/*! 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 4 Determine the dimensions, in bot... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 4

Determine the dimensions, in both the \(F L T\) system and the \(M L T\) system, for (a) the product of force times acceleration, (b) the product of force times velocity divided by area, and (c) momentum divided by volume.

Short Answer

Expert verified
(a) F L T: \( F \cdot L \cdot T^{-2} \), M L T: \( M L^2 T^{-4} \); (b) F L T: \( F L^{-1} T^{-1} \), M L T: \( M T^{-3} \); (c) F L T: \( F T L^{-3} \), M L T: \( M L^{-2} T^{-1} \).

Step by step solution

01

Understand the Variables

Identify the relevant physical quantities. Force ( F ) has units of newtons (N), acceleration ( a ) has units of meters per second squared (m/s^2) , velocity ( v ) has units of meters per second (m/s) , area ( A ) has units of square meters (m^2) , momentum ( p ) has units of kilogram-meter per second (kg路m/s) , and volume ( V ) has units of cubic meters (m^3) .
02

Dimensional Analysis in the F L T System

In the F L T (Force-Length-Time) system: - Force (F) is a fundamental dimension. - Length (L) and Time (T) are fundamental dimensions. Apply these to each quantity:(a) Product of force (\(F\)) and acceleration (\(a\)): \[ [F][a] = F \cdot \frac{L}{T^2} = F \cdot L \cdot T^{-2} \](b) Product of force and velocity divided by area: \[ \frac{[F] [v]}{[A]} = \frac{F \cdot \frac{L}{T}}{L^2} = F \cdot L^{-1} \cdot T^{-1} \](c) Momentum divided by volume:\[ \frac{[p]}{[V]} = \frac{F \cdot T}{L^3} = F \cdot T \cdot L^{-3} \]
03

Dimensional Analysis in the M L T System

In the M L T (Mass-Length-Time) system:- Mass (M), Length (L), and Time (T) are fundamental dimensions. Force (F) is derived from these.First, express force in M L T units:\[ [F] = M \cdot L \cdot T^{-2} \]Then, apply to:(a) Product of force and acceleration:\[ [F][a] = (M \cdot L \cdot T^{-2}) \cdot L \cdot T^{-2} = M \cdot L^2 \cdot T^{-4} \](b) Product of force and velocity divided by area:\[ \frac{[F][v]}{[A]} = \frac{(M \cdot L \cdot T^{-2}) \cdot \frac{L}{T}}{L^2} = M \cdot T^{-3} \](c) Momentum divided by volume:\[ \frac{[p]}{[V]} = \frac{M \cdot L \cdot T^{-1}}{L^3} = M \cdot L^{-2} \cdot T^{-1} \]
04

Summarize the Results

Combine dimensional analysis results from both the F L T and M L T systems:(a) Product of force and acceleration:- In F L T: \( F \cdot L \cdot T^{-2} \)- In M L T: \( M \cdot L^2 \cdot T^{-4} \)(b) Product of force and velocity divided by area:- In F L T: \( F \cdot L^{-1} \cdot T^{-1} \)- In M L T: \( M \cdot T^{-3} \)(c) Momentum divided by volume:- In F L T: \( F \cdot T \cdot L^{-3} \)- In M L T: \( M \cdot L^{-2} \cdot T^{-1} \)

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.

