Question

A homogeneous, 4-ft-wide, 16-ft-long rectangular gate weighing 1000 lb is held in place by a horizontal flexible cable as shown in the Fig. P2.87. Water acts against the gate which is hinged at point A. Friction in the hinge is negligible. Determine the tension in the cable.

A homogeneous, 4-ft-wide, 16-ft-long rectangular gate weighing 1000 lb is held in place by a horizontal flexible cable as shown in the Fig. P2.87. Water acts against the gate which is hinged at points A. Friction in the hinge is negligible. Determine the tension in the cable. Figure P2.87 T = lb

Answer 1

Concepts and reason

The concept of hydrostatic force on a plane surface is used to calculate the resultant force acting on rectangular gate. The point of application of this resultant force from the free surface of water is also calculated .

Moment equilibrium of hydrostatic force, weight of the gate, tension induced in the cable about the hinge are considered to calculate the tension in the cable.

The concept of hydrostatic force is used in the design of ships, dams, and storage tanks.This is used in calculation of force acting on the surface of submerged object.

Fundamentals

The magnitude of the resultant force of fluid is equal to the product of pressure acting at centroid of the area and total area.

${F_R} = \gamma {h_c}A$

Here, $\gamma$ is specific weight, ${h_c}$ is the distance from free surface of water to the centroid of area of the object, and A is area of the object.

The relation for distance of hydrostatic force from free water surface is as follows:

${y_R} = \frac{{{I_{xc}}}}{{{y_c}A}} + {y_c}$

Here, ${I_{xc}}$ is the second moment of area with respect to an axis passing through centroid and parallel to x axis, ${y_c}$ is the distance of the centroid from the free surface of water.

The distance of hydrostatic force from free surface is always below the centroid.

Magnitude of moment can be calculated by multiplying the magnitude of force with the perpendicular distance between the point where the force is applied and the point where the moment to be found.It is

$M = Fd$

Here, F is the force and d is the perpendicular distance.

A body is in equilibrium if vector sum of all the forces is equal to zero or moment of all forces about any point is equal to zero. That is,

Summation of all forces in x direction is zero, $\sum {{F_x}} = 0$

Summation of all forces in y direction is zero, $\sum {{F_y}} = 0$

Moment about point any point is zero, $\sum {{M_A}} = 0$

General sign convection for moment: The moment is considered positive in counter-clockwise direction and negative in clockwise direction.

Draw the free body diagram of the rectangular gate.

Calculate the cross sectional area of the submerged gate.

$\begin{array}{c}\\A = 12 \times 4\\\\ = 48\,{\rm{f}}{{\rm{t}}^2}\\\end{array}$

Calculate the vertical distance from the fluid surface to the centroid of the gate.

${h_c} = \left( {{y_c}} \right){\rm{sin 60}}^\circ$

Here, ${y_c}$ is the centroid of the area.

Substitute $\left( {\frac{{12}}{2}\,{\rm{ft}}} \right)\,$ for ${y_c}$ .

$\begin{array}{c}\\{h_c} = \frac{{12}}{2} \times {\rm{sin 60}}^\circ \\\\ = 5.196\,{\rm{ft}}\\\end{array}$

Calculate resultant force exerted on the gate due to the pressure of water.

${F_R} = \gamma {h_c}A$

Here, $\gamma$ is specific weight of the water.

Substitute $48\,{\rm{f}}{{\rm{t}}^2}\,$ for A, $5.196\,\,{\rm{ft}}$ for ${h_c}$ , and $62.4\,{\rm{lb/f}}{{\rm{t}}^3}\,$ for $\gamma$ .

$\begin{array}{c}\\{F_R} = 62.4 \times 48 \times 5.196\\\\ = 15563.06{\rm{ lb}}\\\end{array}$

Calculate the location of the resultant pressure force from the free surface of the water.

$\begin{array}{c}\\{y_R} = \frac{{{I_{xc}}}}{{{y_c}A}} + {y_c}\\\\ = \frac{{\frac{{bh_{}^3}}{{12}}}}{{{y_c}A}} + {y_c}\\\end{array}$

Here, b is the width of submerged gate and h is the length of the submerged gate.

Substitute $\left( {\frac{{12}}{2}\,{\rm{ft}}} \right)\,$ for ${y_c}$ , ${\rm{4}}\,{\rm{ft}}$ for b, $48\,{\rm{f}}{{\rm{t}}^2}$ for A, and $12\,{\rm{ft}}$ for h.

$\begin{array}{c}\\{y_R} = \frac{{\frac{{4 \times 12_{}^3}}{{12}}}}{{\frac{{12}}{2} \times 48}} + \frac{{12}}{2}\\\\ = 2 + 6\\\\ = 8{\rm{ft}}\\\end{array}$

Apply moment equilibrium equation about point A.

$\begin{array}{l}\\\sum {{M_A}} = 0\\\\T\left( {16\sin 60^\circ } \right) - W\left( {\frac{{16}}{2}\cos 60^\circ } \right) - {F_R}\left( {12 - {y_R}} \right) = 0\\\end{array}$

Substitute ${\rm{8ft}}$ for ${y_R}$ , ${\rm{1000}}\,{\rm{lb}}$ for W, and $15563.06{\rm{ lb}}$ for ${F_R}$ .

$\begin{array}{l}\\T\left( {16\sin 60^\circ } \right) - 1000\left( {8\cos 60^\circ } \right) - 15563.06\left( {12 - 8} \right) = 0\\\\13.8564T = 66252.2\\\\T = 4781.34\,{\rm{lb}}\\\end{array}$

Ans:

The tension in the cable is $4781.34\,{\rm{lb}}$ .