To determine the total pressure on a vertically placed circular plate submerged in water and to find the position of the center of pressure, follow these steps:
### Given Information:
- Depth of the center of the plate below the free surface, [tex]\( h_{\text{center}} = 3 \text{ meters} \)[/tex].
- Diameter of the plate, [tex]\( d = 1.5 \text{ meters} \)[/tex].
- Water density, [tex]\( \rho = 1000 \text{ kg/m}^3 \)[/tex].
- Acceleration due to gravity, [tex]\( g = 9.81 \text{ m/s}^2 \)[/tex].
### Step-by-Step Solution:
1. Calculate the Radius of the Circular Plate:
- The radius [tex]\( r \)[/tex] is half of the diameter.
[tex]\[
r = \frac{d}{2} = \frac{1.5 \, \text{m}}{2} = 0.75 \, \text{m}
\][/tex]
2. Calculate the Area of the Circular Plate:
- The area [tex]\( A \)[/tex] of a circle is given by [tex]\( \pi r^2 \)[/tex].
[tex]\[
A = \pi r^2 = \pi (0.75 \, \text{m})^2 = 1.767145867 \, \text{m}^2
\][/tex]
3. Calculate the Pressure at the Center of the Plate:
- The pressure at depth [tex]\( h_{\text{center}} \)[/tex] can be calculated using the hydrostatic pressure formula [tex]\( P = \rho g h \)[/tex].
[tex]\[
P_{\text{center}} = \rho g h_{\text{center}} = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 3 \, \text{m} = 29430 \, \text{Pa}
\][/tex]
4. Calculate the Total Force on the Plate:
- The total force [tex]\( F_{\text{total}} \)[/tex] on the plate is the pressure at the center times the area of the plate.
[tex]\[
F_{\text{total}} = P_{\text{center}} \times A = 29430 \, \text{Pa} \times 1.767145867 \, \text{m}^2 = 52007.10 \, \text{N}
\][/tex]
5. Calculate the Position of the Center of Pressure:
- The center of pressure [tex]\( y_p \)[/tex] for a vertical plate submerged in a fluid is given by:
[tex]\[
y_p = h_{\text{center}} + \frac{I_g}{h_{\text{center}} \cdot A}
\][/tex]
where [tex]\( I_g \)[/tex] is the second moment of area (or moment of inertia) about the centroidal axis, which for a circular plate is [tex]\( \frac{\pi r^4}{4} \)[/tex].
[tex]\[
I_g = \frac{\pi r^4}{4} = \frac{\pi (0.75 \, \text{m})^4}{4} = 0.049087 \, \text{m}^4
\][/tex]
- Substituting into the formula for [tex]\( y_p \)[/tex]:
[tex]\[
y_p = 3 \, \text{m} + \frac{0.049087 \, \text{m}^4}{3 \, \text{m} \times 1.767145867 \, \text{m}^2} = 3.046875 \, \text{m}
\][/tex]
### Final Answers:
- The total force on the plate is [tex]\( 52007.10 \, \text{N} \)[/tex].
- The position of the center of pressure is [tex]\( 3.046875 \, \text{m} \)[/tex] below the free surface.
