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Q16P

Expert-verifiedFound in: Page 117

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Some insects can walk below a thin rod (such as a twig) by hanging from it. Suppose that such an insect has mass m and hangs from a horizontal rod as shown in Figure, with angle ${\theta}{=}{40}{\xb0}$**

(a) Ratio of the tension in each tibia to the insect’s weight is 0.26 .

(b) Tension will decrease in each tibia when the insect straightens out its legs.

** **

The angle of the leg joint, $\theta =40\xb0$

**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

$\underset{}{\overset{}{\sum {F}_{Y}=0}}\phantom{\rule{0ex}{0ex}}T\mathrm{sin}\theta -mg=0(\because fromfreebodydiagram)\phantom{\rule{0ex}{0ex}}T\mathrm{sin}\theta =mg$

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

$6\times T\times \mathrm{sin}40\xb0=mg\phantom{\rule{0ex}{0ex}}\frac{T}{mg}=\frac{1}{6\times \mathrm{sin}40\xb0}\phantom{\rule{0ex}{0ex}}=0.259\phantom{\rule{0ex}{0ex}}\approx 0.26$

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 $\mathrm{sin}\theta $ is inversely proportional to the tension, as value of$\theta $increases, the tension would decrease.

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