Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 8, Problem 91P

(a)

To determine

The maximum wind speed that a blowfly can stand.

(a)

Expert Solution
Check Mark

Answer to Problem 91P

The blowfly can stand a wind speed up to 9.6m/s.

Explanation of Solution

The cross-sectional area of the blowfly is 0.10cm2. The mass of the blowfly is 0.070g.

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.

Physics, Chapter 8, Problem 91P

Write the formula for the condition to stand the wind.

Fwrsinθmgrcosθ=0 (I)

Here, Fw is the force due to wind, r is the distance from the tip of the left foot to the center of gravity, θ 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.

Fw=cAv2 (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).

cAv2rsinθmgrcosθ=0

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

v=mgcAtanθ (III)

Conclusion:

Substitute 0.070g for m, 9.80m/s2 for g, 1.3Ns2/m4 for c, 0.10cm2 for A, 30.0° for θ in equation (III).

v=(0.070g)(9.80m/s2)(1.3Ns2/m4)(0.10cm2)tan30.0°=(0.070g)(9.80m/s2)(1.3Ns2/m4)(0.10×104m2)tan30.0°=9.6m/s

The blowfly can stand a wind speed up to 9.6m/s.

(b)

To determine

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

(b)

Expert Solution
Check Mark

Answer to Problem 91P

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

Explanation of Solution

The cross-sectional area of the blowfly is 0.10cm2. The mass of the blowfly is 0.070g.

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.

Fwrsinθmgrcosθ=0 (I)

Here, Fw is the force due to wind, r is the distance from the tip of the left foot to the center of gravity, θ 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.

Fw=cAv2 (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).

cAv2rsinθmgrcosθ=0

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

v=mgcAtanθ (III)

Conclusion:

Substitute 0.070g for m, 9.80m/s2 for g, 1.3Ns2/m4 for c, 0.10cm2 for A, 80.0° for θ in equation (III).

v=(0.070g)(9.80m/s2)(1.3Ns2/m4)(0.10cm2)tan80.0°=(0.070g)(9.80m/s2)(1.3Ns2/m4)(0.10×104m2)tan80.0°=3.1m/s

The blowfly can stand a wind speed up to 3.1m/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 0.030m2 and weight is 10.0kg.

(c)

Expert Solution
Check Mark

Answer to Problem 91P

The blowfly can stand a wind speed up to 21m/s when the angle is 80.0°, cross-sectional area is 0.030m2 and weight is 10.0kg.

Explanation of Solution

The cross-sectional area of the blowfly is 0.030m2. The mass of the blowfly is 10.0kg.

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.

Fwrsinθmgrcosθ=0 (I)

Here, Fw is the force due to wind, r is the distance from the tip of the left foot to the center of gravity, θ 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.

Fw=cAv2 (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).

cAv2rsinθmgrcosθ=0

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

v=mgcAtanθ (III)

Conclusion:

Substitute 10.0kg for m, 9.80m/s2 for g, 1.3Ns2/m4 for c, 0.030m2 for A, 30.0° for θ in equation (III).

v=(10.0kg)(9.80m/s2)(1.3Ns2/m4)(0.030m2)tan80.0°=(10.0kg)(9.80m/s2)(1.3Ns2/m4)(0.030m2)tan80.0°=21m/s

The blowfly can stand a wind speed up to 21m/s when the angle is 80.0°, cross-sectional area is 0.030m2 and weight is 10.0kg.

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Chapter 8 Solutions

Physics

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