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Q88P

Expert-verifiedFound in: Page 353

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

** The leaning Tower of Pisa is 59.1 m high and 7.44 m** ** in diameter. The top of the tower is displaced ${\mathbf{4}}{\mathbf{.01}}{\mathbf{\text{\hspace{0.17em}m}}}$** ** from the vertical.Treat the tower as a uniform, circular cylinder. (a) What additional displacement, measured at the top, would bring the tower to the verge of toppling? (b) What angle would the tower then make with thevertical?**

(a) The additional displacement before the tower topples, $d=3.37\text{\hspace{0.17em}m}$ .

(b) Angle made by a tower with the vertical,$\theta =7.18\xb0$ .

The height of the Tower of Pisa,$\text{h}=59.1\text{\hspace{0.17em}m}$ .

The diameter of the Tower of Pisa,$D=7.44\text{\hspace{0.17em}m}$ .

The displacement of the Tower was measured from the top, $\text{x}=4.01\text{\hspace{0.17em}m}$ .

**Using the concept of center of gravity, you can calculate the maximum displacement allowed for tilting of the tower building. Any object can be steady until the center of gravity has the support of the base. The formulas used are given below.**

$\begin{array}{l}\mathbf{x}\mathbf{=}\mathbf{h}\mathbf{\times}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta}\\ \mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\theta}\mathbf{=}\mathbf{}\frac{\mathbf{r}}{\mathbf{y}}\end{array}$

We need to calculate the radius of the Tower, so

$\begin{array}{c}r=\frac{D}{2}\\ =\frac{7.44\text{\hspace{0.17em}}m}{2}\\ =3.72\text{\hspace{0.17em}m}\end{array}$

Any building will remain steady until the center of mass has the support of the base. The center of mass of the Tower can be assumed at the midpoint of the Tower.

So, we have

$\begin{array}{c}y=\frac{h}{2}\\ =\frac{59.1\text{\hspace{0.17em}}m}{2}\\ =29.55\text{\hspace{0.17em}m}\end{array}$

And,

x = r = 3.72 m

The Tower will be on the verge of toppling when the center of mass is at the side of the

Cylinder.

So, we can write as,

$tan\theta =\frac{r}{y}$

Using this equation, we can find the angle as,

$\begin{array}{c}\theta =(\frac{3.72\text{\hspace{0.17em}}m}{29.55\text{\hspace{0.17em}}m})\\ =7.18\xb0\end{array}$

Now, the distance measured from the top of the Tower at the verge of toppling will be,

$\begin{array}{c}x=h\times sin\theta \\ =59.1\times sin7.18\xb0\\ =7.38\text{\hspace{0.17em}m}\end{array}$

So, the additional distance before the tower can fall, is,

$\begin{array}{c}d=7.38\text{\hspace{0.17em}}m-4.01\text{\hspace{0.17em}}m\\ =3.37\text{\hspace{0.17em}m}\end{array}$

As calculated in (a), the angle made with the vertical is,

$\theta =7.18\xb0$

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