Fluid Force (Calculus 2)

Question

Set up the integral to compute the magnitude of the fluid force (as a multiple of ρg) on the semicircular face of the half cylindrical tank with cross section given by under the following conditions:

a. vertical axis runs downward (i.e. increasing positive values of the variable lie along the bottom half of the axis) with origin at the center of the tank.

b. vertical axis runs downward with origin at the fluid surface (12ft. above the bottom of the tank).

c. vertical axis runs upward (i.e. increasing positive values of the variable lie along the top half of the axis), with origin at the fluid surface.

d. vertical axis runs upward with origin at the bottom of the tank. To ‘show your work’, draw a carefully detailed sketch or two that indicates where the terms a, b, D(h) and W(h) in your integral come from based on the given picture (see Background on the next page). You may also use written comments.

Answer 1

Let r= radius of semicircle, w= width of tank at h and h = depth of fluid. Thus dA=2wdh ... dh= differential of depth. Pressure = rho*g*h. Force = P*A and dF= Pdh. The width of the tank at h will be 2sqrt(r^2-(r-h)^2)=2sqrt(2rh-h^2). Thus dF = 2rho*g*hsqrt(2rh-h^2)dh