Chapter 3, Problem 87P

Find the force applied by support BC to the gate AB. The width of the gate and support is 3 m and the weight of the gate is 1500N.

To determine

The force applied by support $BC$ to the gate $AB$.

Answer to Problem 87P

The force applied by support $BC$ to the gate $AB$ is $57770.45\text{N}$.

Explanation of Solution

Given information:

The width of the gate is $3\text{m}$, the weight of the gate is $1000\text{N}$, the density of the water is $1000\text{kg}/{\text{m}}^3$.

Write the expression of force applied by the water on the gate.

  $F_W=\rho gA\overline{h}$...... (I)

Here, the density of the water is $\rho$, the acceleration due to gravity is $g$, the area of the gate immersed in the water is $A$ and the depth of centre of gravity of the immersed area from free surface of water is $\overline{h}$.

Write the expression for depth of centre of gravity of the immersed area from free surface of water.

  $\overline{h}=\frac{h}{2}$...... (II)

Here, the width of the gate is $h$.

Write the expression for area of the gate immersed in the water.

  $A=AD\times h$...... (III)

Here, the $AD$ is the distance from $A$ to $D$.

The below figure shows the forces acting on the gate.

  

Figure-(1)

The below figures shows the triangle $ADE$.

  

Figure-(2)

Write the expression for the length $AD$.

  $\begin{array}{l}\cos \theta =\frac{AE}{AD}\\ AD=\frac{AE}{\cos \theta }\end{array}$...... (IV)

Write the expression for distance of centre of pressure from free surface of liquid.

  $h^{\ast }=\frac{I_G{\sin }^2\theta }{A\overline{h}}+\overline{h}$...... (V)

Here, the moment of inertia of area about an axis passing through centre of gravity of the area and parallel to the free surface of liquid is $I_G$.

Write the expression for moment of inertia.

  $I_G=\frac{h\times {\left(AD\right)}^3}{12}$...... (VI)

Write the expression for cosine rule in the triangle $ABC$.

  ${\left(BC\right)}^2={\left(AC\right)}^2+{\left(AB\right)}^2-2\left(AB\right)\left(AC\right)\cos \alpha$...... (VII)

Write the expression for $\sin \beta$ in triangle $ABC$.

  $\sin \beta =\frac{BM}{BC}$...... (VIII)

The figure below shows the triangle $ABC$.

  

Figure-(3)

Write the expression for the angle $ABC$.

  $\angle ABC=180-\left(\beta +\alpha \right)$...... (IX)

The below figure shows the moment about point $A$.

  

Figure-(4)

Write the expression for taking moment of all forces about point $A$.

  $\begin{array}{l}{\displaystyle \sum M_A=0}\\ F\sin \left(\angle ABC\right)\times AB-\left(W\times OA\right)-F_W\times AH=0\end{array}$...... (X)

The below figure shows the triangle $AHG$.

  

Figure-(5)

Write the expression for $\sin \alpha$ in triangle $AHG$.

  $\begin{array}{l}\sin \alpha =\frac{HG}{AH}\\ AH=\frac{HG}{\sin \alpha }\end{array}$...... (XI)

The below figure shows the triangle $AOQ$.

  

Figure-(6)

Write the expression for $\cos \alpha$ in triangle $AOQ$.

  $\begin{array}{l}\cos \alpha =\frac{OA}{AQ}\\ OA=AQ\cos \alpha \end{array}$...... (XII)

Calculation:

Substitute $3\text{m}$ for $h$ in Equation (II).

  $\begin{array}{c}\overline{h}=\frac{3\text{m}}{2}\\ =1.5\text{m}\end{array}$

Substitute $3\text{m}$ for $AE$ in Equation (IV).

  $\begin{array}{c}AD=\frac{3\text{m}}{\cos 40°}\\ =3.916\text{m}\end{array}$

Substitute $3.916\text{m}$ for $AD$ and $3\text{m}$ for $h$ in Equation (III).

