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Chapter 1: Problem 4

If \(P\) is a force and \(x\) a length, what are the dimensions (in the \(F L T \text { system })\) of \((a) d P / d x,(b) d^{3} P / d x^{3},\) and (c) \(\int P d x ?\)

Short Answer

Expert verified
(a) \([F][L]^{-1}\), (b) \([F][L]^{-3}\), (c) \([F][L]\).

Step by step solution

01

Understanding Dimensions of Force and Length

In the \(FLT\) system, dimensions are expressed in terms of Force \(F\), Length \(L\), and Time \(T\). The dimension of force \(P\) is \([F]\), and the dimension of length \(x\) is \([L]\). These are the basic dimensions that will be used to find the dimensions of the given expressions.
02

Finding Dimensions of \(\frac{dP}{dx}\)

The expression \(\frac{dP}{dx}\) indicates the rate of change of force with respect to length. Its dimensions are calculated by dividing the dimensions of \(P\) by \(x\). Therefore, \(\frac{dP}{dx}\) has the dimensions \(\[\frac{[F]}{[L]} = [F][L]^{-1}\].\) These represent force per unit length.
03

Finding Dimensions of \(\frac{d^3P}{dx^3}\)

The expression \(\frac{d^3P}{dx^3}\) corresponds to the third derivative of force with respect to length. This is calculated as the force divided by length cubed. Therefore, the dimensions of \(\frac{d^3P}{dx^3}\) are \(\[\frac{[F]}{[L]^3} = [F][L]^{-3}\].\) These denote force change over length cubed.
04

Finding Dimensions of \(\int P\,dx\)

The integral \(\int P\,dx\) signifies the integral of force over length. This operation effectively multiplies \(P\) by \(x\), resulting in the dimensions \([F][L]\). These are dimensions of force multiplied by length, similar to work or energy.

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Key Concepts

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

Force
Force is a fundamental concept in physics, defined as any interaction that, when unopposed, will change the motion of an object. In the scientific community, it is typically measured in newtons (N), where one newton equates to the amount of force needed to accelerate a one-kilogram mass by one meter per second squared.
In dimensional analysis using the FLT system, the dimension of force is represented simply as
  • [F]
  • Force often results from interactions such as gravitational, electromagnetic, or mechanical forces.
  • It is crucial in evaluating how objects interact with their environment.
Understanding the dimensions related to force enables one to analyze the effects and consequences of different forces in various physical situations.
Length
Length is one of the most fundamental physical dimensions, used to measure how far apart points are from each other. It is usually measured in meters in the metric system or in feet in the imperial system. In dimensional analysis within the FLT system, length has the simple representation of
  • [L]
  • It is a core component in calculating volume and area.
  • Length is integral in determining velocities when coupled with time.
The ability to manipulate and understand the dimension of length is crucial for measuring distances in physical space and contributing to the calculation of integrals and derivatives that involve spatial displacement.
Derivatives and Integrals
Derivatives and integrals are mathematical tools used to analyze how quantities change or accumulate over a certain domain, often time or space. A derivative represents how a function changes at any point, often expressed as
  • dP/dx
  • The process of taking derivatives helps reveal the rate of change of force with respect to length, providing insights into how quickly or slowly a force acts across a distance. In this case, the dimension would be [F][L]^{-1}.
  • The third derivative, d^3P/dx^3, indicates the rate at which the rate of change of the rate of change of force occurs, expressed dimensionally as [F][L]^{-3}.

Integrals aggregate amounts over a domain, often representing sums or areas beneath curves. The integral
  • int P dx gives a dimension of [F][L], relating closely to work, calculated by multiplying force over a defined length.
Dimensionless Quantities
Dimensionless quantities arise in scenarios where inputs to a problem result in outputs that do not have defined physical dimensions. They are pure numbers and often occur as ratios of two quantities with the same dimension, effectively canceling out the units. Beyond ratios, dimensionless numbers frequently appear in physical equations as parameters or constants.
  • They simplify complex equations and experiments by reducing variables to their core relationships.
  • Common examples include Reynolds number and Pi in fluid dynamics and geometry, respectively.
This concept is crucial in ensuring consistency across different physical scenarios and is instrumental in scaling laws and similarity analyses in various engineering and scientific applications.

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Most popular questions from this chapter

An important dimensionless parameter concerned with very high-speed flow is the Mach number, defined as \(\mathrm{V} / \mathrm{c}\), where V is the speed of the object such as an airplane or projectile, and \(c\) is the speed of sound in the fluid surrounding the object. For a projectile traveling at 800 mph through air at \(50^{\circ} \mathrm{F}\) and standard atmospheric pressure, what is the value of the Mach number?

An open, clean glass tube \(\left(\theta=0^{\circ}\right)\) is inserted vertically into a pan of water. What tube diameter is needed if the water level in the tube is to rise one tube diameter (due to surface tension)?

Often the assumption is made that the flow of a certain fluid can be considered as in compressible flow if the density of the fluid changes by less than \(2 \%\). If air is flowing through a tube such that the air pressure at one section is 9.0 psi (gage) and at a downstream section it is 8.6 psi (gage) at the same temperature, do you think that this flow could be considered an in compressible flow? Support your answer with the necessary calculations. Assume standard atmospheric pressure.

The following torque-angular velocity data were obtained with a rotating cylinder viscometer of the type described in Problem 1.61 For this viscometer \(R_{n}=2.50\) in., \(\quad R_{i}=2.45\) in. \(,\) and \(\ell=5.00\) in. Make use of these data and a standard curve-fitting program to determine the viscosity of the liquid contained in the viscometer.

The volume rate of flow, \(Q\), through a pipe containing a slowly moving liquid is given by the equation $$Q=\frac{\pi R^{4} \Delta p}{8 \mu \ell}$$ where \(R\) is the pipe radius, \(\Delta p\) the pressure drop along the pipe, \(\mu\) a fluid property called viscosity \(\left(F L^{-2} T\right),\) and \(\ell\) the length of pipe. What are the dimensions of the constant \(\pi / 8 ?\) Would you ciassify this equation as a general homogeneous equation? Explain.
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