/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 A solid piece of aluminum \(\lef... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 50

A solid piece of aluminum \(\left(\rho=2.70 \mathrm{~g} / \mathrm{cm}^{3}\right)\) has a mass of \(8.35 \mathrm{~g}\) when measured in air. If it is hung from a thread and submerged in a vat of oil \(\left(\rho=0.75 \mathrm{~g} / \mathrm{cm}^{3}\right)\), what will be the tension in the thread?

Short Answer

Expert verified
The tension in the thread when the aluminum is submerged in oil is approximately 81.81 dyn.

Step by step solution

01

Calculate the Volume of the Aluminum

We know the mass and density of the aluminum. We can calculate its volume using the formula \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \). Substituting in the given values, \( \text{Volume} = \frac{8.35 \, \text{g}}{2.70 \, \text{g/cm}^3} = 3.09 \, \text{cm}^3 \).
02

Calculate the Buoyant Force Exerted by the Oil

The buoyant force is equal to the weight of the oil displaced by the aluminum. To find this, use the formula \( \text{Buoyant Force} = \text{Density}_{\text{oil}} \times \text{Volume} \times g \), where \( g \) is the gravitational acceleration (approximately \( 9.8 \, \text{m/s}^2 \)). Thus, \( \text{Buoyant Force} = 0.75 \, \text{g/cm}^3 \times 3.09 \, \text{cm}^3 \times 9.8 \, \text{m/s}^2 = 22.65 \, \text{dyn} \).
03

Convert Buoyant Force to Grams

Since 1 dyn = \( 10^{-5} \) N and using the conversion 1 N = 1000 g, convert the buoyant force to grams: \( 22.65 \, \text{dyn} = 22.65 \times 10^{-5} \times 1000 = 0.002265 \, \text{g} \).
04

Calculate the Tension in the Thread

The tension in the thread when submerged in oil is the actual weight of the aluminum reduced by the buoyant force. This can be calculated as \( \text{Tension} = \text{Weight in Air} - \text{Buoyant Force} \). Thus, \( \text{Tension} = 8.35 \, \text{g} - 22.65 \, \text{dyn} = 8.35 \, \text{g} - 0.002265 \, \text{g} = 8.347735 \, \text{g} \).
05

Convert Tension Back to Force Units

Convert this mass of tension back to force in dyn or N for consistency. For small values, it may be appropriate to keep the answer in mass units for ease of interpretation, but accurate converting yields \( 8.347735 \, \text{g} \times 9.8 \, \text{m/s}^2 = 81.807799 \, \text{dyn} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Density
Density is a fundamental concept when dealing with buoyancy or any material analysis. It is defined as the mass per unit volume of a substance. The formula to calculate density is:\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]For example, in the provided exercise, the density of aluminum is given as 2.70 g/cm鲁. This means each cubic centimeter of aluminum has a mass of 2.70 grams. Knowing the density of a material helps in understanding how it will interact with other substances, such as when submerged in a fluid.Here are a few important aspects:
  • Objects with a higher density than the fluid they are in will generally sink.
  • Objects with a lower density than the fluid will float.
When solving problems, always remember to check if mass and volume units are consistent in order to avoid calculation errors.
Buoyant Force
Buoyant force is the upward force exerted by a fluid that opposes the weight of an object immersed in the fluid. This concept is essential for understanding how objects behave when submerged. The principle is explained by Archimedes' principle which states that the buoyant force on an object is equal to the weight of the fluid displaced by the object.The formula to calculate buoyant force is:\[ \text{Buoyant Force} = \text{Density of Fluid} \times \text{Volume of Displaced Fluid} \times g \]Where:
  • \( \text{Density of Fluid} \) is the density of the fluid the object is submerged in.
  • \( \text{Volume of Displaced Fluid} \) is equivalent to the volume of the object submerged.
  • \( g \) is the acceleration due to gravity, approximately 9.8 m/s虏.
In the exercise, the aluminum piece is submerged in oil. The buoyant force is calculated using the density of the oil and the volume of the aluminum. This force effectively reduces the weight of the object when measured while submerged.
Tension
Tension in a thread or cable refers to the force exerted along it when it is used to support an object. When an object is submerged in a fluid, tension is influenced by both the object's weight and the buoyant force acting on it.The net tension can be described as:\[ \text{Tension} = \text{Actual Weight} - \text{Buoyant Force} \]Where:
  • \( \text{Actual Weight} \) is the weight of the object in air.
  • \( \text{Buoyant Force} \) is subtracted because it acts in the opposite direction.
In our sample exercise, the tension in the thread decreases when the aluminum is submerged in oil due to the upward acting buoyant force. The outcome shows how buoyancy can significantly affect the tension required to hold the object steady. Understanding these interactions is crucial for designing systems involving fluids and maintaining stability when suspending objects.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 2.0-cm cube of metal is suspended by a fine thread attached to a scale. The cube appears to have a mass of \(47.3 \mathrm{~g}\) when measured submerged in water. What will its mass appear to be when submerged in glycerin, sp gr = 1.26? [Hint: Find \(\rho\) too.]

A cube of wood floating in water supports a 200-g mass resting on the center of its top face. When the mass is removed, the cube rises \(2.00 \mathrm{~cm}\). Determine the volume of the cube.

The density of ice is \(917 \mathrm{~kg} / \mathrm{m}^{3}\). What fraction of the volume of a piece of ice will be above the liquid when floating in fresh water? The piece of ice will float in the water, since its density is less than \(1000 \mathrm{~kg} / \mathrm{m}^{3}\), the density of water. As it does, \(F_{B}=\) Weight of displaced water = Weight of piece of ice But the weight of the ice is \(\rho_{\mathrm{ice}} g V\), where \(V\) is the volume of the piece. In addition, the weight of the displaced water is \(\rho_{w} g V^{\prime}\) ' where \(V^{\prime}\) is the volume of the displaced water. Substituting into the above equation $$ \begin{aligned} \rho_{\text {ice }} g V &=\rho_{w} g V^{\prime} \\ V^{\prime} &=\frac{\rho_{\text {ice }}}{\rho_{w}} V=\frac{917}{1000} V=0.917 V \end{aligned} $$ The fraction of the volume that is above water is then $$ \frac{V-V^{\prime}}{V}=\frac{V-0.917 V}{V}=1-0.917=0.083 \text { or } 8.3 \% $$

When a submarine dives to a depth of \(120 \mathrm{~m}\), to how large a total pressure is its exterior surface subjected? The density of seawater is about \(1.03 \mathrm{~g} / \mathrm{cm}^{3}\) \(P=\) Atmospheric pressure \(+\) Pressure of wate $$ \begin{array}{l} =1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\rho g h=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\left(1030 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(120 \mathrm{~m}) \\ =1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+12.1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=13.1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.31 \mathrm{MPa} \end{array} $$

At a height of \(10 \mathrm{~km}\) (33 \(000 \mathrm{ft}\) ) above sea level, atmospheric pressure is about \(210 \mathrm{~mm}\) of mercury. What is the net resultant normal force on a \(600 \mathrm{~cm}^{2}\) window of an airplane flying at this height? Assume the pressure inside the plane is \(760 \mathrm{~mm}\) of mercury. The density of mercury is \(13600 \mathrm{~kg}\).
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Turning Points in Physics

Read Explanation

Rotational Dynamics

Read Explanation

Electricity

Read Explanation

Energy Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.