Question

(II)A 180 km/h wind blowing over the flat roof of a house causes the roof to lift off the house. If the house is\(6.2\;{\rm{m}} \times {\rm{12}}{\rm{.4}}\;{\rm{m}}\)in size, estimate the weight of the roof. Assume the roof is not nailed down.

Short answer

The weight of the roof is \(1.25 \times {10^5}\;{\rm{N}}\).

Step-by-step solution

Understanding the concept of Bernoulli’s equation reduced form for obtaining pressure variation with fluid velocity

Bernoulli’s equation reduced form for obtaining pressure variation with fluid velocity states that the variation in pressure is a function of fluid density and its velocity.

Given data

The speed of wind blowing over the roof is \(v = 180\;{\rm{km/h}}\).

The area of the house is \(A = 6.2\;{\rm{m}} \times {\rm{12}}{\rm{.4}}\;{\rm{m}}\).

The standard value for the density of air is \(1.3\;{\rm{kg/}}{{\rm{m}}^3}\).

Evaluating the weight of roof by using Bernoulli’s equation reduced form for pressure variation

The weight of the roof by using Bernoulli’s equation is calculated below:

\(\begin{array}{c}P = \frac{1}{2}\rho {v^2}\\\frac{W}{A} = \frac{1}{2}\rho {v^2}\\W = \frac{1}{2}\rho {v^2}A\end{array}\)

Here, \(\rho \) is the density of air, P is the pressure and W is the weight of roof.

Substitute the values in the above equation.

\(\begin{array}{c}W = \frac{1}{2}\left( {1.3\;{\rm{kg/}}{{\rm{m}}^3}} \right){\left[ {180\;{\rm{km/h}} \times \left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)\left( {\frac{{1\;{\rm{h}}}}{{3600\;{\rm{s}}}}} \right)} \right]^2}\left( {6.2\;{\rm{m}} \times {\rm{12}}{\rm{.4}}\;{\rm{m}}} \right)\\W = 1.25 \times {10^5}\;{\rm{N}}\end{array}\)

Hence, the weight of the roof is \(1.25 \times {10^5}\;{\rm{N}}\).