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The true weight of an object can be measured in a vacuum, where buoyant forces are absent. A measurement in air, however, is disturbed by buoyant forces. An object of volume \(V\) is weighed in air on an equal-arm balance with the use of counterweights of density \(\rho\). Representing the density of air as \(\rho_{\text {air }}\) and the balance reading as \(F_{g}^{\prime}\), show that the true weight \(F_{g}\) is $$ F_{g}=F_{g}^{\prime}+\left(V-\frac{F_{g}^{\prime}}{\rho g}\right) \rho_{\text {air }} g $$

Short Answer

Expert verified
The true weight \(F_{g}\) of an object weighed in air can be found using the equation \(F_{g} = F_{g}^{\prime} + (V - F_{g}^{\prime} / \rho) \rho_{\text {air }} g\).

Step by step solution

01

Understand the problem

An object's true weight can be measured accurately in a vacuum, where there are no buoyant forces. But in air, the buoyant force interferes with this measurement. We are given the volume of the object \(V\), the density of the counterweights \(\rho\), and the inaccurate weight measure \(F_{g}^{\prime}\). We are asked to find an equation for the true weight \(F_{g}\) in terms of these variables, the density of air \(\rho_{\text {air }}\), and the gravitational acceleration \(g\).
02

Understand buoyant force

The buoyant force on an object is equal to the weight of the fluid that the object displaces. It can be represented as \(F_{B} = V \rho_{\text {air }} g\). This is the additional upward force that the object experiences when weighed in air.
03

Apply the principle of balance

On the balance, the weight of the object (which is \(F_g\)) is balanced by the combined effects of the balance reading \(F_{g}^{\prime}\) and the buoyant force. As a result, we have \(F_{g} = F_{g}^{\prime} + F_{B}\).
04

Substitute for buoyant force in the equation

Substituting the expression for the buoyant force from step 2 into the equation from step 3 gives \(F_{g} = F_{g}^{\prime} + V \rho_{\text {air }} g\). But oceanographers usually have their balances calibrated using weights made of a known material, therefore the counterweights have a known density \(\rho\) and the buoyant force acting on the counterweights is also \(V \rho_{\text {air }} g = F_{g}^{\prime} / \rho\). Therefore, the adjusted factor for buoyant force measured in air is \(V - F_{g}^{\prime} / \rho\). So, \(F_{g} = F_{g}^{\prime} + (V - F_{g}^{\prime} / \rho) \rho_{\text {air }} g\) as required.

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Key Concepts

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

Buoyant Force
The buoyant force is a fundamental concept in understanding how objects behave in a fluid, such as air or water. When an object is placed in a fluid, it displaces a volume of that fluid. According to the principles of physics, the fluid exerts an upward force on the object equal to the weight of the fluid displaced. This force is what we call buoyant force.

What Determines Buoyant Force?

The magnitude of buoyant force depends on two primary factors: the volume of the displaced fluid (\(V\text{, in cubic meters or liters}\)) and the fluid’s density (\(rho_{text{air}}\text{, in kilograms per cubic meter}\)). The gravitational acceleration (\(g\text{, in meters per second squared}\)) is also a crucial factor because it influences the weight of the displaced fluid.

In mathematical terms, the buoyant force (\(F_B\text{, in newtons}\)) can be calculated using the equation: \[F_{B} = V \rho_{text{air}} g\] This means that for any object weighed in a fluid, we must account for the buoyant force to determine the object's true weight.
Gravitational Acceleration
Gravitational acceleration is the acceleration due to gravity experienced by any object on or near the surface of a celestial body. On Earth, the standard value used for gravitational acceleration is approximately 9.81 meters per second squared (\(g = 9.81 m/s^2\text{, approximated for simplicity}\)).

Impact on Weighing Objects

Gravitational acceleration plays an essential role when measuring weights, as it directly affects the force of gravity acting on an object's mass. The true weight (\(F_g\text{, in newtons}\)) of an object is its mass multiplied by the gravitational acceleration: \[ F_g = m \times g \] When considering the buoyant force and gravitational acceleration together, we can understand why objects weigh less in a fluid than in a vacuum where no buoyant force acts on them.
Density
Density is a measure of how much mass is contained in a given volume and is a critical concept in the context of buoyancy and Archimedes' principle. It is defined as the mass per unit volume of a substance and is usually expressed in kilograms per cubic meter (\(kg/m^3\text{, for solids and liquids}\)) or grams per cubic centimeter (\(g/cm^3\text{, another common unit}\)).

