The bottom of the tank is defined by the ODE dy/dx = x + 0.5y from a coordinate system that is rotated downward 45° from the horizontal. The gate exists between two vertical walls 2 m apart as shown, and the tank is 5 m long and contains water that is 5 m deep at the left wall. Find the vertical

Question

The bottom of the tank is defined by the ODE dy/dx = x + 0.5y from a coordinate system that is rotated downward 45° from the horizontal. The gate exists between two vertical walls 2 m apart as shown, and the tank is 5 m long and contains water that is 5 m deep at the left wall. Find the vertical component of force (and its location) on the bottom.

Accepted Answer

The program below obtains the solution. Because of the rotated coordinate system, relationships must be established between the movement in the x direction and in the horizontal direction, denoted by x′ in the above sketch. The differential area to integrate is dA = h dx′, where h is the distance from the bottom of the tank to the water surface, and from the above sketch is seen to be h = H + x sinθ − y cosθ, where H is the depth of water at the origin. Also from the sketch x′ = x cosθ + y sinθ. Since for 45° the sine = cosine, one value will be used as the solution is programmed. While there are alternative approaches, for this problem the ODE will be solved first to obtain the shape of the bottom, and the results stored in an array, YY, and when needed the results from this array will be interpolated. The reason for doing this is that the numerical integration is in terms of x′, and the corresponding value of x (along the rotated axis) is needed, but the only way to determine this x is to note that x′/sin45° = x + y. For any x′, therefore, the table is searched for the entry where x + y is just larger than x′/sin45°, and then interpolated (linearly) to find x. This is done by noting that between two entries in the table y = [latex]y_{o}[/latex] + (Δy/Δx)(x − [latex]x_{o}[/latex]), where subscript o denotes first entry. Thus, x′/sin45° = [latex]y_{o}[/latex] + (Δy/Δx)(x − [latex]x_{o}[/latex]) + x, or solving for x, x = (x′/sin45° − [latex]y_{o}[/latex] + (Δy/Δx)[latex]x_{o}[/latex])/(Δy/Δx + 1). Once x is determined, y is interpolated as y = [latex]y_{o}[/latex] + (x − [latex]x_{o}[/latex])/Δx([latex]y_{1} − y_{o}[/latex]). Note in the program below that the ODE is now solved first in the MAIN program, and that the function EQUAT does the necessary interpolation to obtain h so that Simpson’s Rule can properly evaluate the area. Giving as input 0 2 5 0. 9.801 1.e − 5 30, the solution gives Area = 9.437 m² per meter of basin length, resulting in a force = 92.491 kN/m, or [latex]F_{V}[/latex] = 5 × 92.491 = 462.46 kN. The first moment of area is determined by changing the statement as shown by the comment, and the result of the integration gives [latex]x_{c}[/latex] A = 8.824, so the position of this vertical force is [latex]x_{c}[/latex]= 8.824/9.437 = 0.935 m.

Question B.4: The bottom of the tank is defined by the ODE dy/dx = x + 0.5...

Related Answered Questions

Integrate the function f(x) = x²(x² − 2)sin(x) between the limits of 0 and π/2 using first the trapezoidal rule and then Simpson’s rule and compare the results with the exact integral. ...

Verified Answer:

Find the vertical component of force (including its location) on the bottom surface that is 3 m long with water standing to a height of 6 m. The surface is defined by the equation y = x + 0.5(1 − cos x). The distance between vertical walls is 3 m. ...

Verified Answer:

Verified Answer:

We want to generate a smooth function for plotting x versus y using cubic splines through the following six pairs of (Xj, Yj) values: (0, 0), (5, 2), (9, 9), (14, 15), (18, 17), and (22, 19). ...

Verified Answer:

Verified Answer: