Some insects can walk below a thin rod (such as a twig) by hanging from it. Suppose that such an insect has mass mand hangs from a horizontal rod as shown in Figure, with angle$\theta =40°$its six legs are all under the same tension, and the leg sections nearest the body are horizontal. (a) What is the ratio of the tension in each tibia (forepart of a leg) to the insect’s weight? (b) If the insect straightens out its legs somewhat, does the tension in each tibia increase, decrease, or stay the same?

The angle of the leg joint, $\theta =40°$

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,

$\sum _{}^{}F=0$ (i)

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

$$\begin{gathered} \underset{}{\overset{}{\sum F_Y=0}} \\ T\sin \theta -mg=0(\because fromfreebodydiagram) \\ T\sin \theta =mg \end{gathered}$$

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

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

Hence, the value of the required ratio is 0.26.

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