Role in Buoyant Force

The density of a fluid, such as air or water, is part of the equation for calculating buoyant force. In the buoyant force equation \[F_{B} = V \rho_{text{air}} g\], density (\(\rho_{text{air}}\text{, in this case, the density of air}\)) helps determine the weight of the fluid displaced by an object. A higher density means a greater weight of the displaced fluid, resulting in a stronger buoyant force.
Archimedes' Principle
Archimedes' principle is a physics law that explains how the buoyant force on an object submerged in a fluid is related to the weight of the fluid displaced by the object. Archimedes, a Greek mathematician and inventor, articulated this principle which can be stated as: Any object, partially or fully submerged in a fluid, is buoyed up by a force equal to the weight of the fluid that the object displaces.

Application in True Weight Calculations

This principle is essential when calculating an object's true weight in a non-vacuum environment, such as air. Since air has weight and density, any object weighed in air will experience a buoyant force. To derive the true weight of the object (\(F_g\text{, the actual gravitational force}\)), we must add the additional force to the apparent weight measured (\(F_g'\text{, the balance reading}\)), taking into account the object's volume and the air's density, as shown in the equation from the step-by-step solution. Using Archimedes' principle, we can thus correct the apparent weight to find the actual weight, ensuring the accuracy of measurements in various scientific and industrial applications.

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Most popular questions from this chapter

A \(1.00-\mathrm{kg}\) beaker containing \(2.00 \mathrm{~kg}\) of oil (density \(=916 \mathrm{~kg} / \mathrm{m}^{3}\) ) rests on a scale. A \(2.00-\mathrm{kg}\) block of iron is suspended from a spring scale and is completely submerged in the oil (Fig. P9.43). Find the equilibrium readings of both scales.

Suppose a distant world with surface gravity of \(7.44 \mathrm{~m} / \mathrm{s}^{2}\) has an atmospheric pressure of \(8.04 \times 10^{4} \mathrm{~Pa}\) at the surface. (a) What force is exerted by the atmosphere on a disk-shaped region \(2.00 \mathrm{~m}\) in radius at the surface of a methane ocean? (b) What is the weight of a \(10.0-\mathrm{m}\) deep cylindrical column of methane with radius \(2.00 \mathrm{~m}\) ? (c) Calculate the pressure at a depth of \(10.0 \mathrm{~m}\) in the methane ocean. Note: The density of liquid methane is \(415 \mathrm{~kg} / \mathrm{m}^{3}\).

The approximate diameter of the aorta is \(0.50 \mathrm{~cm}\); that of a capillary is \(10 \mu \mathrm{m}\). The approximate average blood flow speed is \(1.0 \mathrm{~m} / \mathrm{s}\) in the aorta and \(1.0 \mathrm{~cm} / \mathrm{s}\) in the capillaries. If all the blood in the aorta eventually flows through the capillaries, estimate the number of capillaries in the circulatory system.

The average human has a density of \(945 \mathrm{~kg} / \mathrm{m}^{3}\) after inhaling and \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) after exhaling. (a) Without making any swimming movements, what percentage of the human body would be above the surface in the Dead Sea (a body of water with a density of about \(1230 \mathrm{~kg} / \mathrm{m}^{3}\) ) in each of these cases? (b) Given that bone and muscle are denser than fat, what physical characteristics differentiate "sinkers" (those who tend to sink in water) from "floaters" (those who readily float)?

Take the density of blood to be \(\rho\) and the distance between the feet and the heart to be \(h_{H}\). Ignore the flow of blood. (a) Show that the difference in blood pressure between the feet and the heart is given by \(P_{F}-P_{H}=\rho g h_{H}\). (b) Take the density of blood to be \(1.05 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) and the distance between the heart and the feet to be \(1.20 \mathrm{~m}\). Find the difference in blood pressure between these two points. This problem indicates that pumping blood from the extremities is very difficult for the heart. The veins in the legs have valves in them that open when blood is pumped toward the heart and close when blood flows away from the heart. Also, pumping action produced by physical activities such as walking and breathing assists the heart.

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