Chapter 7, Problem 98P

To determine

(a)

The primary dimensions of pressure or stress in F-L-T system.

Answer to Problem 98P

The dimensions of stress is $\left\{\text{σ}\right\}\text{=}\left\{\left({\text{F}}^{\text{1}}{\text{L}}^{\text{-2}}\right)\right\}.$

Explanation of Solution

In M-L-T system,

  $\begin{array}{l}\text{force}\text{=}\text{mass×acceleration}\\ \text{F}\text{=}\text{m}\text{×}\text{a}\\ \left\{\text{F}\right\}\text{=}\left\{\text{mass}\text{×}\frac{\text{meter}}{{\text{sec}}^{\text{2}}}\right\}\\ \left\{\text{F}\right\}\text{=}\left\{\text{M}\text{×}\frac{\text{L}}{{\text{t}}^{\text{2}}}\right\}\\ \left\{\text{F}\right\}\text{=}\left\{\frac{\text{M}\text{L}}{{\text{t}}^{\text{2}}}\right\}\\ \left\{\text{F}\right\}\text{=}\left\{{\text{M}}^{\text{1}}{\text{ L}}^{\text{1}}{\text{ t}}^{\text{-2}}\right\}\end{array}$

Now, write primary dimension of mass in F-L-T system,

  $\begin{array}{l}\left\{\text{F}\right\}\text{=}\left\{{\text{M}}^{\text{1}}{\text{L}}^{\text{1}}{\text{t}}^{\text{-2}}\right\}\\ \left\{\text{M}\right\}\text{=}\left\{\frac{{\text{Ft}}^{\text{2}}}{\text{L}}\right\}\\ \left\{\text{M}\right\}\text{=}\left\{{\text{F}}^{\text{1}}{\text{L}}^{\text{-1}}{\text{t}}^{\text{2}}\right\}\mathrm{.....}\text{(i)}\end{array}$

Stress is defined as force per unit area.

  $\begin{array}{l}\text{stree}\text{=}\frac{\text{force}}{\text{area}}\\ \text{σ}\text{=}\frac{\text{force}}{\text{area}}\\ \left\{\text{σ}\right\}\text{=}\left\{\frac{\text{force}}{\text{area}}\right\}\\ \left\{\text{σ}\right\}\text{=}\left\{\frac{\text{mass×length}}{{\text{time}}^{\text{2}}}\text{×}\frac{\text{1}}{{\text{length}}^{\text{2}}}\right\}\\ \left\{\text{σ}\right\}\text{=}\left\{\frac{\text{m×L}}{{\text{t}}^{\text{2}}}\text{×}\frac{\text{1}}{{\text{L}}^{\text{2}}}\right\}\\ \left\{\text{σ}\right\}\text{=}\left\{\frac{\text{m}}{{\text{t}}^{\text{2}}\text{L}}\right\}\end{array}$

In exponential form,

  $\left\{\text{σ}\right\}\text{=}\left\{{\text{m}}^{\text{1}}{\text{L}}^{\text{-1}}{\text{t}}^{\text{-2}}\right\}$

In F-L-T system,

On substituting the value from equation (i),

  $\left\{\text{σ}\right\}\text{=}\left\{{\text{m}}^{\text{1}}{\text{L}}^{\text{-1}}{\text{t}}^{\text{-2}}\right\}$

  $\begin{array}{l}\left\{\text{σ}\right\}\text{=}\left\{\left({\text{F}}^{\text{1}}{\text{L}}^{\text{-1}}{\text{t}}^{\text{2}}\right){\text{L}}^{\text{-1}}{\text{t}}^{\text{-2}}\right\}\\ \left\{\text{σ}\right\}\text{=}\left\{\left({\text{F}}^{\text{1}}{\text{L}}^{\text{-2}}\right)\right\}\end{array}$

To determine

(b)

The primary dimensions of moment or torque in F-L-T system.

Answer to Problem 98P

The primary dimension of energy is $\left\{\text{E}\right\}\text{=}\left\{{\text{F}}^{\text{1}}{\text{L}}^1\right\}.$

Explanation of Solution

In M-L-T system,

  $\begin{array}{l}\text{force}\text{=}\text{mass×acceleration}\\ \text{F}\text{=}\text{m}\text{×}\text{a}\\ \left\{\text{F}\right\}\text{=}\left\{\text{mass}\text{×}\frac{\text{meter}}{{\text{sec}}^{\text{2}}}\right\}\\ \left\{\text{F}\right\}\text{=}\left\{\text{M}\text{×}\frac{\text{L}}{{\text{t}}^{\text{2}}}\right\}\\ \left\{\text{F}\right\}\text{=}\left\{\frac{\text{M}\text{L}}{{\text{t}}^{\text{2}}}\right\}\\ \left\{\text{F}\right\}\text{=}\left\{{\text{M}}^{\text{1}}{\text{ L}}^{\text{1}}{\text{ t}}^{\text{-2}}\right\}\end{array}$

Now, write primary dimension of mass in F-L-T system,

  $\begin{array}{l}\left\{\text{F}\right\}\text{=}\left\{{\text{M}}^{\text{1}}{\text{L}}^{\text{1}}{\text{t}}^{\text{-2}}\right\}\\ \left\{\text{M}\right\}\text{=}\left\{\frac{{\text{Ft}}^{\text{2}}}{\text{L}}\right\}\\ \left\{\text{M}\right\}\text{=}\left\{{\text{F}}^{\text{1}}{\text{L}}^{\text{-1}}{\text{t}}^{\text{2}}\right\}\mathrm{.....}\text{(i)}\end{array}$

Moment is defined as the product of force and length of arm vector.

