Chapter 9, Problem 62GP

To determine

To Find: The torque about the potential pivot point and whether the building would topple or not.

Answer to Problem 62GP

The torque about the potential pivot point is $-3.53\times {10}^9\text{ N}\cdot \text{m}$ . As torque is negative, building will not topple and it will retain its position.

Explanation of Solution

Given info:

Height, $h=200\text{ m}$ .

Mass of building, $m=1.8\times {10}^7\text{ kg}$ .

Weight of building, $W=1.8\times {10}^8\text{ N}$ .

Base of building $\text{= }40\text{ m}\times 70\text{ m}$ .

Force exerted by wind, $F_A=950{\text{N/m}}^{\text{2}}$ .

Wind speed, $v=200\text{ km / h}$ .

Formula used:

Torque at the potential pivot point of the building that is at the bottom left corner.

  $\sum \tau =(F_A\times \text{height up to mid point) - (}W\times \text{distance up to CG)}$

Here,

  $\tau$ is the torque,

  $F_A$ is wind force,

  $W$ is the weight of building.

Calculation:

Substitute the given values in the above equation, the torque about pivot point is,

  $\begin{array}{l}\sum \tau =(950\times 100)-(1.8\times {10}^8\times 20)\\ \sum \tau =-3.59\times {10}^9\text{N}\text{.m}\end{array}$

Conclusion:

Thus, the torque about the potential pivot point is $-3.53\times {10}^9\text{ N}\cdot \text{m}$ . As torque is negative building will not topple and it will retain its position.