Find the expression for hydrostatic force on a square

Question

Find the expression for hydrostatic force on a square as it rotated under the water with one vertex always on the surface of water. Break down the square in three parts and find the force as a function of angle theta then add them up to find the TOTAL FORCE ON THE PLATE.

 

This question regards integral in Calculus. Need help immediately!!!!!

Answer 1

I cannot be sure this is correct or if this makes sense but I will give it a try.

 

Draw an xy coordinate system with the y axis pointing down and the x-axis pointing left.  The x axis is the surface of the water.  Draw a square in the 1st quadrant whose sides are of length a.  Label the vertices of the square A, B, C and D.  A is the vertex at the surface of the water that the square rotates about.  Side AB should point to left.  The vertices have coordinates of A(0,0), B(a,0), C(a,a) and D(0,a).  We can use a rotational matrix to rotate these points in a plane CCW through an angle theta or by noting that C is 45 degrees from B and D is 90 degrees from B.  The new coordinates are B(acos(θ), asin(θ)), C(acos(θ)-asin(θ), asin(θ)+acos(θ)) and D(-asin(θ), acos(θ)) where theta is CCW from the positive x-axis.  Redraw the square with side AB rotated slightly.  Notice the position of vertex B and D which parts the square up into three regions.  The depth to a section in each of these regions is y_i.  Note that the variable of integration is y not theta.  Now write the equations of the lines AB and AD using the new coordinates.  For line AB we have m=(asin(θ)-0)/(acos(θ)-0) = tan(θ).  Note that the slope of line CD is the same.  Writing the equation of the line AB we have y-0 = tan(θ)(x-0) giving y = xtan(θ).  Solve this equation for x we have x = ycot(θ).  Now doing the same for line AD we have y = -xcot(θ) and solving for x we have x = -ytan(θ).  Consider only the top part, from A to B and let y be the depth.  We find the width of the ith section as w_i = x_AB - x_AD = ycot(θ)+ytan(θ) = y(tan(θ)+cot(θ)).  Next find the area of the ith section as Ai = w_iΔy =  y(tan(θ)+cot(θ))Δy.  Find the pressure as P = δd = δy.  Find the force on the ith section as F = PA = δy(y(tan(θ)+cot(θ)).  Now sum up all the sections from A to B to find the total force for the top half.  We have Integral from 0 to asin(θ) {which is the y-coordinate of B} of  δy²(tan(θ)+cot(θ))dy gives

 

Ftop = δa³sin³(θ)(cot(θ) + tan(θ))/3

 

For the middle use line BC and AD and do a similar process as above.  You will integrate from the y-coordinate of vertex B to the y-coordinate of vertex D.  For the last section use lines BC and CD and integrate from the y-coordinate of D to the y-coordinate of C.  The other integrals I came up with is as follows

 

Middle.  Integral from asin(θ) to acos(θ) of δasec(θ)ydy,  I simplified to get sec(θ)

 

Bottom:  Integral from acos(θ) to asin(θ)+acos(θ) of δ(y2(-tan(θ)-cot(θ))+ay(sec(θ)+csc(θ)))dy

 

My final answer is

 

F = δa³sin³(θ)(cot(θ)+tan(θ))/3+δa³sec(θ)(cos²(θ)-sin²(θ))/2-δa3(tan(θ)+cot(θ))( (asin(θ)+acos(θ))³-cos³(θ))/3+δa3(sec(θ)+csc(θ))(( asin(θ)+acos(θ))²-cos²(θ))/2