Chapter 8, Problem 91P

(a)

To determine

The maximum wind speed that a blowfly can stand.

Answer to Problem 91P

The blowfly can stand a wind speed up to $9.6\text{m/s}$.

Explanation of Solution

The cross-sectional area of the blowfly is $0.10{\text{cm}}^{\text{2}}$. The mass of the blowfly is $0.070\text{g}$.

For the blowfly to stand the wind, the net torque on it will be zero. Consider the axis of the rotation to pass through the tip of the left leg.

The figure below shows the blowfly and different forces.

Write the formula for the condition to stand the wind.

$F_{w}r\sin \theta -mgr\cos \theta =0$ (I)

Here, $F_w$ is the force due to wind, $r$ is the distance from the tip of the left foot to the center of gravity, $\theta$ is the angle made by $r$ with the horizontal, $m$ is the mass, $g$ is the acceleration due to gravity.

Write the formula for the force due to wind.

$F_w=cA{v}^2$ (II)

Here, $c$ is a constant, $A$ is the area of cross-section, $v$ is the speed of the wind.

Substitute equation (II) in equation (I).

$cA{v}^{2}r\sin \theta -mgr\cos \theta =0$

Re-write the above equation to get an expression for $v$.

$v=\sqrt{\frac{mg}{cA\tan \theta }}$ (III)

Conclusion:

Substitute $0.070\text{g}$ for $m$, $9.80{\text{m/s}}^2$ for $g$, $1.3\text{N}\cdot {\text{s}}^{\text{2}}{\text{/m}}^{\text{4}}$ for $c$, $0.10{\text{cm}}^{\text{2}}$ for $A$, $30.0°$ for $\theta$ in equation (III).

$\begin{array}{c}v=\sqrt{\frac{\left(0.070\text{g}\right)\left(9.80{\text{m/s}}^2\right)}{\left(1.3\text{N}\cdot {\text{s}}^{\text{2}}{\text{/m}}^{\text{4}}\right)\left(0.10{\text{cm}}^{\text{2}}\right)\tan 30.0°}}\\ =\sqrt{\frac{\left(0.070\text{g}\right)\left(9.80{\text{m/s}}^2\right)}{\left(1.3\text{N}\cdot {\text{s}}^{\text{2}}{\text{/m}}^{\text{4}}\right)\left(0.10\times {10}^{-4}{\text{m}}^{\text{2}}\right)\tan 30.0°}}\\ =9.6\text{m/s}\end{array}$

The blowfly can stand a wind speed up to $9.6\text{m/s}$.

(b)

To determine

The maximum wind speed that a blowfly can stand when the angle is $80.0°$.

Answer to Problem 91P

The blowfly can stand a wind speed up to $3.1\text{m/s}$ when the angle is $80.0°$.

Explanation of Solution

The cross-sectional area of the blowfly is $0.10{\text{cm}}^{\text{2}}$. The mass of the blowfly is $0.070\text{g}$.

For the blowfly to stand the wind, the net torque on it will be zero. Consider the axis of the rotation to pass through the tip of the left leg.

Write the formula for the condition to stand the wind.

$F_{w}r\sin \theta -mgr\cos \theta =0$ (I)

Here, $F_w$ is the force due to wind, $r$ is the distance from the tip of the left foot to the center of gravity, $\theta$ is the angle made by $r$ with the horizontal, $m$ is the mass, $g$ is the acceleration due to gravity.

Write the formula for the force due to wind.

$F_w=cA{v}^2$ (II)

Here, $c$ is a constant, $A$ is the area of cross-section, $v$ is the speed of the wind.

Substitute equation (II) in equation (I).

$cA{v}^{2}r\sin \theta -mgr\cos \theta =0$

Re-write the above equation to get an expression for $v$.