  $\begin{array}{c}A=3.916\text{m}\times 3\text{m}\\ =11.748{\text{m}}^2\\ \approx 11.75{\text{m}}^2\end{array}$

Substitute $11.75{\text{m}}^2$ for $A$

  $1.5\text{m}$ for $\overline{h}$

  $9.81\text{m}/{\text{s}}^2$ for $g$ and $1000\text{kg}/{\text{m}}^3$ in Equation (I).

  $\begin{array}{c}{F}_W=1000\text{kg}/{\text{m}}^3\times 9.81{\text{m}/\text{s}}^2\times 11.75\text{m}\times 1.5\text{m}\\ =9810\text{kg}/{\left(\text{ms}\right)}^2\times 17.625{\text{m}}^2\\ =172901.25\text{N}\end{array}$

Substitute $3\text{m}$ for $h$ and $3.916\text{m}$ for $AD$ in Equation (VI).

  $\begin{array}{c}{I}_G=\frac{3\text{m}\times {\left(3.916\text{m}\right)}^3}{12}\\ =\frac{180.156{\text{m}}^4}{12}\\ =15.013{\text{m}}^4\end{array}$

Substitute $15.013{\text{m}}^4$ for $I_G$, $50°$ for $\theta$

  $11.75{\text{m}}^2$ for $A$ and $1.5\text{m}$ for $\overline{h}$ in Equation (V).

  $\begin{array}{c}{h}^{\ast }=\frac{\left(15.013{\text{m}}^4\times {\sin }^{2}50°\right)}{\left(11.75{\text{m}}^2\times 1.5\text{m}\right)}+1.5\text{m}\\ \text{=}\frac{\text{8}\text{.81}{\text{m}}^4}{17.625{\text{m}}^3}+1.5\text{m}\\ =1.999858\text{m}\\ \approx \text{2}\text{m}\end{array}$

Substitute $4\text{m}$ for $AC$, $5\text{m}$ for $AB$ and $50°$ for $\alpha$ in Equation (VII).

  $\begin{array}{c}{\left(BC\right)}^2={\left(4\text{m}\right)}^2+{\left(5\text{m}\right)}^2-2\times \left(5\text{m}\right)\times \left(4\text{m}\right)\cos 50°\\ =41{\text{m}}^2-25.7115{\text{m}}^2\\ =15.289{\text{m}}^2\\ =3.91\text{m}\end{array}$

Substitute $3.83\text{m}$ for $BM$ and $3.91\text{m}$ for $BC$ in Equation (VIII).

  $\begin{array}{c}\sin \beta =\frac{3.83\text{m}}{3.91\text{m}}\\ \sin \beta =0.97953\\ \beta ={\sin }^{-1}\left(0.97953\right)\\ =78.389°\end{array}$

Substitute $78.389°$ for $\beta$ and $50°$ for $\alpha$ in Equation (IX).

  $\begin{array}{c}\angle ABC=180°-\left(78.389°+1850°\right)\\ =180°-128.389°\\ =51.61°\end{array}$

Substitute $1\text{m}$ for $HG$ and $50°$ for $\alpha$ in Equation (XI).

  $\begin{array}{l}AH=\frac{1\text{m}}{\sin 50°}\\ AH=1.3\text{m}\end{array}$

Substitute $2.5\text{m}$ for $AQ$ and $50°$ for $\alpha$ in Equation (XII).

  $\begin{array}{c}OA=2.5\text{m}\times \cos 50°\\ 2.5\text{m}\times 0.642\\ =1.6\text{m}\end{array}$

Substitute $5\text{m}$ for $AB$, $1000\text{N}$ for $W$, $1.6\text{m}$ for $OA$

  $172901.25\text{N}$ for $F_W$ and $1.3\text{m}$ for $AH$ in Equation (X).

  $\begin{array}{l}F\sin \left(51.6°\right)\times 5\text{m}-\left(1000\text{N}\times 1.6\text{m}\right)-\left(172901.25\text{N}\times 1.3\text{m}\right)=0\\ F\sin \left(51.6°\right)\times 5\text{m}=226371.625\text{N}\cdot \text{m}\\ F\sin \left(51.6°\right)=45274.325\text{N}\\ F=57770.45\text{N}\end{array}$

Conclusion:

The force applied by support $BC$ to the gate $AB$ is $57770.45\text{N}$.