Chapter 3, Problem 87P

To determine

(a)

The mud height at which the block will overcome the friction.

Answer to Problem 87P

The mud height at which the block will overcome the friction is $\underset{\_}{0.86\text{m}}$.

Explanation of Solution

Given information:

Height of concrete blocks is $1.2\text{m}$, width of concrete block is $0.4\text{m}$, density of concrete is $2700\text{kg}/{\text{m}}^{\text{3}}$, coefficient of friction is $0.4$, density of mud is $1400\text{kg}/{\text{m}}^{\text{3}}$.

  Figure below shows the free body diagram of wall and mud system.

  Figure-(1)

Write the expression for the volume of the wall.

   $V_{wall}=hwt$    ...... (I)

Here, height of the wall is $h$, width of the wall is $w,$ and thickness of the wall is $w$.

Write the expression for the weight of the wall.

   $W_{wall}={\rho }_{wall}g{V}_{wall}$    ...... (II)

Here, density of the wall is ${\rho }_{wall}$ and acceleration due to gravity is $g$.

Write the expression for the equilibrium equation to the system.

   $\begin{array}{c}N-W_{wall}=0\\ N=W_{wall}\end{array}$    ...... (III)

Here, normal reaction exerted by the ground on the wall is $N$.

Write the expression for the friction force between the wall and the ground.

   $F_f=fN$    ...... (IV)

Here, coefficient of friction is $f$.

Write the expression for the location of the centre of pressure.

   $h_c=\frac{h_m}{2}$    ...... (V)

Here, the height of mud is $h_m$.

Write the expression for the horizontal hydrostatic force acting on the vertical wall.

   $F_H=P_{avg}{A}_H$    ...... (VI)

Here, the average pressure acting on the centre of pressure is $P_{avg}$ an area of the wall resisting the horizontal hydrostatic force is $A_H$.

Write the expression for the average pressure acting on the centre of pressure.

   $P_{avg}={\rho }_{mud}g{h}_c$    ...... (VII)

Here, density of mud is

   ${\rho }_{mud}$.

Write the expression for the area of wall resisting horizontal hydrostatic force.

   $A_H=t{h}_m$    ...... (VIII)

Write the expression for the equilibrium equation of horizontal forces.

   $\begin{array}{c}{F}_f-F_H=0\\ F_H=F_f\end{array}$    ...... (IX)

Calculation:

Substitute $1.2\text{m}$ for $h$, $0.4\text{m}$ for $w$ and $1\text{m}$ for $t$ in Equation (I).

   $\begin{array}{c}{V}_{wall}=1.2\text{m}\times 0.4\text{m}\times 1\text{m}\\ \text{=0}\text{.48}{\text{m}}^3\end{array}$

Substitute $2700\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{wall}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.48{\text{m}}^3$ for $V_{wall}$ in Equation (II).

   $\begin{array}{c}{W}_{wall}=2700\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 0.48{\text{m}}^3\\ =26487\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\times 0.48{\text{m}}^3\\ =12713.76\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =12713.76\text{N}\end{array}$

Substitute $12713.76\text{N}$ for $W_{wall}$ in Equation (III).

   $N=12713.76\text{N}$

Substitute $0.4$ for $f$ and $12713.76\text{N}$ for $N$ in Equation (IV).

   $\begin{array}{c}{F}_f=0.4\times 12713.76\text{N}\\ =\text{5085}\text{.504}\text{N}\end{array}$

Substitute $1400\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{mud}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $\frac{h_m}{2}$ for $h_c$ in Equation (VII).

   $\begin{array}{c}{P}_{avg}=1400\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times \frac{h_m}{2}\\ =700\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times h_m\\ =h_{m}6867\text{kg}/{\text{m}}^2\cdot {\text{s}}^2\end{array}$

Substitute $1\text{m}$ for $t$ in Equation (VIII).

   $\begin{array}{c}{A}_H=1\text{m}\times h_m\\ =h_m\text{m}\end{array}$

Substitute $h_{m}6867\text{kg}/{\text{m}}^2\cdot {\text{s}}^2$ for $P_{avg}$ and $h_m\text{m}$ for $A_H$ in Equation (VI).

   $\begin{array}{c}{F}_H=h_{m}6867\text{kg}/{\text{m}}^2\cdot {\text{s}}^2\times h_m\text{m}\\ =h_m{}^{2}6867\text{kg}/\text{m}\cdot {\text{s}}^2\end{array}$

Substitute $h_m{}^{2}6867\text{kg}/\text{m}\cdot {\text{s}}^2$ for $F_H$ and $\text{5085}\text{.504}\text{N}$ for $F_f$ in Equation (IX).

   $\begin{array}{l}{h}_m{}^{2}6867\text{kg}/\text{m}\cdot {\text{s}}^2=\text{5085}\text{.504}\text{N}\\ h_m{}^{2}6867\text{kg}/\text{m}\cdot {\text{s}}^2=\text{5085}\text{.504}\text{N}\left(\frac{\text{1}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{\text{1}\text{N}}\right)\\ h_m{}^{2}6867\text{kg}/\text{m}\cdot {\text{s}}^2=\text{5085}\text{.504}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\\ h_m{}^2=0.7405{\text{m}}^2\end{array}$

   $\begin{array}{c}{h}_m=\sqrt{0.7405{\text{m}}^2}\\ =0.86\text{m}\end{array}$

Conclusion:

The mud height at which the block will overcome the friction is $\underset{\_}{0.86\text{m}}$.

To determine

(b)

The mud height at which the blocks will tip over.

Answer to Problem 87P

The mud height at which the blocks will tip over is $\underset{\_}{0.885\text{m}}$.

Explanation of Solution

  Figure below shows the free body diagram of wall and mud system.

  Figure-(2)

Write the expression for the height at which force of mud acts on the wall.

   $h_{c2}=\frac{h_m}{3}$    ...... (X)

Write the expression for the moment equilibrium equation about point $A$.

   $\begin{array}{l}{W}_{wall}\left(\frac{t}{2}\right)-F_H\left(h_{c2}\right)=0\\ W_{wall}\left(\frac{t}{2}\right)=F_H\left(h_{c2}\right)\end{array}$    ...... (XI)

Calculation:

Substitute $12713.76\text{N}$ for $W_{wall}$, $0.25\text{m}$ for $t$, $h_m{}^{2}6867\text{kg}/\text{m}\cdot {\text{s}}^2$ for $F_H$ and $\frac{h_m}{3}$ for $h_{c2}$ in Equation (XI).

   $\begin{array}{l}12713.76\text{N}\left(\frac{0.25\text{m}}{2}\right)=h_m{}^{2}6867\text{kg}/\text{m}\cdot {\text{s}}^2\left(\frac{h_m}{3}\right)\\ 1589.22\text{N}\cdot \text{m}=h_m{}^3\text{2289}\text{kg}/\text{m}\cdot {\text{s}}^2\\ h_m{}^3=\frac{1589.22\text{N}\cdot \text{m}}{\text{2289}\text{kg}/\text{m}\cdot {\text{s}}^2\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}\right)}\\ h_m{}^3=\frac{1589.22\text{N}\cdot \text{m}}{\text{2289}\text{N}/{\text{m}}^{\text{2}}}\end{array}$

   $\begin{array}{c}{h}_m={\left(0.69428{\text{m}}^3\right)}^{\frac{1}{3}}\\ =0.885\text{m}\end{array}$

Conclusion:

The mud height at which the blocks will tip over is $\underset{\_}{0.885\text{m}}$.