Question

The zoo is building a new polar bear exhibit, and wants to put a semi-circular window in the concrete wall of the swimming tank. (Note: picture is not to scale)

If the semi-circle has diameter 70 centimeters, and the bottom of the window is at a depth of 2.5 meters, find the hydrostatic force on the window.

Answer 1

The hydrostatic force on the semi-circular window can be found by calculating the pressure exerted by the water at a depth of 2.5 meters and then multiplying it by the area of the window. So, the value hydrostatic force is 9432500000 N.

Step-by-step explanation:

To find the hydrostatic force on the semi-circular window, we need to calculate the pressure exerted by the water at a depth of 2.5 meters. The pressure exerted by a fluid at a certain depth is given by the formula P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth. In this case, the fluid is water, which has a density of approximately 1000 kg/m³, and g is approximately 9.8 m/s².

First, we convert the depth from meters to centimeters: 2.5 meters = 250 centimeters.

Next, we substitute the values into the formula:

P = (1000 kg/m³)(9.8 m/s²)(250 cm) = 2,450,000 g cm²/s².

Finally, we convert the pressure from g cm²/s² to N/m²:

1 N/m² = 1 Pascal (Pa), so the pressure is 2,450,000 Pa.

To find the hydrostatic force on the semi-circular window, we need to calculate the area of the window and then multiply it by the pressure. The area of a semi-circle is given by the formula A = πr²/2, where A is the area and r is the radius. In this case, the radius is half the diameter, so r = 70 cm/2 = 35 cm.

Substituting the value into the formula:

A = π(35 cm)²/2 = 3850 cm².

Finally, we multiply the pressure by the area to find the hydrostatic force:

F = PA = (2,450,000 Pa)(3850 cm²) = 9432500000 N.

Answer 2

The hydrostatic force acting on the semi-circular window at a depth of 2.5 meters is approximately
`\(4689 \, \text{N}\)`.

To find the hydrostatic force on the window, we can use the formula for hydrostatic pressure:

`\[ P = \rho \cdot g \cdot h \]`

Where:

- P is the pressure at a certain depth

-
`\( \rho \)` is the density of the fluid (assumed to be water, with a density of
`\(1000 \, \text{kg/m}^3\)`)

- g is the acceleration due to gravity
`(\(9.81 \, \text{m/s}^2\))`

- h is the depth of the fluid

First, let's convert the depth of the window into meters (since the density of water is typically given in kilograms per cubic meter):

Depth of window = 2.5 meters

The hydrostatic pressure at the depth of the window will act over the entire surface area of the window facing the water.

The area of a semi-circle with diameter 70 centimeters is
`\( (\pi \cdot d^2)/(8) \)`, where d is the diameter.

Given the diameter is 70 centimeters, the radius ( r ) is half of that:
`\( r = \frac{70 \, \text{cm}}{2} = 35 \, \text{cm} = 0.35 \, \text{m} \)`.

So, the area of the semi-circle window facing the water is:

`\[ A = (\pi \cdot 0.35^2)/(2) = (\pi \cdot 0.1225)/(2) \approx 0.191 \, \text{m}^2 \]`

Now, we can calculate the hydrostatic force acting on this area at a depth of 2.5 meters:

`\[ \text{Pressure} = \rho \cdot g \cdot h \]`

`\[ \text{Force} = \text{Pressure} * \text{Area} \]`

`\[ \text{Pressure} = 1000 \, \text{kg/m}^3 * 9.81 \, \text{m/s}^2 * 2.5 \, \text{m} = 24525 \, \text{Pa} \]`

`\[ \text{Force} = 24525 \, \text{Pa} * 0.191 \, \text{m}^2 \approx 4689 \, \text{N} \]`