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Chapter 2: Problem 105

It is said that Archimedes discovered the buoyancy laws when asked by King Hiero of Syracuse to determine whether his new crown was pure gold (SG = 19.3 ). Archimedes measured the weight of the crown in air to be \(11.8 \mathrm{N}\) and its weight in water to be \(10.9 \mathrm{N}\). Was it pure gold?

Short Answer

Expert verified
The crown is not made of pure gold; its density is less than that of pure gold.

Step by step solution

01

Understand the Problem

We need to determine if the crown is made of pure gold by using Archimedes' principle, which involves comparing the specific gravity (SG) of the crown with that of pure gold (SG = 19.3).
02

Calculate the Buoyant Force

The buoyant force can be determined by the difference in the weight of the crown in air and in water: \( F_b = W_{air} - W_{water} = 11.8 \mathrm{N} - 10.9 \mathrm{N} = 0.9 \mathrm{N} \)
03

Determine the Volume of the Crown

The volume of the crown \( V \) can be calculated using the buoyant force and the formula: \( F_b = \rho_{water} \cdot V \cdot g \), where \( \rho_{water} = 1000 \mathrm{kg/m^3} \) and \( g = 9.81 \mathrm{m/s^2} \). So, \( 0.9 = 1000 \cdot V \cdot 9.81 \) leads to \( V = \frac{0.9}{1000 \cdot 9.81} = 9.17 \times 10^{-5} \mathrm{m^3} \).
04

Calculate the Density of the Crown

The mass of the crown in air \( m \) can be derived from its weight: \( W_{air} = m \cdot g \rightarrow 11.8 = m \cdot 9.81 \rightarrow m = \frac{11.8}{9.81} \approx 1.20 \mathrm{kg} \). Then the density \( \rho_{crown} \) is \( \rho_{crown} = \frac{m}{V} = \frac{1.20}{9.17 \times 10^{-5}} = 13086 \mathrm{kg/m^3} \).
05

Compare with Pure Gold's Density

The density of pure gold is approximately \( 19300 \mathrm{kg/m^3} \). The calculated density of the crown (\( 13086 \mathrm{kg/m^3} \)) is less than that of pure gold, indicating that the crown is not pure gold.

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Key Concepts

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

Buoyancy
Buoyancy is a force that acts upwards on an object when it is submerged in a fluid, such as water. This force is what makes objects feel lighter when they are underwater. The famous scientist Archimedes discovered that the buoyant force on an object is equal to the weight of the fluid that the object displaces. This is known as Archimedes' Principle.

To understand this better, think of placing a rock in a jug full of water. The water spills out equivalent to the volume of the rock submerged. The force that pushes the rock upwards is equal to the weight of this displaced water.
  • A completely submerged object experiences a buoyant force that opposes gravity.
  • The difference in weight of an object in air and water helps calculate the buoyant force.
In the exercise, the crown weighs 11.8 N in air and 10.9 N in water. The difference, 0.9 N, is the buoyant force experienced by the crown.
Specific Gravity
Specific gravity (SG) is a measure of the density of a substance compared to the density of a reference material, usually water for a liquid or solid. It is a unitless number because it is the ratio of two similar units.

Specific gravity can help determine if an object is made of a particular material. In the case of the crown, Archimedes was tasked with finding out if it was made of pure gold.
  • For gold, the SG is 19.3, meaning it is 19.3 times denser than water.
  • This value helps differentiate gold from other substances by comparing densities.
Knowing the density of the crown and comparing it to the SG of gold tells us whether the crown is pure gold. In our problem, after calculating the crown's density and comparing it, we found it does not match pure gold, indicating the crown is not pure.
Density Calculation
Density is a crucial concept in understanding buoyancy and specific gravity. It is defined as the mass of an object divided by its volume, represented as \( \rho = \frac{m}{V} \). In the exercise, we calculate the crown's density to determine its composition.

Here are the steps involved:
  • First, determine the mass by using the weight of the crown in air: \( m = \frac{W_{air}}{g} \).
  • Then calculate the volume from the buoyant force: \( V = \frac{F_b}{\rho_{water} \cdot g} \).
  • Finally, calculate the density: \( \rho_{crown} = \frac{m}{V} \).
In the exercise, these steps calculated a density of \( 13086 \mathrm{kg/m^3} \) for the crown. Since pure gold has a density of \( 19300 \mathrm{kg/m^3} \), the crown's lower density shows it is not pure gold.

This detailed approach highlights how critical careful density calculations are in material identification and validation.

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Most popular questions from this chapter

Consider a homogeneous right circular cylinder of length \(L,\) radius \(R,\) and specific gravity \(\mathrm{SG}=0.5,\) floating in water \((\mathrm{SG}=1) .\) Show that the body will be stable with its axis horizontal if \(L / R>2.0\)

A cubical tank is \(3 \times 3 \times 3 \mathrm{m}\) and is layered with 1 meter of fluid of specific gravity 1.0,1 meter of fluid with \(\mathrm{SG}=0.9,\) and 1 meter of fluid with \(\mathrm{SG}=0.8 .\) Neglect atmospheric pressure. Find \((a)\) the hydrostatic force on the bottom and \((b)\) the force on a side panel.

A child is holding a string onto which is attached a heliumfilled balloon. (a) The child is standing still and suddenly accelerates forward. In a frame of reference moving with the child, which way will the balloon tilt, forward or backward? Explain. (b) The child is now sitting in a car that is stopped at a red light. The helium-filled balloon is not in contact with any part of the car (seats, ceiling, etc.) but is held in place by the string, which is in turn held by the child. All the windows in the car are closed. When the traffic light turns green, the car accelerates forward. In a frame of reference moving with the car and child, which way will the balloon tilt, forward or backward? Explain. (c) Purchase or borrow a helium-filled balloon. Conduct a scientific experiment to see if your predictions in parts \((a)\) and (b) above are correct. If not, explain.

A small submarine, with a hatch door 30 in in diameter, is submerged in seawater. \((a)\) If the water hydrostatic force on the hatch is 69,000 lbf, how deep is the sub? (b) If the sub is 350 ft deep, what is the hydrostatic force on the hatch?

A closed tank contains 1.5 \(\mathrm{m}\) of SAE 30 oil, 1 \(\mathrm{m}\) of water, \(20 \mathrm{cm}\) of mercury, and an air space on top, all at \(20^{\circ} \mathrm{C}\) The absolute pressure at the bottom of the tank is \(60 \mathrm{kPa}\) What is the pressure in the air space?
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