Understanding Center of Pressure in Fluid Mechanics

The center of pressure is the point on a submerged surface where the total hydrostatic force acts. For a plane submerged surface, the depth of the center of pressure below the free surface is given by the formula:

\[ h_{cp} = \bar{y} + \frac{I_G \sin^2 \theta}{A \bar{y}} \]

Where:

• \(h_{cp}\) is the depth of the center of pressure from the free surface.
• \(\bar{y}\) is the depth of the centroid of the submerged area from the free surface.
• \(I_G\) is the moment of inertia of the submerged area about an axis passing through its centroid and parallel to the free surface.
• \(\theta\) is the angle between the plane of the submerged surface and the free surface of the liquid.
• \(A\) is the area of the submerged surface.

Analyzing the Beaker and Water Height

In this problem, a beaker contains water up to a height of \(h/3\). We assume the beaker has vertical walls, so the submerged surface is a vertical rectangle.

• The height of the water (and the submerged surface) is \(H = h/3\).
• Let the width of the beaker (and the submerged surface) be \(b\).
• The submerged surface is vertical, so the angle \(\theta\) with the free surface is \(90^\circ\). Therefore, \(\sin^2 \theta = \sin^2 90^\circ = 1\).

Calculating Parameters for the Formula

We need to find \(\bar{y}\), \(A\), and \(I_G\) for the vertical rectangular submerged area of height \(H = h/3\) and width \(b\).

• Depth of the centroid (\(\bar{y}\)): For a rectangle, the centroid is at the middle of its height. Since the top of the submerged area is at the free surface, the depth of the centroid is half the height of the water column.

  \[ \bar{y} = \frac{H}{2} = \frac{h/3}{2} = \frac{h}{6} \]

• Area of the submerged surface (\(A\)): The area of a rectangle is width multiplied by height.

  \[ A = b \times H = b \times \frac{h}{3} = \frac{bh}{3} \]

• Moment of inertia (\(I_G\)): For a rectangle of width \(b\) and height \(H\) about its centroidal axis parallel to the base, the moment of inertia is \(I_G = \frac{bH^3}{12}\). Substituting \(H = h/3\):

  \[ I_G = \frac{b (h/3)^3}{12} = \frac{b (h^3/27)}{12} = \frac{b h^3}{12 \times 27} = \frac{b h^3}{324} \]

Calculating the Center of Pressure (\(h_{cp}\))

Now substitute the calculated values into the formula for \(h_{cp}\):

\[ h_{cp} = \bar{y} + \frac{I_G \sin^2 \theta}{A \bar{y}} \]

\[ h_{cp} = \frac{h}{6} + \frac{\left(\frac{b h^3}{324}\right) \times 1}{\left(\frac{bh}{3}\right) \times \left(\frac{h}{6}\right)} \]

Simplify the denominator:

\[ A \bar{y} = \left(\frac{bh}{3}\right) \times \left(\frac{h}{6}\right) = \frac{bh^2}{18} \]

Substitute back into the formula:

\[ h_{cp} = \frac{h}{6} + \frac{\frac{b h^3}{324}}{\frac{b h^2}{18}} \]

To simplify the fraction, multiply the numerator by the reciprocal of the denominator:

\[ h_{cp} = \frac{h}{6} + \frac{b h^3}{324} \times \frac{18}{b h^2} \]

Cancel out \(b\) and \(h^2\) from the second term:

\[ h_{cp} = \frac{h}{6} + \frac{18 h}{324} \]

Simplify the fraction \(\frac{18}{324}\): \(\frac{18}{324} = \frac{18 \div 18}{324 \div 18} = \frac{1}{18}\).

\[ h_{cp} = \frac{h}{6} + \frac{h}{18} \]

To add these fractions, find a common denominator, which is 18:

\[ \frac{h}{6} = \frac{h \times 3}{6 \times 3} = \frac{3h}{18} \]

\[ h_{cp} = \frac{3h}{18} + \frac{h}{18} = \frac{3h + h}{18} = \frac{4h}{18} \]

Simplify the final fraction:

\[ h_{cp} = \frac{4h \div 2}{18 \div 2} = \frac{2h}{9} \]

Conclusion

The location of the center of pressure on the vertical wall of the beaker, measured from the free surface of the water, is \(2h/9\).