The figure shows two fluids held by a hinged gate. The atmospheric pressure is Pa = 100 kPa. The moment per unit width about the base of the hinge is ____________ kNm/m. (Rounded off to one decimal place) Take the acceleration due to gravity to be g = 9.8 m/s² .

The correct answer is 57.2.

1. Analysis of the Given Figure and Data:
Based on the provided diagram, a vertical gate holds two layers of fluids:
• Upper fluid layer: height $h_1=1\text{m}$ and density ${\rho }_1=1000{\text{kg/m}}^3$
• Lower fluid layer: height $h_2=2\text{m}$ and density ${\rho }_2=2000{\text{kg/m}}^3$
• The atmospheric pressure $P_a=100\text{kPa}$ acts on both the free surface of the upper fluid and the dry right-hand side of the gate. Thus, the atmospheric pressure cancels out, and we can perform all calculations using gauge pressure.
• The gate is hinged at its base. Let us set a vertical coordinate axis $y$ originating at the hinge ($y=0$ at the base) and pointing vertically upwards. The gate spans from $y=0$ to the top of the fluid at $y=3\text{m}$.

2. Pressure Distribution (Gauge Pressure):
For the upper fluid layer ($2\le y\le 3$):
$P\left(y\right)={\rho }_{1}g(3-y)$
$P\left(y\right)=1000\times 9.8\times (3-y)=9800(3-y)\text{Pa}$

At the interface ($y=2\text{m}$), the gauge pressure is:
$P_{\text{interface}}=9800\times (3-2)=9800\text{Pa}$

For the lower fluid layer ($0\le y\le 2$):
$P\left(y\right)=P_{\text{interface}}+{\rho }_{2}g(2-y)$
$P\left(y\right)=9800+2000\times 9.8\times (2-y)=9800+19600(2-y)\text{Pa}$

3. Moment per Unit Width about the Hinge:
The moment per unit width about the base ($y=0$) is given by the integral:
$M={\int }_0^{3}yP\left(y\right)dy=I_1+I_2$

Where:
$I_1={\int }_2^{3}y\times 9800(3-y)dy$
$I_2={\int }_0^{2}y\left[9800+19600(2-y)\right]dy$

Evaluating $I_1$:
$I_1=9800{\int }_2^3(3y-y^2)dy=9800{\left[\frac{3y^2}{2}-\frac{y^3}{3}\right]}_2^3$
$I_1=9800\left[\left(\frac{27}{2}-9\right)-\left(6-\frac{8}{3}\right)\right]=9800\left[4.5-3.3333\right]=9800\times \frac{7}{6}\approx 11433.33\text{N·m/m}$

Evaluating $I_2$:
Simplify the integrand first:
$P\left(y\right)=9800+39200-19600y=49000-19600y$
$I_2={\int }_0^2(49000y-19600y^2)dy={\left[24500y^2-\frac{19600y^3}{3}\right]}_0^2$
$I_2=24500\left(4\right)-\frac{19600\left(8\right)}{3}=98000-52266.67\approx 45733.33\text{N·m/m}$

Total Moment:
$M=11433.33+45733.33=57166.67\text{N·m/m}=57.1667\text{kNm/m}$

Rounding to one decimal place, the moment per unit width is 57.2 kNm/m.