Question

A 2-ft-high and 6-ft-wide rectangular plate is submerged vertically in water so that the top is 1 ft below the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)

Answer 1

Answer: F = 211312.5ft/lb.

Step-by-step explanation: Hydrostatic pressure is pressure caused by a fluid due to the force of gravity. It is calculated following the formula:

P = ρgh.

ρ is the fluid density (62.5 lb/ft³);

g is the gravitacional acceleration( 32.2ft/s²);

h is the height of the fluid columm;

P = 62.5.32.2.h

Pressure at h = 1ft:

P = 62.5.32.2.1

P = 2012.5 psi.

Now, hydrostatic force is F = $\frac{P}{A}$

The pressure exerted on the surface is not constant. It depends on the height. So, to evaluate the force:

dF = P(h) dA

Since it a rectangular plate, dA = w.dh, where w is its width and equals 6 and dh is its height element.

That gives: dF = P(h)w.dh

$\int\limits^a_b {2012.5hw} \, dh$ = 2012.5.6. $\int\limits^6_1 {h} \, dh$ = 2012.5.6($\frac{h^{2} }{2}$) = 12075($\frac{6^{2} }{2} - \frac{1^{2} }{2}$) = 211312.5ft/lb.