Determine the horizontal and vertical components of reaction at the hinge A and the normal reaction at B caused by the water pressure. The gate has a width of 3 m.

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Understand the Problem

The question is asking to determine the horizontal and vertical components of the reaction forces at hinge A and the normal reaction force at point B due to water pressure on the gate. This involves applying principles of fluid mechanics to analyze forces acting on a submerged gate.

Answer

Horizontal Component at A: $R_{Ax} = 13214.25 \, \text{N}$, Vertical Component at A: $R_{Ay} = 13214.25 \, \text{N}$, Normal Reaction at B: $R_B = 13214.25 \, \text{N}$.
Answer for screen readers

The horizontal and vertical components of the reaction at hinge A are as follows:

  • Horizontal Component, $R_{Ax} = 13214.25 , \text{N}$
  • Vertical Component, $R_{Ay} = F - R_B = 13214.25 , \text{N}$

The normal reaction at point B is equal to the hydrostatic force exerted by the water: $$ R_B = \text{Total force} = 13214.25 , \text{N} $$

Steps to Solve

  1. Determine the hydrostatic pressure The hydrostatic pressure at a depth $h$ in a fluid is given by the equation: $$ P = \rho g h $$ where $\rho$ is the density of water (approximately $1000 , \text{kg/m}^3$), $g$ is the acceleration due to gravity (approximately $9.81 , \text{m/s}^2$), and $h$ is the depth ($3 , \text{m}$ in this case).

  2. Calculate the total force exerted by the water The total force acting on the gate due to water pressure can be calculated by integrating the pressure over the area: $$ F = \int_0^{3} P , dA $$ Given that $A = b \cdot h$, with $b$ as the width of the gate ($3 , \text{m}$): $$ F = \int_0^{3} \rho g h \cdot b , dh = \rho g b \int_0^{3} h , dh $$

  3. Evaluate the integral The integral evaluates to: $$ \int_0^{3} h , dh = \left[ \frac{h^2}{2} \right]_0^{3} = \frac{3^2}{2} - 0 = \frac{9}{2} $$ Thus, $$ F = \rho g b \cdot \frac{9}{2} $$

  4. Substituting known values Substituting the known values into the total force equation: $$ F = 1000 , \text{kg/m}^3 \cdot 9.81 , \text{m/s}^2 \cdot 3 \cdot \frac{9}{2} $$

  5. Calculate the force Now calculating the force: $$ F = 1000 \cdot 9.81 \cdot 3 \cdot \frac{9}{2} = 13214.25 , \text{N} $$

  6. Determine the location of the center of pressure The location of the center of pressure from the water surface is given by: $$ z_{cp} = \frac{I_G}{A \cdot \bar{y}} + \bar{y} $$ where $I_G$ is the second moment of area about the surface, $A$ is the area, and $\bar{y}$ is the centroid distance from the surface.

  7. Calculate the moment about hinge A To find the horizontal and vertical components of the reaction forces at hinge A, set up moment equilibrium equations about hinge A and include vertical and horizontal reactions at points A and B.

  8. Resolve the reactions From the free body diagram, resolve horizontal and vertical components using:

  • Vertical equilibrium: $\sum F_y = R_A - F + R_B = 0$
  • Horizontal equilibrium: $\sum F_x = R_{Ax} - F_{h} = 0$

The horizontal and vertical components of the reaction at hinge A are as follows:

  • Horizontal Component, $R_{Ax} = 13214.25 , \text{N}$
  • Vertical Component, $R_{Ay} = F - R_B = 13214.25 , \text{N}$

The normal reaction at point B is equal to the hydrostatic force exerted by the water: $$ R_B = \text{Total force} = 13214.25 , \text{N} $$

More Information

The problem involves calculating the forces acting on a submerged gate due to hydrostatic pressure. The hydrostatic pressure increases linearly with depth, affecting the reactions at the hinges and support points.

Tips

  • Forgetting to account for the width of the gate: Ensure you include the width when calculating total force.
  • Misapplying the pressure formula: Remember that pressure depends on the depth of the fluid.
  • Ignoring moments about the hinge: Always consider the moments about hinge points to solve for reactions accurately.
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