The Figure represents an insect caught at the midpoint of a spider-web thread. The thread breaks under a stress of $\mathbf{8}\mathbf{.20}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathit{N}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$and a strain of $\mathbf{2}\mathbf{.00}$. Initially, it was horizontal and had a length of $\mathbf{2}\mathbf{.00}\mathit{c}\mathit{m}$ and a cross-sectional area of $\mathbf{8}\mathbf{.00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}{\mathit{m}}^{\mathbf{2}}$. As the thread was stretched under the weight of the insect, its volume remained constant. If the weight of the insect puts the thread on the verge of breaking, what is the insect’s mass? (A spider’s web is built to break if a potentially harmful insect, such as a bumble bee, becomes snared in the web.)

Thread breaks under a stress of$8.20\times {10}^8\text{\hspace{0.17em}N}/m^2$and strain of $2.00$.

Initial length of the thread is $2.00cm$ and a cross-sectional area is $8.00\times {10}^{-12}{m}^2$.

The stress is equal force per unit area. The strain is equal to change in length per original length. Using the condition for equilibrium,and equation for stress and strain, we can find the insect’s mass, if the weight of the insect puts the spider’s web on the verge of breaking.

Consider the required equation as follows:

$mg=2Tsin\theta$

The length of thread on the verge of breaking is

$L=L_0+L$

Solve further as:

$\begin{array}{c}L=L_0(1+\frac{L}{L_0})\\ L=L_0(1+2)\\ =3L_0\end{array}$

Consider the value of the length is:

$\begin{array}{c}{L}_0=2.00cm\\ =0.0200m\end{array}$

Solve for the strain as:

$strain=L/L_0$

But, according to the conditions of equilibrium,

$mg=2Tsin\theta$

From the free body diagram, we can see that the vertical components of the tensions are in the same direction.So, they will be added and weight mg is balanced by $2T\text{sin}\theta$.

$\begin{array}{c}T=A\left(stress\right)\\ =A_0\left(\frac{L_0}{L}\right)\\ =\frac{A_0}{3}\end{array}$

Consider by the Pythagoras theorem find the vertical distance of the fly:

$\begin{array}{c}y=\sqrt{\frac{9L_0^2}{4}-\frac{L_0^2}{4}}\\ =\sqrt{2}{L}_0\end{array}$

Therefore, the formula for the mass of the insect is as follows:

$\begin{array}{c}m=\frac{2T\text{sin}\theta }{g}\\ =\frac{2\left(\frac{A_0}{3}\right)\left(stress\right)\text{sin}\theta }{g}\\ =\frac{2A_0\left(stress\right)y}{3g\left(\frac{3L_0}{2}\right)}\\ =\frac{4\sqrt{2}{A}_0\left(stress\right)}{9g}\end{array}$

This equation is the final expression for the mass of insect.

Substitute the given values in the expression derived in step 4 to calculate the mass of the insect.

$\begin{array}{c}m=\frac{4\sqrt{2}(8.00\times {10}^{-12}{m}^2)(8.20\times {10}^8\text{N}/{\text{m}}^2)}{9(9.8\text{m}/{\text{s}}^2)}\\ =0.421g\end{array}$

Therefore, the insect’s mass, if the weight of the insect puts the spider’s web on the verge of breaking, is $0.421g$