Static Equilibrium of a building

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A 50-story building, 200 m high and weighing approximately 1.8 x 10^8 N, is analyzed for static equilibrium against wind forces. A wind speed of 200 km/h exerts a pressure of 1000 N/m² on the building's 70 m wide face, creating a torque about the rear edge. The torque from the wind force is calculated at the midpoint of the building's face, while the weight of the building acts downward at its center of mass. If the torque from the wind exceeds that from the building's weight, the structure risks toppling. The analysis concludes that maintaining stability requires careful consideration of these forces and torques.
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A 50 story building is being planned. It is to be 200 m high with a base 56 m by 70 m. Its total mass will be about 1.8 X10^7 kg and its weight therefore about 1.8 108 N. Suppose a 200 km/h wind exerts a force of 1000 N/m2 over the 70 m wide face (Fig. 12-86). Calculate the torque about the potential pivot point, the rear edge of the building (where FE acts in Fig. 12-86), and determine whether the building will topple. Assume the total force of the wind acts at the midpoint of the building's face, and that the building is not anchored in bedrock. [Hint: FE in Fig. 12-86 represents the force that the Earth exerts on the building in the case where the building is just beginning to tip.]

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Calculate the torques from weight of building and wind force about the edge.
If the weight "wins", the building does not topple; if the torque from the wind force wins, the building topples.
 
Treat the building as what it is, a rectangle, the center of gravity or center of mass (assuming uniform gravitational field and uniform mass distribution) will be where the diagonals intersect, use the info given and apply the torque equilibrium condition.
 
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