Author Archives: Engineering Math

Hibbeler Statics 14E P1.14 — Evaluation of expression to correct SI Units

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (212 mN)2, (b) (52800 ms)2, and (c) [548(106)]1/2 ms.

Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-14

Solution:

Part A

(212 mN)2=[212×10−3 N]2=0.0449 N2=4.49×10−2 N2\begin{align*} \left( 212 \ \text{mN} \right)^2 & = \left[ 212\times 10^{-3} \ \text{N} \right]^2 \\ & = 0.0449 \ \text{N}^2 \\ & = 4.49\times 10^{-2} \ \text{N}^2\\ \end{align*}

Part B

(52800 ms)2=[52800×10−3 s]2=2788 s2=2.79×103 s2\begin{align*} \left( 52800 \ \text{ms} \right)^2 & = \left[ 52800\times 10^{-3} \ \text{s} \right]^2 \\ & =2788 \ \text{s}^2 \\ & = 2.79 \times 10^3 \ \text{s}^2 \end{align*}

Part C

[548(106)]1/2 ms=23409 ms=23409×10−3 s=23.4×103×10−3 s=23.4 s\begin{align*} \left[ 548\left( 10^6 \right) \right]^{1/2} \ \text{ms}& =23409 \ \text{ms}\\ & =23409\times 10^{-3}\ \text{s}\\ & = 23.4\times 10^3\times 10^{-3} \ \text{s}\\ & = 23.4 \ \text{s} \end{align*}
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Hibbeler Statics 14E P1.13 — Conversion of Density from Slug per Cubic Foot to Appropriate SI Unit

The density (mass volume) of aluminum is 5.26 slug/ft3. Determine its density in SI units. Use an appropriate prefix.

Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-13

Solution:

5.26 slug/ft3=(5.26 slugft3)(1 ft0.3048 m)3(14.59 kg1 slug)=2710 kg/m3=2.71×103 kg/m3=2.71 Mg/m3\begin{align*} 5.26 \ \text{slug/ft}^3 & =\left( \frac{5.26 \ \text{slug}}{\text{ft}^3} \right)\left( \frac{1 \ \text{ft}}{0.3048\ \text{m}} \right)^3\left( \frac{14.59\ \text{kg}}{1\ \text{slug}} \right)\\ & = 2710\ \text{kg/m}^3\\ & = 2.71\times 10^3 \ \text{kg/m}^3\\ & = 2.71\ \text{Mg/m}^3 \end{align*}
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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=684×10−6 m43×10−3 s=15.9×10−3 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)=[28×10−3 s][45.8×10−3×106 m]348×10−3×10−3 kg=3.69×106 m⋅skg=3.69 Mm⋅s/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)=[2.68×10−3 m][426×103 kg]=1.14×103 m⋅kg=1.14 km⋅kg\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 P1.11 — Representing Measurements with SI units Having Appropriate Prefix

Represent each of the following with SI units having an appropriate prefix: (a) 8653 ms, (b) 8368 N, (c) 0.893 kg.

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-11

Solution:

Part A

8653 ms=8653 (10−3) s=8.653 s\begin{align*} 8653 \ \text{ms} & = 8653 \ \left( 10^{-3} \right) \ \text{s} \\ & =8.653 \ \text{s} \end{align*}

Part B

8368 N=8.368×103 N=8.368 kN\begin{align*} 8368 \ \text{N} & = 8.368\times 10^3 \ \text{N}\\ & = 8.368 \ \text{kN}\\ \end{align*}

Part C

0.893 kg=0.893×103 g=893 g\begin{align*} 0.893 \ \text{kg} & = 0.893\times 10^3 \ \text{g} \\ & =893 \ \text{g} \end{align*}
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Determining if a given Differential Equation is Separable or Not

Determine whether each of the following differential equations is or is not separable, and, if it is separable, rewrite the equation in the form dy/dx=f(x) g(y).
a)dydx=xy−3x−2y+6\qquad \textbf{a}) \quad \frac{dy}{dx}=xy-3x-2y+6
b)dydx=sin⁡(x+y)\qquad \textbf{b})\quad \frac{dy}{dx}=\sin \left( x+y \right)
c)ydydx=ex−3y2\qquad \textbf{c}) \quad y\frac{dy}{dx}=e^{x-3y^2}

