2021 A/L Physics Past Paper · Question 25 (English Medium)

Lifting force of cyclone wind on a roof

Wind generated by cyclone Burevi passed across a region in Sri Lanka with speed of 30 m s−130\ \text{m s}^{-1}. If the wind had passed over a roof of a house with effective cross-sectional area of 100 m2100\ \text{m}^2, what was the lifting force on the roof due to wind? (Assume that air inside the house was still and the density of air is 1.3 kg m−31.3\ \text{kg m}^{-3})

  1. (1) 5.85×102 N5.85 \times 10^2\ \text{N}
  2. (2) 5.85×104 N5.85 \times 10^4\ \text{N}
  3. (3) 7.61×104 N7.61 \times 10^4\ \text{N}
  4. (4) 1.17×105 N1.17 \times 10^5\ \text{N}
  5. (5) 1.95×105 N1.95 \times 10^5\ \text{N}
Answer and explanation

Answer: (2) 5.85×104 N5.85 \times 10^4\ \text{N}

Given information

We know:

The wind speed over the roof is v=30 m s−1v = 30\ \text{m s}^{-1}.

The roof has an effective area A=100 m2A = 100\ \text{m}^2.

The density of air is ρ=1.3 kg m−3\rho = 1.3\ \text{kg m}^{-3}.

The air inside the house is still.

What the question asks

We want: the lifting force FF on the roof.

The air above the roof moves fast. The air below it is still.

The plan: find the pressure difference, then multiply by the area.

Relevant theory

He has shown that in a streamline flow, the useful work done by pressures on a unit volume of the fluid is equal to the sum of the increase in its potential energy and the increase in kinetic energy.

Roofs of certain closed houses are blown off during gales.

Required equations

p+12ρv2+ρgh=Kp + \tfrac{1}{2}\rho v^2 + \rho g h = K

pin=pout+12ρv2p_{\text{in}} = p_{\text{out}} + \tfrac{1}{2}\rho v^2

F=(pin−pout) AF = (p_{\text{in}} - p_{\text{out}})\,A

Applying it to this question

All the quantities are in SI units.

kg m−3×m2s−2=Pa\text{kg m}^{-3} \times \text{m}^2\text{s}^{-2} = \text{Pa}, and Pa×m2=N\text{Pa} \times \text{m}^2 = \text{N}.

585585

Δp=12ρv2=12×1.3×302\Delta p = \tfrac{1}{2}\rho v^2 = \tfrac{1}{2} \times 1.3 \times 30^2

5850058500

F=Δp×A=585×100F = \Delta p \times A = 585 \times 100

Reasoning and calculation

The pressure under the roof is 585 Pa585\ \text{Pa} higher than the pressure above it.

That pressure pushes up on each of the 100 m2100\ \text{m}^2.

F=58 500 N=5.85×104 NF = 58\,500\ \text{N} = 5.85 \times 10^4\ \text{N}

The answer is (2).

Checks

0.5∗rho∗v∗v∗A0.5 * rho * v * v * A

Does it make sense?

The force is about 5.9×104 N5.9 \times 10^4\ \text{N}. That is the weight of about 66 tonnes.

A light roof of 100 m2100\ \text{m}^2 can weigh less than that. So a roof can be lifted off in a strong wind. This is why roofs are blown off in gales.

Conclusion

F=12ρv2A=5.85×104 NF = \tfrac{1}{2}\rho v^2 A = 5.85 \times 10^4\ \text{N}

The pressure difference is 12ρv2=585 Pa\tfrac{1}{2}\rho v^2 = 585\ \text{Pa}.

The lifting force is 585×100=5.85×104 N585 \times 100 = 5.85 \times 10^4\ \text{N}. The answer is (2).

Fast air means low pressure.

Between still air and moving air at the same height, the pressure difference is 12ρv2\tfrac{1}{2}\rho v^2. Multiply it by the area.

Exam tip

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