Question

1 lets continue to practice learning the dimensions of common engineering terms. then, solve by completing tables without using your notes or peers.
characteristics dimension si unit equation
length l m -
mass m kg -
time t s -
area m² example: a = l w
volume l or m³
velocity m/s
acceleration m/s²
force n
work j
power w
pressure pa
*information should be memorized; it will not be given on a quiz/test.
2 you will also need to be able to determine the dimensions of unfamiliar terms based solely upon units. to practice this skill, write the following units in dimensional form.
units dimensional equivalents
(kg s)/m³
(pa s)/m
kg/m²
w h/m²
kg/(m s²)

Explanation:

Step1: Recall base - dimension relationships

Length has dimension $L$, mass $M$, time $T$. Force $F = ma$, so $[F]=MLT^{- 2}$ (where $a$ has dimension $LT^{-2}$ and $m$ has dimension $M$). Work $W = Fd$ (where $d$ is distance with dimension $L$), so $[W]=ML^{2}T^{-2}$. Power $P=\frac{W}{t}$, so $[P]=ML^{2}T^{-3}$. Pressure $P=\frac{F}{A}$, so $[P]=ML^{-1}T^{-2}$.

Step2: Find dimensions for $(kg\cdot s)/m^{3}$

Since $kg = M$, $s=T$ and $m = L$, the dimensional form is $ML^{-3}T$.

Step3: Find dimensions for $(Pa\cdot s)/m$

Since $Pa=ML^{-1}T^{-2}$, then $(Pa\cdot s)/m=(ML^{-1}T^{-2}\cdot T)/L=ML^{-2}T^{-1}$.

Step4: Find dimensions for $kg/m^{2}$

Since $kg = M$ and $m = L$, the dimensional form is $ML^{-2}$.

Step5: Find dimensions for $W\cdot h/m^{2}$

Since $W = ML^{2}T^{-3}$ and $h = 3600s$ (time), $W\cdot h=ML^{2}T^{-3}\cdot T=ML^{2}T^{-2}$, so $W\cdot h/m^{2}=ML^{0}T^{-2}$.

Step6: Find dimensions for $kg/(m\cdot s^{2})$

Since $kg = M$, $m = L$ and $s=T$, the dimensional form is $ML^{-1}T^{-2}$.

Answer:

Units	Dimensional Equivalents
$(Pa\cdot s)/m$	$ML^{-2}T^{-1}$
$kg/m^{2}$	$ML^{-2}$
$W\cdot h/m^{2}$	$ML^{0}T^{-2}$
$kg/(m\cdot s^{2})$	$ML^{-1}T^{-2}$