  $\begin{array}{l}\text{<mover accent="true"><mo>M</mo><mo>→</mo></mover>=}\text{<mover accent="true"><mo>F</mo><mo>→</mo></mover><mo>×</mo>;<mover accent="true"><mo>r</mo><mo>→</mo></mover>}\\ \left\{\text{M}\right\}\text{=}\left\{\left(\text{force}\right)\text{×}\left(\text{length}\right)\right\}\\ \left\{\text{M}\right\}\text{=}\left\{\frac{\text{mass×length}}{{\text{time}}^{\text{2}}}\text{×length}\right\}\\ \left\{\text{M}\right\}\text{=}\left\{\frac{\text{mL}}{{\text{t}}^{\text{2}}}\text{×L}\right\}\\ \left\{\text{M}\right\}\text{=}\left\{\frac{{\text{mL}}^{\text{2}}}{{\text{t}}^{\text{2}}}\right\}\\ \left\{\text{M}\right\}\text{=}\left\{{\text{mL}}^{\text{2}}{\text{t}}^{\text{-2}}\right\}\end{array}$

In exponential form, it can be expressed as,

  $\left\{\text{M}\right\}\text{=}\left\{{\text{mL}}^{\text{2}}{\text{t}}^{\text{-2}}\right\}$

In F-L-T system,

On substituting the value from equation (i),

  $\begin{array}{l}\left\{\text{M}\right\}\text{=}\left\{{\text{mL}}^{\text{2}}{\text{t}}^{\text{-2}}\right\}\\ \left\{\text{E}\right\}\text{=}\left\{\left({\text{F}}^{\text{1}}{\text{L}}^{\text{-1}}{\text{t}}^{\text{2}}\right){\text{L}}^{\text{2}}{\text{t}}^{\text{-2}}\right\}\\ \left\{\text{E}\right\}\text{=}\left\{{\text{F}}^{\text{1}}{\text{L}}^1\right\}\end{array}$

To determine

(c)

The primary dimensions of work or energy in F-L-T system.

Answer to Problem 98P

The primary dimension of energy is $\left\{\text{E}\right\}\text{=}\left\{{\text{F}}^{\text{1}}{\text{L}}^1\right\}.$

Explanation of Solution

In M-L-T system,

  $\begin{array}{l}\text{force}\text{=}\text{mass×acceleration}\\ \text{F}\text{=}\text{m}\text{×}\text{a}\\ \left\{\text{F}\right\}\text{=}\left\{\text{mass}\text{×}\frac{\text{meter}}{{\text{sec}}^{\text{2}}}\right\}\\ \left\{\text{F}\right\}\text{=}\left\{\text{M}\text{×}\frac{\text{L}}{{\text{t}}^{\text{2}}}\right\}\\ \left\{\text{F}\right\}\text{=}\left\{\frac{\text{M}\text{L}}{{\text{t}}^{\text{2}}}\right\}\\ \left\{\text{F}\right\}\text{=}\left\{{\text{M}}^{\text{1}}{\text{ L}}^{\text{1}}{\text{ t}}^{\text{-2}}\right\}\end{array}$

Now, write primary dimension of mass in F-L-T system,

  $\begin{array}{l}\left\{\text{F}\right\}\text{=}\left\{{\text{M}}^{\text{1}}{\text{L}}^{\text{1}}{\text{t}}^{\text{-2}}\right\}\\ \left\{\text{M}\right\}\text{=}\left\{\frac{{\text{Ft}}^{\text{2}}}{\text{L}}\right\}\\ \left\{\text{M}\right\}\text{=}\left\{{\text{F}}^{\text{1}}{\text{L}}^{\text{-1}}{\text{t}}^{\text{2}}\right\}\mathrm{.....}\text{(i)}\end{array}$

Energy: It is defined as the capacity of doing work.

