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A paperweight, when weighed in air, has a weight of \(W=6.9 \mathrm{N}\). When completely immersed in water, however, it has a weight of \(W_{\text {in water }}=\) \(4.3 \mathrm{N} .\) Find the volume of the paperweight.

Short Answer

Expert verified
The volume of the paperweight is approximately \( 2.65 \times 10^{-4} \text{ m}^3 \).

Step by step solution

01

Understanding Buoyant Force

To find the volume of the paperweight, we need to consider the buoyant force. The buoyant force is the difference between the weight of the paperweight in air and the weight in water. It is given by:\[ F_b = W - W_{\text{in water}} = 6.9 \text{ N} - 4.3 \text{ N} = 2.6 \text{ N} \]
02

Applying Archimedes' Principle

Archimedes' principle states that the buoyant force on an object submerged in liquid is equal to the weight of the liquid displaced by the object. The weight of the displaced water can be expressed as: \[ F_b = V \cdot \rho \cdot g \] where \( V \) is the volume of the displaced water (which equals the volume of the paperweight), \( \rho \) is the density of water (approximately \( 1000 \text{ kg/m}^3 \)), and \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \)).
03

Solving for Volume

Rearranging the equation \( F_b = V \cdot \rho \cdot g \) for \( V \), we have:\[ V = \frac{F_b}{\rho \cdot g} \]Substituting the known values, we get:\[ V = \frac{2.6 \text{ N}}{1000 \text{ kg/m}^3 \cdot 9.81 \text{ m/s}^2} = \frac{2.6}{9810} \text{ m}^3 \]\[ V \approx 2.65 \times 10^{-4} \text{ m}^3 \]
04

Conclusion

The volume of the paperweight is approximately \( 2.65 \times 10^{-4} \text{ m}^3 \). This volume represents the amount of water displaced by the paperweight when it is submerged.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Archimedes' principle
Archimedes' principle is a fundamental concept in fluid mechanics that helps us understand the behavior of objects in liquids. According to this principle, an object submerged in a liquid experiences an upward force, known as the buoyant force. This force is exactly equal to the weight of the fluid that the object displaces.

Using this insight, you can determine how an object will behave when placed in a fluid.
  • If the object’s weight is greater than the buoyant force, it will sink.
  • If the weight is less, it will float.
  • If the forces are equal, the object will remain suspended, neither rising nor sinking.

Archimedes' principle not only helps in finding the buoyant force but also allows us to calculate the volume of an object that is submerged in the fluid, as seen in our initial problem. Understanding this principle is crucial for solving problems related to floating and submerged objects.
Density of Water
The density of water is a vital factor when solving problems involving buoyancy and fluid displacement. Density is defined as the mass per unit volume of a substance, and for water, it is approximately \( 1000 \text{ kg/m}^3 \) at room temperature.

This constant allows us to calculate the weight of water displaced by an object submerged in it.
  • The formula \( \rho = \frac{m}{V} \) describes the relationship, where \( \rho \) is density, \( m \) is mass, and \( V \) is volume.
  • Using this with Archimedes' principle, the density of water helps determine the volume of an otherwise irregularly shaped object.

In situations where objects are submerged in other liquids, knowing the density of the fluid becomes essential. Without accurate density values, calculating buoyant forces and displaced volumes would be impractical.
Acceleration Due to Gravity
The acceleration due to gravity, usually denoted as \( g \), is an important constant in many physics-related calculations involving weight and buoyancy. On Earth, its value is approximately \( 9.81 \text{ m/s}^2 \). This value plays a crucial role when calculating forces like weight and the buoyant force.

In the context of buoyancy:
  • The force of gravity acting on a immersed object is counteracted by the buoyant force.
  • This balance of forces dictates whether an object floats, sinks, or remains neutrally buoyant.
  • The equation \( F = m \cdot g \) is frequently used to calculate the gravitational force on an object.

Therefore, knowing the value of \( g \) is indispensable for performing accurate computations when dealing with objects under the influence of gravity within a fluid medium.

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Most popular questions from this chapter

One of the concrete pillars that support a house is \(2.2 \mathrm{m}\) tall and has a radius of \(0.50 \mathrm{m}\). The density of concrete is about \(2.2 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) Find the weight of this pillar in pounds \((1 \mathrm{N}=0.2248 \mathrm{lb})\)

A mercury barometer reads \(747.0 \mathrm{mm}\) on the roof of a building and \(760.0 \mathrm{mm}\) on the ground. Assuming a constant value of \(1.29 \mathrm{kg} / \mathrm{m}^{3}\) for the density of air, determine the height of the building.

An airplane wing is designed so that the speed of the air across the top of the wing is \(251 \mathrm{m} / \mathrm{s}\) when the speed of the air below the wing is \(225 \mathrm{m} / \mathrm{s} .\) The density of the air is \(1.29 \mathrm{kg} / \mathrm{m}^{3} .\) What is the lifting force on a wing of area \(24.0 \mathrm{m}^{2} ?\)

A hand-pumped water gun is held level at a height of \(0.75 \mathrm{m}\) above the ground and fired. The water stream from the gun hits the ground a horizontal distance of \(7.3 \mathrm{m}\) from the muzzle. Find the gauge pressure of the water gun's reservoir at the instant when the gun is fired. Assume that the speed of the water in the reservoir is zero and that the water flow is steady. Ignore both air resistance and the height difference between the reservoir and the muzzle.

A water bed for sale has dimensions of \(1.83 \mathrm{m} \times 2.13 \mathrm{m} \times 0.229 \mathrm{m}\) The floor of the bedroom will tolerate an additional weight of no more than \(6660 \mathrm{N}\). Find the weight of the water in the bed and determine whether the bed should be purchased.

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