91Ó°ÊÓ

The torque depends on the force and the radius of rotation.

The height of the wind force application is \({\rm{r = 20}}\;{\rm{cm}}\).

The horizontal distance between the center of gravity and the leg is\({\rm{r' = 9}}\;{\rm{cm}}\)

The diagram is shown below:

The Diagram of the Forces on the Chicken

The torque due to the weight of chicken is,

\(\begin{array}{c}{{\rm{\tau }}_{\rm{w}}}{\rm{ = mgr'}}\\{\rm{ = 2}}{\rm{.5 \times 9}}{\rm{.8 \times 9 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}\\{\rm{ = 2}}{\rm{.205}}\;{\rm{N \times m}}\end{array}\)

The torque due to the wind force is,

\(\begin{array}{c}{{\rm{\tau }}_{{\rm{wind}}}}{\rm{ = }}{{\rm{F}}_{{\rm{wind}}}}{\rm{ \times r}}\\{\rm{ = }}{{\rm{F}}_{{\rm{wind}}}}{\rm{ \times 20 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}\end{array}\)

For equilibrium, both the torque will be the same.

So, the force by the wind will be,

\(\begin{array}{c}{{\rm{F}}_{{\rm{wind}}}}{\rm{ = }}\frac{{\rm{\tau }}}{{{{\rm{r}}_{{\rm{wind}}}}}}\\{\rm{ = }}\frac{{{\rm{2}}{\rm{.205}}}}{{{\rm{20 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}}}\\{\rm{ = 11}}{\rm{.025}}\;{\rm{N}}\end{array}\)

Hence, the force is \({\rm{11}}{\rm{.025}}\;{\rm{N}}\).

\(\begin{array}{c}\frac{{{{\rm{F}}_{{\rm{wind}}}}}}{{{\rm{mg}}}}{\rm{ = }}\frac{{{\rm{11}}{\rm{.025}}}}{{{\rm{2}}{\rm{.5 \times 9}}{\rm{.8}}}}\\{\rm{ = 0}}{\rm{.45}}\end{array}\)

Hence, the ratio is \({\rm{0}}{\rm{.45}}\).

In this case, we consider the chicken as point object. If the chicken does not have a stable construction, we cannot solve the problem. Thus, the chicken has a stable construction