91Ó°ÊÓ

The angle of the leg joint, $\mathrm{θ}=40°$

From the free-body diagram of the insect, we can calculate the tension in both cases. According to the conservation law, all forces acting on the body are to be balanced for the body to be at equilibrium. So, the total force in a given direction is zero.

Formula:

Since, the total forces acting on the body are balanced in the given direction,

$\underset{}{\overset{}{∑}}F=0$ (i)

We take the summation of all forces in y direction to be zero, so from equation (i), we get

$$\begin{gathered} \underset{}{\overset{}{∑F_Y=0}} \\ T\sin \mathrm{θ}-mg=0(âˆ\mu fromfreebodydiagram) \\ T\sin \mathrm{θ}=mg \end{gathered}$$

As there are 6 legs of the insect, we get the force from 6 sine components. Hence, we get,

$$\begin{gathered} 6×T×\sin 40°=mg \\ \frac{T}{mg}=\frac{1}{6×\sin 40°} \\ =0.259 \\ ≈0.26 \end{gathered}$$

Hence, the value of the required ratio is 0.26.

As the insect straightens out its legs, the value of angle$\mathrm{θ}$would increase. Since the $\sin \mathrm{θ}$is inversely proportional to the tension, as value of$\mathrm{θ}$increases, the tension would decrease.