Question

(III) Four bricks are to be stacked at the edge of a table, each brick overhanging the one below it, so that the top brick extends as far as possible beyond the edge of the table. (a) To achieve this, show that successive bricks must extend no more than (starting at the top) \(\frac{{\bf{1}}}{{\bf{2}}}{\bf{,}}\frac{{\bf{1}}}{{\bf{4}}}{\bf{,}}\frac{{\bf{1}}}{{\bf{6}}}\) and \(\frac{{\bf{1}}}{{\bf{8}}}\)of their length beyond the one below (Fig. 9–75a). (b) Is the top brick completely beyond the base? (c) Determine a general formula for the maximum total distance spanned by n bricks if they are to remain stable. (d) A builder wants to construct a corbeled arch (Fig. 9–75b) based on the principle of stability discussed in (a) and (c) above. What minimum number of bricks, each 0.30 m long and uniform, is needed if the arch is to span 1.0 m?

Short answer

1. The maximum distance at which the edge of four successive bricks are placed with respect to one beyond it starting from the top is\(\frac{1}{2},\frac{1}{4},\frac{1}{6}\), and\(\frac{1}{8}\)of their length.
2. Yes, the top brick is completely beyond the base.
3. The general formula for the maximum total distance spanned by the n bricks is\({x_{\max .}} = \sum\limits_{i = 1}^n {\frac{l}{{2i}}} \).
4. A total of 35 bricks is needed to form an arch that spans a distance of 1.0 m.

Step-by-step solution

Center of mass

The center of mass of a group of objects is a point at which net force should act in order to determine the translational motion of a group of objects as a whole. For example, if four objects of masses \({m_1},\;{m_2},\;{m_3}\), and \({m_4}\)are grouped, the x-component of center of mass of the system of four objects is given by

\({x_{{\rm{CM}}}} = \frac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3} + {m_4}{x_4}}}{{{m_1} + {m_2} + {m_3} + {m_4}}}\).

Here, \({x_1},{x_2},{x_3},{\rm{ and }}{x_4}\) are the distances of objects from a certain reference point.

(a) Determination of maximum distances at which successive bricks must be placed

Let us name the four successive bricks as A, B, C, and D, where the top one being A and the lowest one is in contact with the table being D. Let the length and mass of each brick be l and m, respectively.

Since the four bricks are uniform, their center of mass lies in the middle, i.e., at a distance\(\frac{l}{2}\)from the edge of each brick. Bricks can be placed on each other by following the steps below:

1. The first brick A will remain balanced until its center of mass is placed at maximum at the edge of the next brick B. Thus, brick A will extend by the length \({x_1} = \frac{l}{2}\) from the brick B, as shown in the figure below:

1. The top two bricks, i.e., A and B, will remain balanced on the brick C until their center of mass is placed at maximum at the edge of the next brick C. The center of mass of the system of top two bricks A and B with respect to the center of brick B lie at point P at a distance

\(x = \frac{{m\left( 0 \right) + m\left( {\frac{l}{2}} \right)}}{{2m}} = \frac{l}{4}\).

Thus, the stack of bricks A and B should be placed on a brick C at a distance \({x_2} = \frac{l}{4}\) from the right end of the brick B, as shown in the figure.

1. Now, the stack of bricks, A, B, and C, remains balanced on brick D if their center of mass is placed at maximum on the edge of the brick D. The center of mass of the stack of bricks A, B, and C with respect to the center of brick C lies at point Q at a distance

\(x = \frac{{m\left( 0 \right) + 2m\left( {\frac{l}{2}} \right)}}{{3m}} = \frac{l}{3}\).

Thus, the stack of bricks, A, B, and C, should be placed on a brick D at a distance \({x_3} = \frac{l}{2} - \frac{l}{3} = \frac{l}{6}\) from the right end of the brick C on the edge of brick D, as shown in the figure.

1. The stack of bricks, A, B, C, and D, will remain balanced on the table if their center of mass is placed at maximum on the edge of the table. The center of mass of the stack of bricks A, B, C, and D with respect to the center of brick D will lie at a point R which is at a distance

\(x = \frac{{m\left( 0 \right) + 3m\left( {\frac{l}{2}} \right)}}{{4m}} = \frac{{3l}}{8}\).

Thus, the stack of bricks A, B, C, and D should be placed on the table at a distance \({x_4} = \frac{l}{2} - \frac{{3l}}{8} = \frac{l}{8}\) from the right end of the brick D on the edge of the table, as shown in the figure.

(b) Determination of distance of top brick from the edge of the table

Distance between the right end of the top brick A and the edge of the table is calculated as follows:

\(\begin{array}{c}x = {x_1} + {x_2} + {x_3} + {x_4}\\ = \frac{l}{2} + \frac{l}{4} + \frac{l}{6} + \frac{l}{8}\\ = \frac{{25}}{{24}}l\\ = \left( {1.04} \right)l\\ > l\end{array}\)

Thus, it is clear that the distance of the top brick from the edge of the table is more than its length. Therefore, you can say that the top brick is completely beyond the base.

(c) Determination of maximum total distance spanned by n bricks

It is clear from the discussion in part (a) that the maximum distance at which edge of the first brick A is placed from the edge of the second brick B is\({x_1} = \frac{l}{2}\); the maximum distance at which edge of the second brick B is placed from the edge of third brick is\({x_2} = \frac{l}{4}\), and maximum distance at which edge of the third brick is placed from the edge of fourth brick is\({x_3} = \frac{l}{6}\).

Thus, similarly, the maximum distance at which the edge of the nth brick is placed from the edge of (n-1)th brick will be \({x_{\rm{n}}} = \frac{l}{{2n}}\).

Therefore, the maximum total distance spanned by n bricks from the edge of the table is calculated as follows:

\(\begin{array}{c}{x_{\max .}} = {x_1} + {x_2} + {x_3} + {x_4} + ....... + {x_n}\\ = \frac{l}{2} + \frac{l}{4} + \frac{l}{6} + \frac{l}{8} + ....... + \frac{l}{{2n}}\\ = \sum\limits_{i = 1}^n {\frac{l}{{2i}}} \end{array}\)

This is the general formula for the maximum total distance spanned by n bricks.

(d) Determination of the minimum number of bricks needed to form an arch that spans a distance of 1.0 m

The length of each brick is l = 0.30 m.

The distance that needs to be spanned by the arch is 1.0 m. Therefore, the stack of bricks on each side of the arch should span a distance equal to or greater than 0.5 m, i.e.,

\(\begin{array}{l}{x_{\max .}} = \sum\limits_{i = 1}^n {\frac{l}{{2i}}} \ge 0.5\;{\rm{m}}\\\sum\limits_{i = 1}^n {\frac{{0.30\;{\rm{m}}}}{{2i}}} \ge 0.5\;{\rm{m}}\end{array}\)

By, hit and trial method, it is found that 16 bricks span the distance, which is approximately equal to 0.5 m. Thus, a total of 32 bricks will form an arch. But, it can be seen from the figure that one extra brick is placed at the top, and one brick is present on the base on both sides of the arch. Thus, you can say that a total of 35 bricks is needed to form an arch that spans a distance of 1.0 m.