The Leaning Tower of Pisa is 59.1 m high and 7.44 m in diameter. The top of the tower is displaced 4.01 m from the vertical. Treat the tower as a uniform, circular cylinder.

What angle would the tower make with the vertical if an additional displacement, measured at the top, would bring the tower to the verge of toppling?

The Leaning Tower of Pisa would make an angle of approximately 3.61 degrees with the vertical when it's on the verge of toppling, determined by calculating the arctangent of the ratio between the radius of the base and the height of the tower.

Explanation:

To determine the angle at which the Leaning Tower of Pisa would be at the verge of toppling, we will use the concepts of torque and the point of contact. For an object not to topple, its center of gravity must project within its base of support. When the tower's center of gravity aligns with the edge of its base, the tower will be at the verge of toppling.

The tower's current displacement from the vertical at the top is 4.01 meters, and we'll assume this is the horizontal distance from the center of the base to the vertical through the center of gravity. The radius of the base, since the diameter is 7.44 meters, is 7.44 / 2 = 3.72 meters. At the toppling point, the horizontal displacement will be equal to the radius.

Using trigonometry, we can calculate the angle θ at the point of toppling as:

tan(θ) = opposite/adjacent = 3.72 m / 59.1 m

θ = arctan(3.72/59.1)

We can now perform the calculation:

θ = arctan(3.72/59.1) ≈ arctan(0.0629)

θ ≈ 3.61 degrees

Thus, the Leaning Tower of Pisa would make an approximately 3.61-degree angle with the vertical at the verge of toppling.