Chapter 8, Problem 6P

To determine

To show: That it is possible to do this so that the top book extends entirely beyond the table, and show that the top book can extend any distance at all beyond the edge of the table if the stack is high enough.

Answer to Problem 6P

The top books can be extended beyond the edge of the table if enough books are added.

Explanation of Solution

Given:

Consider the given information.

Calculation:

Sketch the schematic diagram of the stack of books.

  

Place the y-axis at the end of he first book of length $L$ at the bottom of the stack.

Calculate the x-coordinate of the center of mass of the first book.

  $x_1=\frac{L}{2}$

Calculate the x-coordinates of the center of the second book.

  $x_2=\frac{L}{2\left(n-1\right)}+\frac{L}{2}$

Calculate the x-coordinates of the center of the third book.

  $x_3=\frac{L}{2\left(n-2\right)}+\frac{L}{2\left(n-1\right)}+\frac{L}{2}$

Calculate the centroid of the stack of the books.

  $\begin{array}{c}\overline{x}=\frac{m{x}_1+m{x}_2+m{x}_3}{mn}\\ =\frac{1}{n}\left(\frac{L}{2}+\frac{L}{2\left(n-1\right)}+\frac{L}{2}+\frac{L}{2\left(n-2\right)}+\frac{L}{2\left(n-1\right)}+\frac{L}{2}\right)\\ =\frac{1}{n}\left(\frac{n-1}{2\left(n-1\right)}+\frac{n-2}{2\left(n-2\right)}+\mathrm{...}+\frac{2}{4}+\frac{1}{2}+\frac{n}{2}\right)\\ =\frac{L}{2}\left(\frac{n-1}{2}+\frac{n}{2}\right)\end{array}$

Simplify further.

  $\overline{x}=\left(\frac{2n-1}{2n}\right)L$

Therefore, the centroid of the stack of books is less than length of each book which shows that no matter how many books are added, the center of mass lies above the table, and series is also a divergent series.

Thus, the top books can be extended beyond the edge of the table if enough books are added.