Dimensional Analysis in the Force-Length-Time System
In the Force-Length-Time (FLT) system, the fundamental dimensions are Force (F), Length (L), and Time (T). These serve as the building blocks for expressing various physical quantities. Force is considered a fundamental dimension here, making it the basis for analysis that involves other dimensions like length and time.
  • **Product of force and acceleration:** In FLT, when we multiply force by acceleration, the dimensions are expressed as: \[ [F][a] = F \cdot \frac{L}{T^2} = F \cdot L \cdot T^{-2} \] This shows how force combined with acceleration looks in the FLT system.
  • **Force times velocity divided by area:** For this calculation, it's about how a force interacts with velocity and area:\[ \frac{[F][v]}{[A]} = \frac{F \cdot \frac{L}{T}}{L^2} = F \cdot L^{-1} \cdot T^{-1} \] Here, dividing by area changes the dimensional expression.
  • **Momentum divided by volume:** This illustrates how momentum is expressed as:\[ \frac{[p]}{[V]} = \frac{F \cdot T}{L^3} = F \cdot T \cdot L^{-3} \] Each term in this expression affects how the output is dimensionally arranged.
By using the FLT system, you can easily break down how force-related quantities interact with other physical properties.
Dimensional Analysis in the Mass-Length-Time System
The Mass-Length-Time (MLT) system, unlike the FLT system, considers Mass (M) as a fundamental dimension, alongside Length (L) and Time (T). This choice affects how force is derived because force is not a primary dimension but a derived one. Here, force is expressed as: \[ [F] = M \cdot L \cdot T^{-2} \]
  • **Product of force and acceleration:** In the MLT framework, force combined with acceleration involves:\[ [F][a] = (M \cdot L \cdot T^{-2}) \cdot L \cdot T^{-2} = M \cdot L^2 \cdot T^{-4} \] This expression tells us how complex interactions between mass, length, and time occur.
  • **Force times velocity divided by area:** Here, analyzing this in MLT shows:\[ \frac{[F][v]}{[A]} = \frac{(M \cdot L \cdot T^{-2}) \cdot \frac{L}{T}}{L^2} = M \cdot T^{-3} \] The manipulation of time here reduces the power of time dimensions significantly.
  • **Momentum divided by volume:** In MLT, momentum appears as:\[ \frac{[p]}{[V]} = \frac{M \cdot L \cdot T^{-1}}{L^3} = M \cdot L^{-2} \cdot T^{-1} \] This showcases simple dimensional interactions derived from mass and other necessary variables.
Understanding the MLT system assists in analyzing forces and related dimensions with a clear focus on mass.
Understanding Fundamental Dimensions
Fundamental dimensions are the baseline measurements used to describe physical quantities in both FLT and MLT systems. Each system selects different base quantities as its fundamentals, like force in FLT and mass in MLT. By specifying which dimensions are fundamental, these systems make it easier to conduct dimensional analysis.
**Relation to Physical Quantities:**
  • **Force-Length-Time system: Aligns with dynamics where force is central.** This system makes computations straightforward when dealing with forces.
  • **Mass-Length-Time system: Commonly used in classical mechanics.** It places mass as a core quantity, reflecting its significance in motion equations.
**Why They Matter?** These dimensions help:
  • Convert complex physical relationships into simpler, readable terms.
  • Ensure equations remain consistent regardless of the unit system used.
  • Allow predictions of how altering one dimension affects others.
Grasping the concept of fundamental dimensions lays the foundation for understanding the principles of dimensional analysis. This understanding is pivotal in multiple areas of physical science and engineering.

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

Water flows from a large drainage pipe at a rate of 1200 gal/min. What is this volume rate of flow in (a) \(\mathrm{m}^{3} / \mathrm{s}\) (b) liters/min, and (c) \(\mathrm{ft}^{3} / \mathrm{s} ?\)

Develop a computer program for calculating the density of an ideal gas when the gas pressure in pascals (abs), the temperature in degrees Celsius, and the gas constant in \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) are specified. Plot the density of helium as a function of temperature from \(0^{\circ} \mathrm{C}\) to \(200^{\circ} \mathrm{C}\) and pressures of \(50,100,150,\) and 200 \(\mathrm{kPa}(\mathrm{abs})\).

A tank of oil has a mass of 25 slugs. (a) Determine its weight in pounds and in newtons at the Earth's surface. (b) What would be its mass (in slugs) and its weight (in pounds) if located on the moon's surface where the gravitational attraction is approximately one- sixth that at the Earth's surface?

The specific weight of a certain liquid is \(85.3 \mathrm{lb} / \mathrm{ft}^{3}\). Determine its density and specific gravity.

There are many fluids that exhibit non-Newtonian behavior (see, for example, Video \(V 1.6\) ). For a given fluid the distinction between Newtonian and non- Newtonian behavior is usually based on measurements of shear stress and rate of shearing strain. Assume that the viscosity of blood is to be determined by measurements of shear stress, \(\tau,\) and rate of shearing strain, duldy, obtained from a small blood sample tested in a suitable viscometer. Based on the data given below, determine if the blood is a Newtonian or non-Newtonian fluid. Explain how you arrived at your answer. $$\begin{array}{l|l|l|l|l|l|l|l|l} \tau\left(\mathrm{N} / \mathrm{m}^{2}\right) & 0.04 & 0.06 & 0.12 & 0.18 & 0.30 & 0.52 & 1.12 & 2.10 \\ \hline d u / d y\left(\mathrm{s}^{-1}\right) & 2.25 & 4.50 & 11.25 & 22.5 & 45.0 & 90.0 & 225 & 450 \end{array}$$
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Turning Points in Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Nuclear Physics

Read Explanation

Electromagnetism

Read Explanation

Astrophysics

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.