Chapter 3, Problem 86P

Gate AB ( $0.6\times 0.9-m$ ) is located at the bottom of a tank filled with methyl alcohol (SG = 0.79), and hinged along its bottom edge. A Knowing that the weight of the gate is 300 N, determine the minimum force that must be applied to the cable (BCD) to open the gate.

To determine

The minimum force that is applied to the cable to open the gate.

Answer to Problem 86P

The tension in the cable is $1471.60\text{N}$.

Explanation of Solution

Given information:

The length of the gate is $0.6\text{m}$, the breadth of the gate is $0.9\text{m}$, the specific gravity of methyl alcohol is $0.79,$ and the weight of the gate is $300\text{N}$.

Write the expression for area of the gate.

  $A=l\times b$...... (I)

Here, the length of the gate is $l$ and the breadth of the gate is $b$.

The figure below shows the forces acting on the tank.

  

Figure-(1)

Write the expression for angle of inclination $\theta$.

  $\sin \theta =\frac{l}{b}$...... (II)

Write the expression for angle of inclination $\alpha$.

  $\tan \alpha =\frac{t_1}{t_2}$...... (III)

Here, the horizontal distance from point $B$ to the free surface is $t_2$ and the vertical distance from point $B$ to the free surface is $t_1$.

Write the expression for centroid of gate from free surface.

  $\overline{h}=t_1+\frac{l}{2}$...... (IV)

Write the expression for centre of pressure.

  $\begin{array}{c}{h}_{CP}=\overline{h}+\frac{I_G{\sin }^2\theta }{hA}\\ =\overline{h}+\frac{\left(\frac{b{h}^3}{12}\right){\sin }^2\theta }{\overline{h}A}\end{array}$...... (V)

Write the expression of $\sin \theta$ for calculation of $x$.

  $\begin{array}{c}\sin \theta =\frac{\left(t_1+l_1\right)-h_{CP}}{x}\\ x=\frac{\left(l_1+t_1\right)-h_{CP}}{\sin \theta }\end{array}$...... (VI)

Write the expression for hydrostatic force on gate.

  $F_{ky}=\left(SG\times {\rho }_w\right)g\times \overline{h}\times A$...... (VII)

Here, the specific gravity of methyl alcohol is $SG$ and the density of methyl alcohol is ${\rho }_w$.

Write the expression for equilibrium of forces.

  $\begin{array}{l}{M}_A=0\\ F\sin \left(\theta +\alpha \right)\left(AB\right)-W\left(\frac{AB}{2}\cos \theta \right)-\left(F_{hy}\times x\right)=0\end{array}$...... (VIII)

Here, the weight of the gate is $W$.

Calculation:

Substitute $0.6\text{m}$ for $l$ and $0.9\text{m}$ for $b$ in Equation (I).

  $\begin{array}{c}A=\left(0.6\text{m}\times 0.9\text{m}\right)\\ =0.54{\text{m}}^2\end{array}$

Substitute $0.6\text{m}$ for $l$ and $0.9\text{m}$ for $b$ in Equation (II).

  $\begin{array}{c}\sin \theta =\frac{0.6\text{m}}{0.9\text{m}}\\ =0.666\text{m}\\ \theta ={\sin }^{-1}\left(0.666\right)\\ =41.81°\end{array}$

Substitute $0.4\text{m}$ for $t_1$ and $0.8\text{m}$ for $t_2$ in Equation (III).

  $\begin{array}{l}\tan \left(\alpha \right)=\frac{0.4\text{m}}{0.8\text{m}}\\ \alpha ={\tan }^{-1}\left(\frac{0.4\text{m}}{0.8\text{m}}\right)\\ \alpha =26.56°\end{array}$

Substitute, $0.4\text{m}$ for $t_1$ and $0.6\text{m}$ for $l$ in Equation (IV).

  $\begin{array}{c}\overline{h}=0.4\text{m}+\left(\frac{0.6\text{m}}{2}\right)\\ =0.7\text{m}\end{array}$

Substitute, $0.7\text{m}$ for $\overline{h}$, $0.6\text{m}$ for $b$, $41.81°$ for $\theta$ and $0.54{\text{m}}^2$ for $A$ in Equation (V).

  $\begin{array}{c}{h}_{CP}=\left(0.7\text{m}\right)+\frac{\left(\frac{0.6\text{m}\times {\left(0.9\text{m}\right)}^3}{12}\right){\sin }^2\left(41.81°\right)}{\left(0.7\text{m}\right)\left(0.54{\text{m}}^2\right)}\\ =\left(0.7\text{m}\right)+\frac{0.03645{\text{m}}^4\times 0.44443}{0.378{\text{m}}^3}\\ =0.7\text{m}+0.04285\text{m}\\ =0.74285\text{m}\end{array}$

Substitute, $0.4\text{m}$ for $t_1$, $0.6\text{m}$ for $l$

  $0.74285\text{m}$ for $h_{CP}$ and $41.81°$ for $\theta$ in Equation (VI).

  $\begin{array}{c}x=\frac{\left(0.4\text{m}+0.6\text{m}\right)-0.74285\text{m}}{\sin 41.81°}\\ x=\frac{0.25715\text{m}}{0.6666}\\ =0.3858\text{m}\end{array}$

Substitute, $0.79$ for $SG$, $1000\text{kg}/{\text{m}}^3$ for ${\rho }_w$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.7\text{m}$ for $\overline{h}$ and $0.54{\text{m}}^{\text{2}}$ for $A$ in Equation (VII).

  $\begin{array}{l}{F}_{ky}=\left(0.79\times 1000\text{kg}/{\text{m}}^3\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(0.7\text{m}\right)\times \left(0.54{\text{m}}^{\text{2}}\right)\\ =\left(790\text{kg}/{\text{m}}^3\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(0.378{\text{m}}^3\right)\\ =2929.46\text{N}\end{array}$

Substitute $0.9\text{m}$ for $AB$, $300\text{N}$ for $W$, $2929.46\text{N}$ for $F_{hy}$, $0.3858\text{m}$ for $x$, $41.81°$ for $\theta$ and $26.56°$ for $\alpha$ in Equation (VII).

  $\begin{array}{c}F=\left[\frac{\left(300\text{N}\right)\left(\frac{0.9\text{m}}{2}\right)\times \left(\cos \left(41.81°\right)+\left(2929.46\text{N}\right)\left(0.3858\text{m}\right)\right)}{\left(0.9\text{m}\right)\sin \left(41.81°+26.56°\right)}\right]\\ =\left[\frac{135\text{N}\cdot \text{m}\times \left(\cos \left(41.81°\right)+1130.18\text{N}\cdot \text{m}\right)}{\left(0.9\text{m}\right)\times 0.9295}\right]\\ =\left[\frac{\left(135\times \text{1129}\text{.618}\right)\text{N}\cdot \text{m}}{0.83662\text{m}}\right]\\ =1471.160\text{N}\end{array}$

Conclusion:

The tension in the cable is $1471.60\text{N}$.