91Ó°ÊÓ

The given data can be listed below as:

• The height of the Leaning Tower of Pisa is\(55{\rm{ m}}\).
• The top of the tower is tilted toward the downward direction whose value is\(4.5{\rm{ m}}\).
• The radius of the leaning tower is\(x = 7.7{\rm{ m}}\).
• The acceleration due to gravity is \(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

When the tower is in uniform composition, the center of gravity will be in the middle of the tower. The center of gravity of the tower is over its base, and then it should be stable.

The base of the tower is thinner. However, when the center of gravity is not vertically above its base, it will be unstable. The force due to gravity develops an unbalanced torque on the tower. Due to this unbalanced torque, the tower will fall downward

The tower can be represented as:

The tower will be in an unstable condition at (x,y) position. Here, x is the maximum distance moved by the center of gravity (CG) toward the right, whose value is 7.7 m, and y is the vertical distance moved by the center of gravity.

The center of gravity of the tower above the ground can be calculated as:

\(\begin{array}{c}y = \frac{{55{\rm{ m}}}}{2}\\ = 27.5{\rm{ m}}\end{array}\)

The farther distance moved by the tower from its top can be given as:

\(\begin{array}{c}{x_1} = \frac{{{\rm{4}}{\rm{.5 m}}}}{2}\\ = 2.25{\rm{ m}}\end{array}\)

The tower has moved or displaced by a distance of 2.25 m.

The farther distance moved from the top can be given as:

\(\begin{array}{c}{x_2} = 7.7{\rm{ m}} - 4.5{\rm{ m}}\\ = 3.2{\rm{ m}}\end{array}\)

The farther distance moved from the center of the tower is:

\(\begin{array}{c}C = \frac{{{x_2}}}{2}\\ = \frac{{3.2{\rm{ m}}}}{2}\\ = 1.6{\rm{ m}}\end{array}\)

Thus, the tower can lean over 1.6 m from its center, or the farther distance it can lean before it becomes unstable is 3.2 m.

The distance moved by the tower from its resting position to become unstable can be given as:

\(\begin{array}{c}d = \left( {7.7{\rm{ m}} \times {\rm{2}}} \right) - 4.5{\rm{ m}}\\ = 10.9{\rm{ m}}\end{array}\)

Thus, the distance moved by the top of the tower from its resting position to become unstable is 10.9 m.

The angle of tilt \(\left( \theta \right)\) of the tower can be expressed as:

The angle of tilt can be expressed as:

\(\begin{array}{c}\tan \theta = \frac{{4.5{\rm{ m}}}}{{55{\rm{ m}}}}\\\theta = {\tan ^{ - 1}}\left( {\frac{{4.5{\rm{ m}}}}{{55{\rm{ m}}}}} \right)\\ = 4.68^\circ \end{array}\)

The farther distance moved by the tower from its top can be given as:

\(\begin{array}{c}\tan 4.68^\circ = \frac{{{x_1}}}{{27.5{\rm{ m}}}}\\{x_1} = 2.25{\rm{ m}}\end{array}\)

The CG has displaced a distance of 2.25 m. So, the tower’s radius (7.7 m) is more than the calculated distance of 2.25 m.

Thus, the tower is in stable equilibrium until the distance of the top edge of the tower is less than 7.7 m from the center of the tower’s base.

The tower will become unstable when the center of gravity exceeds the base distance of the tower. So, the distance of CG to become unstable must exceed the distance of 7.7 m.

The angle of unstableness can be expressed as:

\(\begin{array}{c}\tan {\theta _1} = \frac{{{\rm{7}}{\rm{.7 m}}}}{{{\rm{27}}{\rm{.5 m}}}}\\{\theta _1} = {\tan ^{ - 1}}\left( {\frac{{{\rm{7}}{\rm{.7 m}}}}{{{\rm{27}}{\rm{.5 m}}}}} \right)\\ = 15.64^\circ \end{array}\)

The maximum angle of tilt of the tower to become unstable can be estimated as:

\(\begin{array}{c}{\theta _2} = {\theta _1} - \theta \\ = 15.64^\circ - 4.68^\circ \\ = 10.96^\circ \end{array}\)

Thus, the maximum angle of \(10.96^\circ \) made by the tower becomes unstable.