Question

Consider a 200-ft-high, 1200-ft-wide dam filled to capacity. Determine (a) the hydrostatic force on the dam and (b) the force per unit area of the dam near the top and near the bottom. Note: we will see that the resultant hydrostatic force will be

Answer 1

Answer:

a)  $F_g=1.5*10^9Ibf$

b)  $F_t=12490Ibf/ft^2$

     $F_b=0$

Explanation:

From the question we are told that:

Height $h=200ft$

Width $w=1200ft$

a)

Generally the equation for Dam's Hydro static force is mathematically given by

$F_g=\rho*g*\frac{h}{2}(w*h)$

Where

$\rho=Density\ of\ water$

$\rho=62.4Ibm/ft^3$

Therefore

$F_g=62.4*32.2*\frac{200}{2}(1200*200)$

$F_g=1.5*10^9Ibf$

b)

Generally the equation for Dam's Force per unit area is mathematically given by

$F=\rho*g*h$

For Top

$F_t=\rho*g*h$

$F_t=62.4*32.2*200$

$F_t=12490Ibf/ft^2$

For bottom

$Here \\H=0 zero$

Therefore

$F_b=0$

Answer 2

The hydrostatic force on the dam is $2.995 \times 10^9 \ lbF$.

The force per unit area near the top is 86.74 psi.

The force per unit area near the bottom is zero.

Hydrostatic force

The hydrostatic force on the dam is the force exerted on the dam by the column of the water.

$F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF$

Force per unit area near the top

The force per unit area is the pressure exerted near the top of the dam.

$P = \rho gh\\\\P = 0.052 \times \rho h$

where;

P is pressure in PSI

ρ is density of water in lb/gal

h is the vertical height in ft

$P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi$

The pressure near the bottom is zero, become the vertical height is zero.

Learn more about hydrostatic pressure here: https://brainly.com/question/11681616