/*! 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 52 A tank containing oil of sp gr \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 52

A tank containing oil of sp gr \(=0.80\) rests on a scale and weighs \(78.6 \mathrm{~N}\). By means of a very fine wire, a \(6.0 \mathrm{~cm}\) cube of aluminum, sp gr \(=2.70\), is submerged in the oil. Find \((a)\) the tension in the wire and \((b)\) the scale reading if none of the oil overflows.

Short Answer

Expert verified
a) 4.0152 N, b) 80.2934 N

Step by step solution

01

Determine weight of aluminum cube

First, we calculate the volume of the cube, using its side length: \ \[ V = (6.0 \text{ cm})^3 = 216 \text{ cm}^3 = 216 \times 10^{-6} \text{ m}^3 \] \ The specific gravity of aluminum is 2.70, therefore its density \( \rho \) in terms of water density (\(1000 \text{ kg/m}^3\)) is \ \[ \rho = 2.70 \times 1000 = 2700 \text{ kg/m}^3 \] \ The weight of the aluminum cube \( W_a \) is calculated as: \ \[ W_a = \rho \times V \times g = 2700 \times 216 \times 10^{-6} \times 9.8 = 5.7086 \text{ N} \] \
02

Calculate buoyant force on aluminum

The buoyant force \( F_b \) is found using the specific gravity of the oil. The density of the oil \( \rho_o \) can be calculated as:\ \[ \rho_o = 0.80 \times 1000 = 800 \text{ kg/m}^3 \] \ The buoyant force is given by:\ \[ F_b = \rho_o \times V \times g = 800 \times 216 \times 10^{-6} \times 9.8 = 1.6934 \text{ N} \] \
03

Calculate tension in the wire

The tension \( T \) is the difference between the weight of the cube and the buoyant force: \ \[ T = W_a - F_b = 5.7086 - 1.6934 = 4.0152 \text{ N} \] \ Tension in the wire is \( 4.0152 \text{ N} \).
04

Calculate new scale reading

The scale initially reads the weight of the tank \( 78.6 \text{ N} \). The apparent buoyant force due to submerged aluminum is added to the scale reading as this force supports part of the cube's weight: \ \[ \text{New scale reading} = 78.6 + F_b = 78.6 + 1.6934 = 80.2934 \text{ N} \] \ The new scale reading is \( 80.2934 \text{ N} \).

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.

Specific Gravity: Understanding Its Role in Buoyancy and Density
Specific gravity is a dimensionless quantity that compares the density of a substance to the density of water at 4掳C (which is approximately 1000 kg/m鲁). It's a useful metric as it helps us understand how different materials will behave when placed in a fluid. In our exercise, the oil has a specific gravity of 0.80, meaning it's less dense than water. Aluminum, with a specific gravity of 2.70, is denser than water.

This knowledge allows us to predict buoyancy: since aluminum is denser, it will sink in oil, but the oil will exert an upward buoyant force against it. Understanding specific gravity simplifies many calculations by turning them into ratios of densities, thereby providing insights into whether substances will float or sink in different fluids.
Density Calculation: Getting to the Core of Mass and Volume
To find the density of a material, divide its mass by its volume. It's a fundamental property that tells how tightly matter is packed in a substance. For our task, we first calculate the volume of the aluminum cube, which is done using the formula for the volume of a cube: side length cubed. The cube with a side of 6.0 cm has a volume of 216 cm鲁.

Then, using the specific gravity given, we convert this to density:
  • Aluminum鈥檚 density: calculate using specific gravity
  • Oil鈥檚 density: apply specific gravity to adjust from the density of water
These calculations are crucial for determining both the weight of the cube and the buoyant force exerted on it by the oil.
Tension Force: Balancing Act Between Weight and Buoyancy
The tension force in this setting is the force experienced by the wire holding the submerged aluminum cube. It's a result of the opposing forces at play: weight pulling it down, versus buoyant force pushing it up.
  • Weight is calculated using density times volume times gravity.
  • Buoyant force uses the same volume but applies the oil鈥檚 density.
Tension is simply the net force required to keep the system in equilibrium. Subtract the buoyant force from the weight of the cube to find the tension: it has to support what the buoyant force doesn't. Understanding this helps in visualizing how substances are held when submerged, balancing different forces to maintain stability.
Scale Reading: Tracking Changes in Weight Perception
The scale reading is essentially what you would observe as the total weight perceived by the scale. Initially, the scale reads the weight of the tank and oil alone. When the aluminum is submerged, the apparent weight seems to increase.

Why is that? Because the buoyant force that counters part of the aluminum's weight adds to the load on the scale. The buoyant force reduces the tension in the wire but transfers its effect to the scale. So, the modified reading on the scale reflects both the original weight and this added buoyant effect. This understanding of scale reactions is essential for interpreting measurements correctly, especially in environments involving varying fluid dynamics.

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 glass tube is bent into the form of a U. A \(50.0-\mathrm{cm}\) height of olive oil in one arm is found to balance \(46.0 \mathrm{~cm}\) of water in the other. What is the density of the olive oil?

How high would water rise in the essentially open pipes of a building if the water pressure gauge shows the pressure at the ground floor to be \(270 \mathrm{kPa}\) (about \(40 \mathrm{lb} / \mathrm{in}^{2}\).)? Water pressure gauges read the excess pressure just due to the water, that is, the difference between the absolute pressure in the water and the pressure of the atmosphere. The water pressure at the bottom of the highest column that can be supported is \(270 \mathrm{kPa}\). Therefore, \(P=\rho_{w} p h\) gives $$ h=\frac{P}{\rho_{w} g}=\frac{2.70 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}}{\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=27.5 \mathrm{~m} $$

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.

A wooden cylinder has a mass \(m\) and a base area \(A\). It floats in water with its axis vertical. Show that the cylinder undergoes SHM if given a small vertical displacement. Find the frequency of its motion. When the cylinder is pushed down a distance \(y\), it displaces an additional volume Ay of water. Because this additional displaced volume has mass \(A y_{\rho w}\), an additional buoyant force \(A y_{\rho w g}\) acts on the cylinder, where \(\rho_{w}\) is the density of water. This is an unbalanced force on the cylinder and is a restoring force. In addition, the force is proportional to the displacement and so is a Hooke's Law force. Therefore, the cylinder will undergo SHM, as described in Chapter 11 . Comparing \(F_{B}=A \rho_{w} g y\) with Hooke's Law in the form \(F=k y\), we see that the elastic constant for the motion is \(k=A \rho_{w} g .\) This, acting on the cylinder of mass \(m\), causes it to have a vibrational frequency of $$ f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{A \rho_{w} g}{m}} $$

The area of a piston of a force pump is \(8.0 \mathrm{~cm}^{2}\). What force must be applied to the piston to raise oil \(\left(\rho=0.78 \mathrm{~g} / \mathrm{cm}^{2}\right)\) to a height of \(6.0 \mathrm{~m}\) ? Assume the upper end of the oil is open to the atmosphere.
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Solid State Physics

Read Explanation

Waves Physics

Read Explanation

Electrostatics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Physics of Motion

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.