Tag Archives: combinations of SI units

Hibbeler Statics 14E P1.12 — Evaluation of Expression to Three Significant Figures with Appropriate SI Units


Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (684 µm)/(43 ms), (b) (28 ms)(0.0458 Mm)/(348 mg), (c) (2.68 mm)(426 Mg).

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-12


Solution:

Part A

(684 μm)/43 ms=684×106 m43×103 s=15.9×103 ms=15.9 mm/s\begin{align*} \left( 684 \ \mu\text{m} \right)/43 \ \text{ms} & =\frac{684\times 10^{-6} \ \text{m}}{43\times 10^{-3} \ \text{s}}\\ & = \frac{15.9\times 10^{-3}\ \text{m}}{\text{s}}\\ & = 15.9 \ \text{mm/s} \end{align*}

Part B

(28 ms)(0.0458 Mm)/(348 mg)=[28×103 s][45.8×103×106 m]348×103×103 kg=3.69×106 mskg=3.69 Mms/kg\begin{align*} \left( 28 \ \text{ms} \right)\left( 0.0458 \ \text{Mm} \right)/\left( 348 \ \text{mg} \right) & = \frac{\left[ 28\times 10^{-3} \ \text{s} \right]\left[ 45.8\times 10^{-3}\times 10^6 \ \text{m} \right]}{348\times 10^{-3}\times 10^{-3} \ \text{kg}} \\ & = \frac{3.69\times 10^6 \ \text{m}\cdot \text{s}}{\text{kg}}\\ & = 3.69 \ \text{Mm}\cdot \text{s}/\text{kg}\\ \end{align*}

Part C

(2.68 mm)(426 Mg)=[2.68×103 m][426×103 kg]=1.14×103 mkg=1.14 kmkg\begin{align*} \left( 2.68 \ \text{mm} \right)\left( 426 \ \text{Mg} \right) & = \left[ 2.68\times 10^{-3} \ \text{m} \right]\left[ 426\times 10^3 \ \text{kg} \right]\\ & = 1.14\times 10^3 \ \text{m}\cdot \text{kg}\\ & = 1.14 \ \text{km}\cdot \text{kg}\\ \end{align*}

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Hibbeler Statics 14E P1.8 — Representing Combinations of Units in the Correct SI Form using Appropriate Prefix


Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg/mm, (b) mN/μs, (c) μm∙Mg.

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-8


Solution:

Part A

Mg/mm=106 g103 m=109 gm=Gg/m\begin{align*} \text{Mg/mm} & = \frac{10^6 \ \text{g}}{10^{-3} \ \text{m}} \\ & = 10^9 \ \frac{\text{g}}{\text{m}}\\ & = \text{Gg/m} \end{align*}

Part B

mN/μs=103 N106 s=103 Ns=kN/s\begin{align*} \text{mN/} \mu \text{s} & = \frac{10^{-3} \ \text{N}}{10^{-6} \ \text{s}} \\ & = 10^3 \ \frac{\text{N}}{\text{s}} \\ & = \text{kN/s} \end{align*}

Part C

μmMg=(106 m)(106 g)=mgThis can also be written as:=mmkg\begin{align*} \mu \text{m}\cdot \text{Mg} & = \left( 10^{-6} \ \text{m} \right)\left( 10^6 \ \text{g} \right) \\ & = \text{m}\cdot \text{g}\\ \text{This can also be written as:} \\ & = \text{mm} \cdot \text{kg} \end{align*}

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Representing combinations of units in correct SI Form


Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) 0.000431 kg, (b) 35.3(10³) N, and (c) 0.00532 km.

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-7


Solution:

Part A

0.000 431 kg=0.431×103 kg=0.431×103×103 g=0.431 g\begin{align*} 0.000 \ 431 \ \text{kg} & = 0.431 \times 10^{-3} \ \text{kg}\\ & = 0.431\times 10^{-3} \times 10^3 \ \text{g} \\ & = 0.431 \ \text{g} \end{align*}

Part B

35.3(103) N=35.3 kN\begin{align*} 35.3\left( 10^3 \right) \ \text{N} & = 35.3 \ \text{kN} \end{align*}

Part C

0.00532 km=0.00532×103 m=5.32×103×103 m=5.32 m\begin{align*} 0.00532 \ \text{km} & = 0.00532 \times 10^3 \ \text{m} \\ & = 5.32 \times 10^{-3} \times 10^3 \ \text{m} \\ & = 5.32 \ \text{m} \end{align*}