  $\begin{array}{l}\left\{\text{E}\right\}\text{=}\left\{\text{force×}\text{length}\right\}\\ \left\{\text{E}\right\}\text{=}\left\{\frac{\text{mass×length}}{{\text{time}}^{\text{2}}}\text{×length}\right\}\\ \left\{\text{E}\right\}\text{=}\left\{\frac{{\text{mL}}^{\text{2}}}{{\text{t}}^{\text{2}}}\right\}\end{array}$

In exponential form, it can be expressed as,

  $\boxed{\left\{\text{E}\right\}\text{=}{\text{m}}^1{\text{L}}^{\text{2}}{\text{t}}^{\text{-2}}}$

In F-L-T system,

On substituting the value from equation (i),

  $\boxed{\left\{\text{E}\right\}\text{=}{\text{m}}^1{\text{L}}^{\text{2}}{\text{t}}^{\text{-2}}}$

  $\begin{array}{l}\left\{\text{E}\right\}\text{=}\left\{\left({\text{F}}^{\text{1}}{\text{L}}^{\text{-1}}{\text{t}}^{\text{2}}\right){\text{L}}^{\text{2}}{\text{t}}^{\text{-2}}\right\}\\ \left\{\text{E}\right\}\text{=}\left\{{\text{F}}^{\text{1}}{\text{L}}^1\right\}\end{array}$

Surface tension,

It is defined as the behavior of fluid to minimize the area by shrinking into the minimum surface area possible.

In M-L-T system,

  $\begin{array}{l}\text{Surface}\text{tension,}\\ \text{γ=}\frac{\text{force}}{\text{length}}\\ \left\{\text{γ}\right\}\text{=}\left\{\frac{{\text{M}}^{\text{1}}{\text{L}}^{\text{1}}{\text{T}}^{\text{-2}}}{\text{L}}\right\}\\ \left\{\text{γ}\right\}\text{=}\left\{{\text{M}}^{\text{1}}{\text{T}}^{\text{-2}}\right\}\end{array}$

Now, write primary dimension of mass in F-L-T system,

  $\begin{array}{l}\left\{\text{γ}\right\}\text{=}\left\{{\text{M}}^{\text{1}}{\text{T}}^{\text{-2}}\right\}\\ \left\{\text{γ}\right\}\text{=}\left\{\left({\text{F}}^{\text{1}}{\text{L}}^{\text{-1}}{\text{T}}^{\text{2}}\right)\left({\text{T}}^{\text{-2}}\right)\right\}\\ \left\{\text{γ}\right\}\text{=}\left\{{\text{F}}^{\text{1}}{\text{L}}^{\text{-1}}\right\}\end{array}$

Viscosity: A physical property of fluid which denotes the resistance to flow.

Viscosity

In M-L-T system,

  $\begin{array}{l}\text{Viscosity,}\\ \text{μ}\text{=}\frac{\text{force}}{\text{area}}\text{×}\frac{\text{distance}}{\text{velocity}}\\ \left\{\text{μ}\right\}\text{=}\left\{\frac{{\text{M}}^{\text{1}}{\text{L}}^{\text{1}}{\text{T}}^{\text{-2}}}{{\text{L}}^{\text{2}}}\text{×}\frac{\text{L}}{{\text{LT}}^{\text{-1}}}\right\}\\ \left\{\text{μ}\right\}\text{=}\left\{{\text{M}}^{\text{1}}{\text{L}}^{\text{-1}}{\text{T}}^{\text{-1}}\right\}\end{array}$

Now, in F-L-T system,

  $\begin{array}{l}\left\{\text{μ}\right\}\text{=}\left\{\left({\text{F}}^{\text{1}}{\text{L}}^{\text{-1}}{\text{T}}^{\text{2}}\right)\left({\text{L}}^{\text{-1}}{\text{T}}^{\text{-1}}\right)\right\}\\ \left\{\text{μ}\right\}\text{=}\left\{{\text{F}}^{\text{1}}{\text{L}}^{\text{-2}}{\text{T}}^{\text{1}}\right\}\end{array}$

In M-L-T system,

  $\begin{array}{l}\text{force}\text{=}\text{mass×acceleration}\\ \text{F}\text{=}\text{m}\text{×}\text{a}\\ \left\{\text{F}\right\}\text{=}\left\{\text{mass}\text{×}\frac{\text{meter}}{{\text{sec}}^{\text{2}}}\right\}\\ \left\{\text{F}\right\}\text{=}\left\{\text{M}\text{×}\frac{\text{L}}{{\text{t}}^{\text{2}}}\right\}\\ \left\{\text{F}\right\}\text{=}\left\{\frac{\text{M}\text{L}}{{\text{t}}^{\text{2}}}\right\}\\ \left\{\text{F}\right\}\text{=}\left\{{\text{M}}^{\text{1}}{\text{ L}}^{\text{1}}{\text{ t}}^{\text{-2}}\right\}\end{array}$

Now, write primary dimension of mass in F-L-T system,

  $\begin{array}{l}\left\{\text{F}\right\}\text{=}\left\{{\text{M}}^{\text{1}}{\text{L}}^{\text{1}}{\text{T}}^{\text{-2}}\right\}\\ \left\{\text{M}\right\}\text{=}\left\{\frac{{\text{FT}}^{\text{2}}}{\text{L}}\right\}\\ \left\{\text{M}\right\}\text{=}\left\{{\text{F}}^{\text{1}}{\text{L}}^{\text{-1}}{\text{T}}^{\text{2}}\right\}\mathrm{.....}\text{(i)}\end{array}$