Solution:

Part A

dydx=xy−3x−2y+6dydx=(xy−3x)−(2y−6)dydx=x(y−3)−2(y−3)dydx=(x−2)(y−3)\begin{align*} \frac{dy}{dx} & = xy-3x-2y+6 \\ \frac{dy}{dx} & = \left( xy-3x \right)-\left( 2y-6 \right)\\ \frac{dy}{dx} & = x\left( y-3 \right)-2\left( y-3 \right)\\ \frac{dy}{dx} & = \left( x-2 \right)\left( y-3 \right) \end{align*}

Since F(x,y) is factorable in the form f(x) g(y), the given differential equation is separable.

Part B

dydx=sin⁡(x+y)dydx=sin⁡(x)cos⁡(y)+cos⁡(x)sin⁡(y)\begin{align*} \frac{dy}{dx} & = \sin\left( x+y \right) \\ \frac{dy}{dx} & = \sin\left( x \right)\cos\left( y \right) +\cos\left( x \right)\sin\left( y \right)\\ \end{align*}

Since F(x,y) is not factorable in the form f(x) g(y), the given differential equation is not separable.

Part C

ydydx=ex−3y2ydydx=exe3y2dydx=exye3y2dydx=ex(1ye3y2)\begin{align*} y \frac{dy}{dx} & = e^{x-3y^2}\\ y \frac{dy}{dx} & = \frac{e^x}{e^{3y^2}}\\ \frac{dy}{dx} & =\frac{e^x}{y e^{3y^2}} \\ \frac{dy}{dx} & = e^x \left( \frac{1}{ye^{3y^2}} \right) \\ \end{align*}

Since F(x,y) is factorable in the form f(x) g(y), the given differential equation is separable.

Hibbeler Statics 14E P1.10 — Representing Combinations of Units in the Correct SI Form

Represent each of the following combinations of units in the correct SI form: (a) GN∙µm, (b) kg/µm, (c) N/ks2, and (d) KN/µs.

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-10

Solution:

Part A

GN⋅μm=(109 N)(10−6 m)=103 N⋅m=kN⋅m\begin{align*} \text{GN} \cdot \mu \text{m} & = \left( 10^9 \ \text{N} \right)\left( 10^{-6} \ \text{m} \right)\\ & = 10^3 \ \text{N} \cdot \text{m}\\ & = \text{kN} \cdot \text{m} \end{align*}

Part B

kg/μm=103 g10−6 m=109 gm=Gg/m\begin{align*} \text{kg/}\mu\text{m} & = \frac{10^3 \ \text{g}}{10^{-6} \ \text{m}} \\ & = 10^9 \ \frac{\text{g}}{\text{m}} \\ & = \text{Gg/m} \end{align*}

Part C

N/ks2=N(103 s)2=N106 s2=10−6 Ns2=μN/s2\begin{align*} \text{N/ks}^2 & = \frac{\text{N}}{\left( 10^3 \ \text{s} \right)^2}\\ & = \frac{\text{N}}{10^6 \ \text{s}^2} \\ & = 10^{-6} \ \frac{\text{N}}{\text{s}^2} \\ & = \mu \text{N}/\text{s}^2 \end{align*}

Part D

kN/μs=103 N10−6 s=109 Ns=GN/s\begin{align*} \text{kN}/ \mu\text{s} & = \frac{10^3 \ \text{N}}{10^{-6} \ \text{s}} \\ & = 10^9 \ \frac{\text{N}}{\text{s}}\\ & = \text{GN/s} \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 g10−3 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=10−3 N10−6 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

μm⋅Mg=(10−6 m)(106 g)=m⋅gThis can also be written as:=mm⋅kg\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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Hibbeler Statics 14E P1.5 — Representing a number between 0.1 and 1000 Using an Appropriate Prefix

Represent each of the following as a number between 0.1 and 1000 using an appropriate prefix: (a) 45 320 kN, (b) 568(105) mm, (c) 0.00563 mg.

