below, a vertical plate is submerged in water. compute the hydrostatic force against one side of the plate…

Explanation:

Step1: Recall hydro - static force formula

The hydro - static force $F$ on a submerged vertical plate is given by $F=\int_{a}^{b}\rho g h(x)w(x)dx$, where $\rho$ is the density of the fluid, $g$ is the gravitational acceleration, $h(x)$ is the depth of the point $x$ and $w(x)$ is the width of the plate at depth $x$. For water, $\rho = 1000\ kg/m^{3}$, $g = 9.8\ m/s^{2}$. The depth $h(x)$ of a point on the plate varies from $h_1 = 4\ m$ to $h_2=4 + 10=14\ m$, and the width $w(x)=5\ m$ (constant).

Step2: Set up the integral

$F=\int_{4}^{14}\rho g h\cdot w\ dh$. Substituting $\rho = 1000\ kg/m^{3}$, $g = 9.8\ m/s^{2}$, and $w = 5\ m$ into the formula, we get $F=1000\times9.8\times5\int_{4}^{14}h\ dh$.

Step3: Integrate

We know that $\int h\ dh=\frac{1}{2}h^{2}+C$. Then $1000\times9.8\times5\int_{4}^{14}h\ dh=49000\left[\frac{1}{2}h^{2}\right]_{4}^{14}$.

Step4: Evaluate the definite integral

$49000\times\frac{1}{2}(h^{2}\big|_{4}^{14}) = 24500(14^{2}-4^{2})=24500(196 - 16)=24500\times180$.

Step5: Calculate the result

$24500\times180 = 4410000\ N$.

Answer:

$4410000\ N$