Problem 22 - Basic Technical Mathematics with Calculus

Question

Find the force on one side of a cubical container \(6.0 \mathrm{cm}\) on an edge if the container is filled with mercury. The density of mercury is \(133 \mathrm{kN} / \mathrm{m}^{3}\).

Step-by-Step Solution

Answer

The force on one side of the container is 282.11 N.

Step 1: Identify the Known Quantities

The container is a cube with an edge of length of 6.0cm. Convert this length into meters: \(6.0 \text{ cm} = 0.06 \text{ m}\). The density of mercury is given as \(133 \text{ kN/m}^3 = 133,000 \text{ N/m}^3\).

Step 2: Determine the Area of One Side

Since the container is a cube, the area of one side is a square with a side length of 0.06 m. The area \(A\) can be calculated as: \[A = (0.06)^2 = 0.0036 \text{ m}^2\].

Step 3: Calculate Pressure Exerted by Mercury

The pressure exerted by a liquid column on the base is given by \(P = \rho \cdot g \cdot h\), where \(\rho\) is the density, \(g\) is the acceleration due to gravity \(9.81 \text{ m/s}^2\), and \(h\) is the height (which is the same as the side length of the cube, 0.06 m). \[P = 133,000 \times 9.81 \times 0.06 = 78,364.2 \text{ N/m}^2\].

Step 4: Calculate the Force on One Side

The force \(F\) on one side of the container is given by \(F = P \cdot A\). Substituting the values from the previous steps: \[F = 78,364.2 \times 0.0036 = 282.11 \text{ N}\].

Key Concepts

Density

Density is a fundamental concept in physics that relates the mass of a substance to the volume it occupies. It’s essentially a measure of how much matter is packed into a given space. The formula for density is:

\[ \rho = \frac{m}{V} \]where

• \( \rho \) is the density,
• \( m \) is the mass, and
• \( V \) is the volume.

In this exercise, mercury's density is integral to determining the pressure it exerts inside the container. Mercury's density is exceptionally high, which is why it's classified as a heavy liquid. This high density translates into a substantial weight even over small volumes, like our cubical container of 6 cm on each side. Thus, understanding density helps us predict how substances will behave under different conditions.

Pressure

Pressure is the force exerted per unit area. It's an important concept when dealing with liquids like mercury in a container. Pressure can be calculated using the equation:

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

• \( P \) is the pressure,
• \( \rho \) is the density of the liquid,
• \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \) on Earth),
• \( h \) is the height of the liquid column.

In this scenario, the pressure exerted by mercury on the container's sides results from its density, gravity, and the cube's height. Because the mercury’s density is given as \( 133,000 \text{ N/m}^3 \), this high value plays a substantial role in calculating the pressure that urges against the walls of the cubical container.

Force Calculation

Calculating the force exerted on an object involves determining the product of pressure and area. Once we've established the pressure from the mercury in the cube, we can easily compute the force using:

\[ F = P \cdot A \]where

• \( F \) is the force,
• \( P \) is the pressure calculated earlier in the exercise,
• \( A \) is the area of one side of the cube.

By converting the side length to meters (0.06 m), the area of one side becomes \( 0.0036 \text{ m}^2 \). Multiplying this area by the pressure we calculated (\( 78,364.2 \text{ N/m}^2 \)), we find the force exerted by the mercury is approximately \( 282.11 \text{ N} \). This calculation ties together density, area, and pressure into one straightforward result: the force on the container's side.

Unit Conversion

Unit conversion is critical in solving physics problems, as it ensures that all quantities are compatible for calculation. In this problem, the side length of the container was initially given in centimeters, a common unit for small measurements. However, it needed to be converted to meters for the calculations.

Since the SI unit for length in physics is the meter, converting 6 cm to 0.06 m was necessary. Similarly, the density of mercury was initially expressed in kN/m\(^3\), which we then converted to N/m\(^3\) for consistency with the pressure and force calculations. Correct unit conversion prevents errors and ensures accuracy, allowing for seamless integration of multiple concepts, like density and force, resulting in accurate and understandable physics calculations.