Chapter 3, Problem 85P

A triangular-shaped gate is hinged at point A, as shown. Knowing that the weight of the gate is 100 N, determine the force needed to keep the gate at its position for unit width. The line of action of the weight of the gate is shown by the dashed line.

To determine

The force needed to keep the gate at its position for unit width.

Answer to Problem 85P

The force needed to keep the gate at its position for unit width is $23070.21\text{N}$.

Explanation of Solution

Given information:

The weight of the gate is $100\text{N}$

The below figure shows the force acting on the gate to keep it at equilibrium position.

  

Figure-(1)

Write the expression for Pythagoras theorem in triangle ABC.

  $BC=\sqrt{A{B}^2-A{C}^2}$...... (I)

Write the expression for hydrostatic force acting on the member $BC$ of the gate.

  $F_{BC}=\rho gA\overline{x}$...... (II)

Here, the density of the water is $\rho$, the gravitational force is $g$, the area of the triangular gate is $A$ and the location of the centroid from the free surface is $\overline{x}$.

Write the expression for $\sin \theta$ from the triangle $ABC$.

  $\sin \theta =\frac{AC}{BC}$...... (III)

Write the expression for hydrostatic force on member $AB$.

  $F_{AB}=\rho gA{\overline{x}}_{AB}$...... (IV)

Here, the location of Centroid of the member $AB$ from the free surface is. ${\overline{x}}_{AB}$.

Write the expression for point of application of force.

  $h_{AB}=\overline{x}+\frac{I}{A\overline{x}}{\sin }^2\theta$...... (V)

Here, the moment of inertia of the triangular gate is $I$.

Write the expression for area of the member $BC$.

  $A_{BC}=BC\times W$

Here, the width of gate to be kept is $1\text{m}$.

Write the expression for area of the member $AB$.

  $A_{AB}=BA\times W$

Here, the width of gate to be kept is $1\text{m}$.

Write the expression for the location of centroid of the member $AB$ from the free surface.

  ${\overline{x}}_{BC}=\left(1+\frac{AB\times \sin \theta }{AC}\right)$...... (VI)

Write the expression for moment of inertia of triangular gate.

  $I=\frac{AB\times W^3}{12}$...... (VII)

Write the expression for moment about point $A$.

  $\begin{array}{l}{\displaystyle \sum M_A=0}\\ F_{BC}\times \frac{BC}{2}+w\times 1+F\times AC-F_{AB}\times \left(\left(AB+\frac{1}{\sin \theta }\right)-\frac{h_{AB}}{\sin \theta }\right)=0\end{array}$..... (VIII)

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

Calculation:

Substitute, $2.4\text{m}$ for $AB$ and $2\text{m}$ for $AC$ in Equation (I).

  $\begin{array}{c}BC=\sqrt{{\left(2.4\text{m}\right)}^2-{\left(2\text{m}\right)}^2}\\ =\sqrt{5.76\text{m}-4\text{m}}\\ =\sqrt{1.76\text{m}}\\ =1.3266\text{m}\end{array}$

Substitute, $1000\text{kg}/{\text{m}}^3$ for $\rho$, $9.81{\text{m}/\text{s}}^2$ for $g$

  $1.3266{\text{m}}^2$ for $A$ and $1\text{m}$ for $\overline{x}$ in Equation (II).

  $\begin{array}{c}{F}_{BC}=1000\text{kg}/{\text{m}}^3\times 9.81\text{m}/{\text{s}}^2\times 1.3266{\text{m}}^2\times 1\text{m}\\ \text{=9810}\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}\times 1.3266{\text{m}}^3\\ =13013.946\text{N}\end{array}$

Substitute, $2\text{m}$ for $AC$ and $2.4\text{m}$ for $BC$ in Equation (III).

  $\sin \theta =\frac{2\text{m}}{2.4\text{m}}$

Substitute, $2.4\text{m}$ for $AB$, $2\text{m}$ and $\sin \theta$ in Equation (VI).

  $\begin{array}{c}{\overline{x}}_{BC}=\left(1+\frac{2.4\text{m}\times \left(\frac{2\text{m}}{2.4\text{m}}\right)}{2\text{m}}\right)\\ =\left(1\text{m}+1\text{m}\right)\\ =2\text{m}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^3$ for $\rho$

  $9.81{\text{m}/\text{s}}^2$ for $g$

  $2.4{\text{m}}^2$ for $A$ and $2\text{m}$ for ${\overline{x}}_{AB}$ in Equation (IV).

  $\begin{array}{c}{F}_{AB}=1000\text{kg}/{\text{m}}^3\times 9.81{\text{m}/\text{s}}^2\times 2.4{\text{m}}^2\times 2\text{m}\\ \text{=9810}\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}\times 4.8{\text{m}}^3\\ =47088\text{N}\end{array}$

Substitute $2.4\text{m}$ for $1\text{m}$ for $W$ in Equation (VII).

  $\begin{array}{c}I=\frac{2.4\text{m}\times {\left(1\text{m}\right)}^3}{12}\\ =0.2{\text{m}}^4\end{array}$

Substitute $2\text{m}$ for $\overline{x}$

  $0.2{\text{m}}^4$ for $I$

  $2.4{\text{m}}^2$ for $A$ and $\sin \theta$ from above in Equation (V).

  $\begin{array}{c}{h}_{AB}=2\text{m}+\frac{0.2{\text{m}}^4}{2.4{\text{m}}^2\times 2\text{m}}\times {\left(\frac{2}{2.4}\right)}^2\\ =2\text{m}+0.0289\text{m}\\ =2.0289\text{m}\end{array}$

Substitute, $1.3266\text{m}$ for $BC$

  $100\text{N}$ for $w$, $47088\text{N}$ for $F_{AB}$, $2.4\text{m}$ for $AB$

  $2.0289\text{m}$ for $h_{AB}$ in Equation (VIII).

  $\begin{array}{l}\left\{\begin{array}{l}13013.946\text{N}\times \frac{1.3266\text{m}}{2}+100\text{N}\times 1+F\times 2\text{m}-47088\text{N}\\ \times \left(\left(2.4\text{m}+\frac{1}{\left(2\text{m}/2.4\text{m}\right)}\right)-\frac{2.0289\text{m}}{\left(2\text{m}/2.4\text{m}\right)}\right)\end{array}\right\}=0\\ 672.55\text{N}\cdot \text{m}+100\text{N}+2F\text{m}-47088\text{N}\left(1.16532\text{m}\right)=0\\ -46140.437\text{N}+2F=0\\ F=23070.21\text{N}\end{array}$

Conclusion:

The force needed to keep the gate at its position for unit width is $23070.21\text{N}$.