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

Short Answer

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

Step by step solution

01

Understanding FLT and MLT Systems

In the FLT system, dimensions are expressed in terms of Force (F), Length (L), and Time (T). In the MLT system, dimensions are expressed in terms of Mass (M), Length (L), and Time (T). We'll convert the given expressions into these systems.
02

Analyzing the Product of Force and Volume (a)

First, identify the dimensions of force and volume: Force in FLT is \([F]\) and in MLT is \([M][L][T]^{-2}\); Volume in both systems is \([L]^3\). Multiply these together: - In FLT: \([F][L]^3\) - In MLT: \([M][L]^4[T]^{-2}\)
03

Analyzing the Product of Pressure and Mass Divided by Area (b)

Identify the dimensions: Pressure is force per area. So in FLT, pressure is \([F][L]^{-2}\) and in MLT, \([M][L]^{-1}[T]^{-2}\). Mass has dimension \([M]\) in MLT and \([F][L]^{-1}[T]^2\) in FLT. Area is \([L]^2\). Combine these:- In FLT: \(\frac{[F][F][L]^{-1}[T]^{2}}{[L]^2} = [F][L]^{-3}[T]^2\)- In MLT: \(\frac{[M][L]^{-1}[T]^{-2}][M]}{[L]^2} = [M]^2[L]^{-3}[T]^{-2}\)
04

Analyzing the Moment of a Force Divided by Velocity (c)

Moment of force, or torque, in FLT is \([F][L]\) and in MLT is \([M][L]^2[T]^{-2}\). Velocity in both systems is distance per time \([L][T]^{-1}\). Divide them:- In FLT: \(\frac{[F][L]}{[L][T]^{-1}} = [F][T]\)- In MLT: \(\frac{[M][L]^2[T]^{-2}}{[L][T]^{-1}} = [M][L][T]^{-1}\)
05

Summary of Dimensions

Now, summarize the dimensions for each expression in both systems:- (a) Product of force and volume: FLT is \([F][L]^3\) and MLT is \([M][L]^4[T]^{-2}\).- (b) Product of pressure and mass divided by area: FLT is \([F][L]^{-3}[T]^2\) and MLT is \([M]^2[L]^{-3}[T]^{-2}\).- (c) Moment of a force divided by velocity: FLT is \([F][T]\) and MLT is \([M][L][T]^{-1}\).

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Key Concepts

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

Force
Force is a key concept in physics, representing a push or pull that can cause an object with mass to change its velocity. It is a vector quantity, meaning it has both magnitude and direction. In the context of dimensional analysis, force has specific dimensions based on the system used:
  • In the FLT (Force-Length-Time) system, force is itself a base dimension represented as \([F]\).
  • In the MLT (Mass-Length-Time) system, force is derived from Newton’s second law of motion, which states that force equals mass times acceleration (\( F = ma \)). The dimensions of force are \([M][L][T]^{-2}\), where \([M]\) represents mass, \([L]\) represents length, and \([T]\) represents time.

Understanding these dimensions allows us to convert and compare different expressions involving force with ease. It also helps in identifying errors in equations, because each side of any physical equation must have matching dimensions. Dimensional consistency is key, as no equation is physically meaningful if its dimensions do not match on each side.
Volume
Volume refers to the amount of space that an object or substance occupies. It is a scalar quantity and is commonly measured in cubic units such as cubic meters or liters.
In dimensional analysis, volume has a straightforward representation in both the FLT and MLT systems:
  • In both systems, volume is expressed using only the dimension of length, denoted as \([L]^3\). This indicates that volume is a factor of space in three dimensions.

When analyzing problems that involve volume, like the product of force and volume, it’s crucial to understand how volume contributes to the overall dimensional expression. For instance, in combining volume with force, one can determine an expression that aids in identifying the physical quantity being described, and whether it makes intuitive physical sense.
By understanding volume’s dimensions clearly, one can better grasp how volume interacts with other physical quantities, such as pressure and density, and simplifies the conversion between different measurement systems.
Torque
Torque, also known as the moment of force, is a measure of the rotational force on an object. It is a pivotal concept in mechanics, especially when studying rotational dynamics.
Here's how torque is understood in dimensional analysis:
  • In the FLT system, torque is the product of force and the perpendicular distance from the axis of rotation, given as \([F][L]\). This multiplication accounts for the rotational capability relative to a pivot point.
  • In the MLT system, torque corresponds to \([M][L]^2[T]^{-2}\). The dimensions indicate that torque involves mass, acceleration, and the distance squared from the axis of rotation.

Torque is essential for understanding how forces cause objects to rotate. By analyzing its dimensions, we can ensure that expressions involving torque are dimensionally accurate, like when computing torque divided by velocity. Such checks help confirm that the calculations obey basic physics principles, allowing engineers and scientists to predict rotational effects accurately.
Simply put, torque plays a crucial role in applications ranging from simple levers to complex machinery, making it a fundamental concept in physics and engineering.

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Most popular questions from this chapter

A formula for estimating the volume rate of flow, \(Q\), over the spillway of a dam is $$Q=C \sqrt{2 g} B\left(H+V^{2} / 2 g\right)^{3 / 2}$$ where \(C\) is a constant, \(g\) the acceleration of gravity, \(B\) the spillway width, \(H\) the depth of water passing over the spillway, and \(V\) the velocity of water just upstream of the dam. Would this equation be valid in any system of units? Explain.

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.

A hydrometer is used to measure the specific gravity of liquids. (See Video \(\vee 2.6 .\) ) For a certain liquid a hydrometer reading indicates a specific gravity of \(1.15 .\) What is the liquid's density and specific weight? Express your answer in SI units.

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula $$h=(0.04 \mathrm{to} 0.09)(\mathrm{D} / \mathrm{d})^{4} \mathrm{V}^{2} / 2 \mathrm{g}$$ where \(h\) is the energy loss per unit weight, \(D\) the hose diameter, \(d\) the nozzle tip diameter, \(V\) the fluid velocity in the hose, and \(g\) the acceleration of gravity. Do you think this equation is valid in any system of units? Explain.

If \(p\) is a pressure, \(V\) a velocity, and \(\rho\) a fluid density, what are the dimensions (in the MLT system) of (a) \(p / \rho,\) (b) \(p V \rho\) and (c) \(p / \rho V^{2} ?\)

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