Question

A block of aluminum that has dimensions \(2.00 \mathrm{cm}\) by \(3.00 \mathrm{cm}\) by \(5.00 \mathrm{cm}\) is suspended from a spring scale. (a) What is the weight of the block? (b) What is the scale reading when the block is submerged in oil with a density of \(850 \mathrm{kg} / \mathrm{m}^{3} ?\)

Short answer

(a) The block's weight in air is 0.79461 N. (b) The scale reads 0.545175 N when submerged in oil.

Step-by-step solution

Calculate the Volume of the Block

The volume of the block is calculated using the formula \( V = ext{length} \times ext{width} \times ext{height} \). Substitute the given dimensions: \( V = 2.00 \, \mathrm{cm} \times 3.00 \, \mathrm{cm} \times 5.00 \, \mathrm{cm} = 30.00 \, \mathrm{cm}^3 \).

Convert Volume to Cubic Meters

To use the density formula in consistent SI units, convert the volume from cubic centimeters to cubic meters. \( 1 \, \mathrm{cm}^3 = 1 \, \times 10^{-6} \, \mathrm{m}^3 \). Therefore, \( V = 30.00 \, \mathrm{cm}^3 \times 1 \, \times 10^{-6} = 3.00 \, \times 10^{-5} \, \mathrm{m}^3 \).

Calculate the Weight of the Block in Air

The weight of the block can be calculated using the formula \( W = m \cdot g \), where \( m \) is the mass and \( g = 9.81 \, \mathrm{m/s}^2 \). To find the mass, use density \( \rho_{\text{aluminum}} = 2700 \, \mathrm{kg/m}^3 \): \( m = \rho \cdot V = 2700 \, \mathrm{kg/m}^3 \times 3.00 \times 10^{-5} \, \mathrm{m}^3 = 0.081 \, \mathrm{kg} \). Therefore, \( W = 0.081 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s}^2 = 0.79461 \, \mathrm{N} \).

Calculate the Buoyant Force when Submerged

When the block is submerged, it experiences a buoyant force given by \( F_b = \rho_{\text{oil}} \times V \times g \). Given \( \rho_{\text{oil}} = 850 \, \mathrm{kg/m}^3 \), \( F_b = 850 \, \mathrm{kg/m}^3 \times 3.00 \times 10^{-5} \, \mathrm{m}^3 \times 9.81 \, \mathrm{m/s}^2 = 0.249435 \, \mathrm{N} \).

Calculate the Scale Reading

The scale reading is the weight of the block in air minus the buoyant force: \( ext{Scale reading} = W - F_b = 0.79461 \, \mathrm{N} - 0.249435 \, \mathrm{N} = 0.545175 \, \mathrm{N} \).

Key concepts

Buoyant Force

Understanding buoyant force is crucial for solving problems involving objects submerged in fluids. Buoyant force is the upward force that a fluid exerts on an object submerged in it. This force is what allows objects to float or seem lighter when they are in a fluid like water or oil.

The principle behind buoyant force is called Archimedes' principle. It states that the buoyant force on an object is equal to the weight of the fluid displaced by that object.

• Buoyant Force Formula: \[ F_b = \rho_f \times V \times g \]
  where:
  • \(F_b\) is the buoyant force.
  • \(\rho_f\) is the fluid density.
  • \(V\) is the volume of the object submerged.
  • \(g\) is the acceleration due to gravity, approximately \(9.81 \, \mathrm{m/s^2}\).

This force plays a significant role in the calculation of the apparent weight of an object in a fluid, as seen in exercises where submerged objects are measured for their scale reading.

Density

Density is a measure of how much mass is contained in a given volume of a substance. It is a key concept in fluid mechanics, particularly when dealing with buoyant forces.

• Density Formula: \[ \rho = \frac{m}{V} \]
  where:
  • \(\rho\) is the density.
  • \(m\) is the mass.
  • \(V\) is the volume.

In practical problems, like our exercise, knowing the density of the fluid and the object (aluminum in this case) allows us to determine whether the object will float or how much buoyant force it will experience.
The density of the oil, in our exercise with a value of \(850 \, \mathrm{kg/m^3}\), directly affects the buoyant force and thus the apparent weight of the aluminum block when submerged.

Volume Calculation

Calculating the volume of an object is an essential step when assessing how much space it occupies and for further calculations like buoyant force or mass.

For a rectangular block, like the aluminum block in the exercise, the volume is calculated by multiplying its length, width, and height:

• Volume Formula: \[ V = \text{length} \times \text{width} \times \text{height} \]

In the example, dimensions are given: \(2.00 \, \mathrm{cm} \times 3.00 \, \mathrm{cm} \times 5.00 \, \mathrm{cm}\), resulting in a volume of \(30.0 \, \mathrm{cm^3}\). This volume is often converted to cubic meters for consistency with SI units, as 1 cm³ is \(1 \times 10^{-6} \, \mathrm{m^3}\). This step ensures our calculations are precise, especially when we are using the density in \(\mathrm{kg/m^3}\).

Weight Calculation

Weight calculation is fundamental when determining how much an object presses downwards under the influence of gravity.

The weight of an object is calculated using its mass and the gravitational pull:

• Weight Formula: \[ W = m \cdot g \]
  where:
  • \(W\) is the weight.
  • \(m\) is the mass.
  • \(g\) is the acceleration due to gravity (\(9.81 \, \mathrm{m/s^2}\)).

In our problem, converting volume to mass through density is key to finding weight. The mass is given by \(m = \rho \cdot V\), where \(\rho\) is the density of the aluminum. By knowing mass and using the gravitational constant, we find the weight of the aluminum block in the air. The calculated weight in the exercise is \(0.79461 \, \mathrm{N}\), demonstrating how density, volume, and gravity interplay to determine an object's weight.