91Ó°ÊÓ

Given data

The number of floors,\(n = 50\).

The height of the building,\(h = 180\;{\rm{m}}\).

The base of the building,\(b = 46\;{\rm{m}}\).

The length of the building,\(l = 76\;{\rm{m}}\).

The total mass,\(m = 1.8 \times {10^7}\;{\rm{kg}}\).

The weight,\(W = 1.8 \times {10^8}\;{\rm{N}}\).

The force exerted by wind,\({F_{\rm{A}}} = 950\;{\rm{N/}}{{\rm{m}}^2}\).

The speed of the wind,\(v = 200\;{\rm{km/h}}\).

The force exerted by Earth is \({F_{\rm{E}}}\).

In the initial stage, the building will be in the vertical position, and the forces due to contact with the Earth are at the bottom left-hand corner.

Determine the net torque about the left band corner by taking a positive direction as counter-clockwise.

The following is the free body diagram.

The relation of net torque can be written as follows:

\(\sum \tau = \left( {{F_{\rm{A}}} \times \frac{h}{2}} \right) - \left( {mg \times \frac{b}{2}} \right)\)

Here, \({F_{\rm{A}}}\) is the horizontal force exerted on the building; \(g\) is the gravitational acceleration.

Plugging in the values in the above relation,

\(\begin{array}{l}\sum \tau = \left( {950\;{\rm{N/}}{{\rm{m}}^2} \times \left( {180\;{\rm{m}} \times 76\;{\rm{m}}} \right)} \right)\left( {\frac{{180\;{\rm{m}}}}{2}} \right) - \left( {\left( {1.8 \times {{10}^7}\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \frac{{46\;{\rm{m}}}}{2}} \right)\\\sum \tau = - 2.887 \times {10^9}\;{\rm{N}} \cdot {\rm{m}}\end{array}\)

The net torque obtained is negative; hence, the building will rotate clockwise, and it will move back down to the ground and not topple.

Thus, \(\sum \tau = - 2.887 \times {10^9}\;{\rm{N}} \cdot {\rm{m}}\)is the net torque on the building.