Chapter 9, Problem 38P

(a)

To determine

ToShow: Successive bricks must extend no more than (starting at the top)

  $\frac{1}{2},\frac{1}{4},\frac{1}{6}\text{ and }\frac{1}{8}$ Of their length beyond the one below.

  

Explanation of Solution

Given data:

System is in equilibrium.

  

Formula used:

To remain in equilibrium, centre of mass of above system must lie just above the right edge.

For one brick over 2

  

To remain in equilibrium, centre of mass of brick $1$ must lie above the right edge of brick $2$ or before. So, for maximum extension overhanging part must be equal to $\frac{L}{2}$ .

Brick is of uniform thickness, so centre of mass is at mid-point.

For two bricks $\left(1,2\right)$ over 3

  

For equilibrium of brick 1 and 2, and for maximum extension,centre of mass of brick 1 and 2 must lie above the edge of $3$ bricks.

  $\begin{array}{c}{x}_{cm}\left(1,2,3\right)=\frac{\frac{3L}{4}\times M+\frac{L}{4}\times M+0.M}{3M}\\ =\frac{L}{3}\end{array}$

Distance of centre of mass of 1, 2 and 3 from the right edge of brick 3:

  $\begin{array}{l}=\frac{L}{2}-\frac{L}{3}\\ =\frac{L}{6}\end{array}$

Centre of mass of brick 1,2 and 3must lie above the right edge of brick 4, so distance between right edges of brick three and $4=\frac{1}{6}$

  

Above system remains in equilibrium when C.M of mass of bricks $1,2,3,4$ lies above the right edge of table.

First find the C.M of brick $1,2,3,4$

  $\text{W}\text{.r}\text{.t}$ C.M of brick $4$ .

Let C.M of brick is origin.

  $x-$ Coordinate of C.M of brick $4:x_{cm}\left(4\right)=0$

  $x-$ Coordinate of C.M of brick $3:x_{cm}\left(3\right)=\frac{L}{6}$

  $x-$ Coordinate of C.M of brick $2:x_{cm}\left(2\right)=\frac{L}{6}+\frac{L}{4}$

  $=\frac{5L}{12}$

  $x-$ Coordinate of C.M of brick $1:x_{cm}\left(1\right)=\frac{L}{6}+\frac{L}{4}+\frac{L}{2}$

  $=\frac{11}{12}L$

With the help of previous solution, you can say that distance between C.M of brick $1\text{ and 2}$ must be equal to $\frac{R}{2}$ . For the center of mass of brick 1 and 2 to lie just abovethe right edge of brick $3$ , centre of mass of brick 1 and 2 must be equidistance to vertical line. Hence the overhanging part of brick $2$ should be equal to $\frac{L}{4}$ .

For equilibrium of $1,2,3\text{ over 4}$ . These three brick remain in equilibrium with maximum overhanging part only when C.M of brick $1$ , brick $2$ and brick $3$ must be above the right edge of brick four.

  

Let C.M of brick $3$ as origin

  $x_{cm}\left(1,2,3\right)=\frac{M{x}_{cm}\left(1\right)+M{x}_{cm}\left(2\right)+M{x}_{cm}\left(3\right)}{3M}$

  $x_{cm}\left(1\right):$ ( x -coordinate of centre of mass of brick 1) $=\frac{3L}{4}$

  $x_{cm}\left(2\right):$ ( x -coordinate of centre of mass of brick 2) $=\frac{L}{4}$

  $x_{cm}\left(3\right):$ ( x -coordinate of centre of mass of brick 3) $=0$

M : Mass of each brick.

  $\begin{array}{c}{x}_{cm}\left(1,2,3,4\right)=\frac{M.x_{cm}\left(1\right)+M{x}_{cm}\left(2\right)+M{x}_{cm}\left(3\right)+M{x}_{cm}\left(4\right)}{4M}\\ =\frac{\left(M\right)\times \left(\frac{11L}{2}\right)+M\left(\frac{5L}{12}\right)+M\left(\frac{L}{6}\right)+M\left(0\right)}{4M}\\ =\frac{3L}{8}\end{array}$

Distance between right edges of brick $4\text{ and }{x}_{cm}\left(1,2,3,4\right)$

  $\begin{array}{l}=\frac{L}{2}-\frac{3L}{8}\\ =\frac{L}{8}\end{array}$

Hence distance between right edge of table and right edge of $4th$ brick $=\frac{L}{8}$

  

Hence hanging part of brick $1,2,3,4th$ are $\frac{L}{2},\frac{L}{4},\frac{L}{6},\frac{L}{8}$ to remain in equilibrium with maximum extend.

