Question

Four bricks of length L, identical and uniform, are stacked on top of oneanother (Fig. 12-71) in such a way that part of each extends beyond the one beneath. Find, in terms of L, the maximum values of (a) ${\text{a}}_{\text{1}}$ , (b) ${\text{a}}_2$, (c) ${\text{a}}_3$ , (d)${\text{a}}_4$ , and (e) $\text{h}$, such that the stack is in equilibrium, on the verge of falling.

Short answer

Maximum values of

$\begin{array}{l}\left(a\right)a_1=\frac{L}{2}\\ \left(b\right)a_2=\frac{L}{4}\\ \left(c\right)a_3=\frac{L}{6}\\ \left(d\right)a_4=\frac{L}{8}\\ \left(e\right)h=\frac{25L}{24}\end{array}$

Step-by-step solution

Understanding the given information

The length of the bricks is L.

Concept and formula used in the given question

For the given equilibrium,you can find the distance of bricks in terms of L. The formulas used are given below.

$\begin{array}{l}\mathbf{\tau }\mathbf{=}\mathbf{0}\\ \mathbf{N}\mathbf{e}\mathbf{t}\mathbf{}\mathbf{f}\mathbf{o}\mathbf{r}\mathbf{c}\mathbf{e}\mathbf{=}\mathbf{0}\end{array}$

(a) Calculation for the a1

You know the condition of equilibrium in which the moment of force, and torque is zero.

The Center of mass of the top brick cannot be extended further to the right$\text{L/2}$; otherwise, it will fall. So maximum value of${\text{a}}_{\text{1}}$is

${\text{a}}_{\text{1}}=\frac{\text{L}}{\text{2}}$

(b) Calculation for the a2

${\text{N}}_{\text{2}}=\text{2mg}$

Moment of force

$\begin{array}{l}\sum \tau =0\\ -a_2\left(N_2\right)+\frac{L}{2}\left(mg\right)=0\\ a_2=\frac{L}{4}\end{array}$

(c) Calculation for the  a3

$N_3=3mg$

Moment of force

$\begin{array}{l}\sum \tau =0\\ -a_3\left(N_3\right)+\frac{L}{2}\left(mg\right)=0\\ a_3=\frac{L}{6}\end{array}$

(d) Calculation for the  a4

${\text{N}}_{\text{4}}=\text{4mg}$

Moment of force

$\begin{array}{l}\sum \tau =0\\ -a_4\left(N_4\right)+\frac{L}{2}\left(mg\right)=0\\ a_4=\frac{L}{8}\end{array}$

(e) Calculation for the  h

$\begin{array}{c}h=a_1+a_2+a_3+a_4\\ =\frac{L}{2}+\frac{L}{4}+\frac{L}{6}+\frac{L}{8}\\ =\frac{25L}{24}\end{array}$