Chapter 9, Problem 37P

To determine

The centre of gravity of the leaning tower of pisa.

Answer to Problem 37P

Solution:

The tower is at stable equilibrium and top needs to be 15.4 m off centre to be unstable

Explanation of Solution

One must determine the centre of gravity of the leaning tower of Pisa. Since the dimensions $L=55$ m and $R=7.7$ m with uniform composition, the centre of gravity matches the centre point of the body. The centre of gravity is at midway along its length which is $Cl=\frac{L}{2}=\frac{55}{2}=27.5$ m and at the central vertical axis.

To remain in equilibrium, a vertical line along the center of gravity must pass through the base. The top is 4.5 m off centre. So, the off-centre, d is 4.5 m.

So, the angle of inclination is

  $\tan \theta =\frac{d}{L}$

  $\tan \theta =\frac{4.5}{55}$

  $\theta ={\tan }^{-1}\frac{4.5}{55}$

  $\theta =4.67°$

The location of the centre of gravity after the tilt is given by the same method with $\theta =4.6$ degrees and substituting the height of the centre of gravity.

  $\tan \theta =\frac{C_r^0}{Cl}$

  $\tan 4.6=\frac{C_r^0}{27.5}$

  $C_r^0=2.25$ m

It is well with the $R=7.7$ m, so the tower is stable.

For the tower to fall or become unstable, the centre of gravity must exceed the dimension of the base. So, the radial offset of the centre of gravity of an unstable tower must exceed $R=7.7$ m. Using this, one can find the angle of unstability.

  $\tan {\theta }^u=\frac{C_r^u}{Cl}$

Substituting the values:

  $\tan {\theta }^u=\frac{7.7}{27.5}$

  $\tan {\theta }^u=0.28$

To find the offset value for an unstable tower, substitute it in

  $\tan {\theta }^u=\frac{d^u}{L}$

  $0.28=\frac{d^u}{55}$

  $d^u=15.4\text{m}$

Conclusion:

The tower is at stable equilibrium and top needs to be 15.4m off centre to be unstable.