Conclusion:

Over hanging part of brick $1,2,3,4th\text{ are }\frac{L}{2},\frac{L}{4},\frac{L}{6},\frac{L}{8}$ , etc. to remain equilibrium and with maximum extend.

(b)

To determine

Whether the top brick is completely beyond the base.

Answer to Problem 38P

Solution:

Yes

Explanation of Solution

(c)

To determine

Ageneral formula for the maximum total distance spanned by $n$ bricks if they are to remain stable.

Answer to Problem 38P

Solution:

The general formula of maximum total spanned by $n$ bricks $={\displaystyle \sum _{i=1}^n\left(\frac{L}{2i}\right)}$ .

Explanation of Solution

Given data:

You can use previous result.

Formula used:

Find out the general term of the series with the help of previous result.

  

Overhanging part of 1^st brick $=\frac{L}{2}=\frac{L}{2\times 1}$

Overhanging part of 2^nd brick $=\frac{L}{4}=\frac{L}{2\times 2}$

Overhanging part of 3^rd brick $\frac{L}{6}=\frac{L}{2\times 3}$

Overhanging part of 4^th brick $=\frac{L}{8}=\frac{L}{2\times 4}$

From above it becomes clear that overhanging part of nth brick = $\frac{L}{2n}$

Hence, maximum span length obtained with the help of $n$ brick.

  $\begin{array}{l}=\frac{L}{2}+\frac{L}{2\times 2}+\frac{L}{2\times 3}+\frac{L}{2\times 4}+\mathrm{...}\frac{L}{2n}\\ ={\displaystyle \sum _{i=1}^n\frac{L}{2i}}\end{array}$

Hence, maximum span length obtained with the help of $n$ brick

  $={\displaystyle \sum _{i=1}^n\left(\frac{L}{2i}\right)}$

Conclusion:

Maximum span length obtained with the help of $n$ brick

  $={\displaystyle \sum _{i=1}^n\left(\frac{L}{2i}\right)}$

(d)

To determine

The minimum number of bricks each $0.30\text{m}$ long and uniform is needed if the arch is to span $1.0\text{m}$ .

Answer to Problem 38P

Solution:

35 bricks

Explanation of Solution

Given data: Span length $=1.\text{0}\text{m}$

Length of brick $=0.30\text{m}$

Formula used:

Maximum span length (overhanging part)

Obtain from $n$ brick

  $={\displaystyle \sum _{i=1}^n\frac{L}{2i}}$

Calculation:

The span length of arch $=1.0\text{m}$

Length of one side span length $=\frac{\text{1}\text{.0}\text{m}}{2}=0.5\text{m}$

Let number of required bricksbe $n$ to develop $0.5\text{m}$ overhanging part.

  $\begin{array}{l}{\displaystyle \sum _{i=1}^n\frac{L}{2i}}\ge 0.5\\ L=0.30\text{m}\left(\text{given}\right)\\ {\displaystyle \sum _{i=1}^n\frac{0.30}{2i}\ge 0.50}\\ {\displaystyle \sum _{i=1}^n\frac{1}{2i}\ge \frac{5}{3}}\end{array}$

  $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\mathrm{...}\text{ to }n\ge \frac{5}{3}$

Do the sum of left side until you obtain the sum is either greater than or equal to $5/3$

  $\begin{array}{l}\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{18}+\frac{1}{20}+\frac{1}{22}+\frac{1}{24}\\ \frac{1}{26}+\frac{1}{28}\\ \left(n=14\right)\\ =1.625\end{array}$

  $\begin{array}{l}\text{If }n=15\\ {\displaystyle \sum _{i=1}^{15}\frac{1}{2i}=1.659}\end{array}$

  $\begin{array}{c}\text{If }n=16\\ {\displaystyle \sum _{i=1}^{16}\frac{1}{2i}=1.690}\\ =\frac{5}{3}=1.666\end{array}$

Form the above solution it becomes clear that

For $n=15,$ overhanging part is less $0.5$

For $n=16,$ overhanging is just greater than $0.5$

Hence, required number of bricks $=2\times 16+\left(1\right)+\left(2\right)$

Multiply 2 both sides

1 is for brick on the top and 2 is for the base of each side.

  $2\times 16+\left(1\right)+\left(2\right)$

  $=35$

Conclusion:

Minimum no of brick to developed span length $1.0\text{m}$ and arch as given in figure $=35$ .