Q61P Question:聽A 50-story building i... [FREE SOLUTION]

Chapter 9: Q61P (page 230)

^{Question:A 50-story building is being planned. It is to be 180.0 m high with a base 46.0 m by 76.0 m. Its total mass will be about\({\bf{1}}{\bf{.8 \times 1}}{{\bf{0}}^{\bf{7}}}\;{\bf{kg}}\)and its weight therefore about\({\bf{1}}{\bf{.8 \times 1}}{{\bf{0}}^{\bf{8}}}\;{\bf{N}}\). Suppose a 200-km/h wind exerts a force of\({\bf{950}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}\)over the 76.0-m-wide face (Fig. 9鈥�80). Calculate the torque about the potential pivot point, the rear edge of the building (where acts in Fig. 9鈥�80), and determine whether the building will topple. Assume the total force of the wind acts at the midpoint of the building鈥檚 face, and that the building is not anchored in bedrock. [Hint:\({\vec F_{\rm{E}}}\)in Fig. 9鈥�80 represents the force that the Earth would exert on the building in the case where the building would just begin to tip.]}

Short Answer

Expert verified

The torque about the potential pivot point is\( - 2.887 \times {10^9}\;{\rm{N}} \cdot {\rm{m}}\),and the building will not topple.

Step by step solution

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 \({F_{\rm{E}}}\).

Understanding net torque on the building

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.

Free body diagram of the building and calculation of net torque on the building

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.