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Found in: Page 316

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (a) What force must be exerted by the wind to support a $${\rm{2}}{\rm{.50}}\;{\rm{kg}}$$chicken in the position shown in Figure $${\rm{9}}{\rm{.33}}$$? (b) What is the ratio of this force to the chicken’s weight? (c) Does this support the contention that the chicken has a relatively stable construction?

(a) The force by the wind is $${\rm{11}}{\rm{.025}}\;{\rm{N}}$$.

(b) The ratio of the wind force and the weight of the chicken is $${\rm{0}}{\rm{.45}}$$.

(c) Yes, the chicken has a stable construction

See the step by step solution

## Step 1: Torque

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

## Step 2: Calculation of Force

1. The mass is $${\rm{m = 2}}{\rm{.50}}\;{\rm{kg}}$$.

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 $${\rm{11}}{\rm{.025}}\;{\rm{N}}$$.

## Step 3: Calculation of the ratio of this force to the chicken’s weight

1. The ratio of the wind force and the weight of the chicken is

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

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

## Step 4: Determination of Stability of the Chicken

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