$v=\sqrt{\frac{mg}{cA\tan \theta }}$ (III)

Conclusion:

Substitute $0.070\text{g}$ for $m$, $9.80{\text{m/s}}^2$ for $g$, $1.3\text{N}\cdot {\text{s}}^{\text{2}}{\text{/m}}^{\text{4}}$ for $c$, $0.10{\text{cm}}^{\text{2}}$ for $A$, $80.0°$ for $\theta$ in equation (III).

$\begin{array}{c}v=\sqrt{\frac{\left(0.070\text{g}\right)\left(9.80{\text{m/s}}^2\right)}{\left(1.3\text{N}\cdot {\text{s}}^{\text{2}}{\text{/m}}^{\text{4}}\right)\left(0.10{\text{cm}}^{\text{2}}\right)\tan 80.0°}}\\ =\sqrt{\frac{\left(0.070\text{g}\right)\left(9.80{\text{m/s}}^2\right)}{\left(1.3\text{N}\cdot {\text{s}}^{\text{2}}{\text{/m}}^{\text{4}}\right)\left(0.10\times {10}^{-4}{\text{m}}^{\text{2}}\right)\tan 80.0°}}\\ =3.1\text{m/s}\end{array}$

The blowfly can stand a wind speed up to $3.1\text{m/s}$ when the angle is $80.0°$.

(c)

To determine

The maximum wind speed that a blowfly can stand when the angle is $80.0°$, cross-sectional area is $0.030{\text{m}}^{\text{2}}$ and weight is $10.0\text{kg}$.

Answer to Problem 91P

The blowfly can stand a wind speed up to $21\text{m/s}$ when the angle is $80.0°$, cross-sectional area is $0.030{\text{m}}^{\text{2}}$ and weight is $10.0\text{kg}$.

Explanation of Solution

The cross-sectional area of the blowfly is $0.030{\text{m}}^{\text{2}}$. The mass of the blowfly is $10.0\text{kg}$.

For the blowfly to stand the wind, the net torque on it will be zero. Consider the axis of the rotation to pass through the tip of the left leg.

Write the formula for the condition to stand the wind.

$F_{w}r\sin \theta -mgr\cos \theta =0$ (I)

Here, $F_w$ is the force due to wind, $r$ is the distance from the tip of the left foot to the center of gravity, $\theta$ is the angle made by $r$ with the horizontal, $m$ is the mass, $g$ is the acceleration due to gravity.

Write the formula for the force due to wind.

$F_w=cA{v}^2$ (II)

Here, $c$ is a constant, $A$ is the area of cross-section, $v$ is the speed of the wind.

Substitute equation (II) in equation (I).

$cA{v}^{2}r\sin \theta -mgr\cos \theta =0$

Re-write the above equation to get an expression for $v$.

$v=\sqrt{\frac{mg}{cA\tan \theta }}$ (III)

Conclusion:

Substitute $10.0\text{kg}$ for $m$, $9.80{\text{m/s}}^2$ for $g$, $1.3\text{N}\cdot {\text{s}}^{\text{2}}{\text{/m}}^{\text{4}}$ for $c$, $0.030{\text{m}}^{\text{2}}$ for $A$, $30.0°$ for $\theta$ in equation (III).

$\begin{array}{c}v=\sqrt{\frac{\left(10.0\text{kg}\right)\left(9.80{\text{m/s}}^2\right)}{\left(1.3\text{N}\cdot {\text{s}}^{\text{2}}{\text{/m}}^{\text{4}}\right)\left(0.030{\text{m}}^{\text{2}}\right)\tan 80.0°}}\\ =\sqrt{\frac{\left(10.0\text{kg}\right)\left(9.80{\text{m/s}}^2\right)}{\left(1.3\text{N}\cdot {\text{s}}^{\text{2}}{\text{/m}}^{\text{4}}\right)\left(0.030{\text{m}}^{\text{2}}\right)\tan 80.0°}}\\ =21\text{m/s}\end{array}$

The blowfly can stand a wind speed up to $21\text{m/s}$ when the angle is $80.0°$, cross-sectional area is $0.030{\text{m}}^{\text{2}}$ and weight is $10.0\text{kg}$.