{\Delta L}{L}=\frac{1.4\text{GPa}}{4.0\text{GPa}}\\ =0.35\end{array}$                                                                            (III)

The original length of the strand is the radius of the strand. Thus, $\frac{\Delta L}{r}=0.35$

Here, $r$ is the radius of the web

Write the expression for angle between the web and the horizontal (Refer figure 1)

$\cos \theta =\frac{r}{r+\Delta L}$                                                                           (IV)

Here, $\theta$ is the angle between the web and the horizontal.

Multiplying both numerator and denominator by $r$ and taking cos^-1 on both sides

$\theta ={\cos }^{-1}\left(\frac{1}{1+\frac{\Delta L}{r}}\right)$                                                                          (V)

Substitute $0.35$ for $\frac{\Delta L}{r}$ in the above equation (V) to find $\theta$

$\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{1}{1+0.35}\right)\\ =42.2°\end{array}$

Thus, the angle between the web and the horizontal is $42.2°$.

(b)

To determine

The mass of the bug given the maximum stress.

Answer to Problem 91P

The mass of the bug is $48\text{g}$.

Explanation of Solution

The cross sectional area of each web strand is $1.0\times {10}^{-11}{\text{ m}}^2$

Write the expression for total force in vertical y-direction at equilibrium position. (Refer figure 2)

$\begin{array}{c}{\displaystyle \sum F_y}=0\\ T\sin \theta -mg=0\end{array}$     

Here, $T$ is the tension on the strand, $m$ is the mass of the bug and $g$ is the acceleration due to gravity.

Rearranging the above equation for $T$

$T=\frac{mg}{\sin \theta }$                                                                            (VII)

Dividing (VII) by $A$

$\frac{T}{A}=\frac{mg}{A\sin \theta }$                                                                     (VIII)

Here, $A$ is the cross sectional area of all the strands.

The tensile force per unit area is the tensile stress i.e. $\frac{T}{A}=\frac{F}{A}$

Rearranging (VIII) for $m$

$m=\frac{F}{A}\frac{\left(A\sin \theta \right)}{g}$                                                   (IX)

Substitute $1.4\text{ GPa}$ for $F/A$, $50\times 1.0\times {10}^{-11}{\text{ m}}^2$ for $A$, $9.8{\text{ms}}^{-2}$ for $g$ and $42.2°$ for $\theta$ in (IX) to find $m$

$\begin{array}{c}m=\frac{1.4\text{ GPa}\left(50\times 1.0\times {10}^{-11}{\text{ m}}^2\right)\sin \left(42.2°\right)}{9.8{\text{ms}}^{-2}}\\ =\frac{1.4\times {10}^9{\text{ Nm}}^{-2}\left(50\times 1.0\times {10}^{-11}{\text{ m}}^2\right)\sin \left(42.2°\right)}{9.8{\text{ms}}^{-2}}\\ =48\text{g}\end{array}$

Thus, the mass of the bug is $48\text{g}$.

(c)

To determine

The extension of the web given the radius of the web.

Answer to Problem 91P

The depth of the web is $9.1\text{cm}$.

Explanation of Solution

The radius of the web is $0.10\text{m}$.

Refer figure 1 and write the expression for the depth of the web

$d=\left(r+\Delta L\right)\sin \theta$                                                   (X)

Here, $d$ is the depth of the web.

Rearrange (X)

$d=r\left(1+\frac{\Delta L}{r}\right)\sin \theta$                                                   (XI)

Substitute $0.10\text{m}$ for $r$, $42.2°$ for $\theta$ and $0.35$ for $\frac{\Delta L}{r}$ in (XI) to find $d$

$\begin{array}{c}d=\left(0.10\text{m}\right)\left(1+0.35\right)\times \sin \left(42.2°\right)\\ =9.1\times {10}^{-2}\text{m}\\ \text{=9}\text{.1}\text{cm}\end{array}$

Thus, the depth of the web is $9.1\text{cm}$.