Statics of Rigid Bodies 14th by RC Hibbeler, Problem 1-5

Solution:

Part A

45 320 kN=45.3×103 kN=45.3×103×103 N=45.3×106 N=45.3 MN\begin{align*} 45 \ 320 \ \text{kN} & = 45.3 \times 10^3 \ \text{kN}\\ & = 45.3\times 10^3\times 10^3 \ \text{N}\\ & = 45.3 \times 10^6 \ \text{N}\\ & = 45.3 \ \text{MN} \end{align*}

Part B

568(105) mm=568(105)×10−3 m=568×102 m=56.8×103 m=56.8 km\begin{align*} 568\left( 10^5 \right)\ \text{mm} & =568\left( 10^5 \right)\times 10^{-3} \ \text{m}\\ & = 568\times 10^2 \ \text{m} \\ & = 56.8 \times 10^3 \ \text{m}\\ & = 56.8 \ \text{km} \end{align*}

Part C

0.00563 mg=5.63×10−3 mg=5.63×10−3×10−3 g=5.63×10−6 g=5.63 μg\begin{align*} 0.00563 \ \text{mg} & = 5.63\times 10^{-3} \ \text{mg}\\ & = 5.63\times 10^{-3}\times 10^{-3} \ \text{g}\\ & = 5.63\times 10^{-6} \ \text{g}\\ & = 5.63 \ \mu\text{g} \end{align*}
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Hibbeler Statics 14E P1.4 — Conversion from English units to Metric Units

Convert: (a) 200 lb·ft to N·m, (b) 350 lb/ft3 to kN/m3, (c) 8 ft/h to mm/s. Express the result to three significant figures. Use an appropriate prefix.

Statics of Rigid Bodies 14th by RC Hibbeler, Problem 1-4

Solution:

Part A

200 lb⋅ft=(200 lb⋅ft)(4.4482 N1 lb)(0.3048 m1 ft)=271 N⋅m\begin{align*} 200\ \text{lb}\cdot \text{ft} & =\left( 200\ \text{lb}\cdot \text{ft} \right)\left( \frac{4.4482\ \text{N}}{1\ \text{lb}} \right)\left( \frac{0.3048\ \text{m}}{1\ \text{ft}} \right)\\ &=271\ \text{N}\cdot \text{m} \end{align*}

Part B

350 lb/ft3=(350 lb1 ft3)(1 ft0.3048 m)3(4.4482 N1 lb)(1 kN1000 N)=55.0 kN/m3\begin{align*} 350 \ \text{lb/ft}^3& = \left( \frac{350\ \text{lb}}{1\ \text{ft}^3} \right)\left( \frac{1\ \text{ft}}{0.3048\ \text{m}} \right)^3\left( \frac{4.4482\ \text{N}}{1\ \text{lb}} \right)\left( \frac{1\ \text{kN}}{1000\ \text{N}} \right)\\ & = 55.0\ \text{kN/m}^3 \end{align*}

Part C

8 ft/hr=(8 ft1 hr)(1 hr3600 s)(0.3048 m1 ft)(1000 mm1 m)=0.677 mm/s\begin{align*} 8 \ \text{ft/hr}& = \left( \frac{8\ \text{ft}}{1\ \text{hr}} \right)\left( \frac{1\ \text{hr}}{3600\ \text{s}} \right)\left( \frac{0.3048\ \text{m}}{1\ \text{ft}} \right)\left( \frac{1000 \ \text{mm}}{1 \ \text{m}} \right)\\ & = 0.677\ \text{mm/s} \end{align*}
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Hibbeler Statics 14E P1.3 — Representing a combination of units in the correct SI form

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

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-3

Solution:

Part A

Mg/ms=103 kg10−3 s=106kg/s=Gg/s\begin{align*} \text{Mg/ms} & = \frac{10^3 \ \text{kg}}{10^{-3} \ \text{s}} \\ & = 10^6 \text{kg/s}\\ & = \text{Gg/s} \end{align*}

Part B

N/mm=1 N10−3 m=103 N/m=kN/m\begin{align*} \text{N/mm} & = \frac{1\ \text{N}}{10^{-3} \ \text{m}}\\ & = 10^3 \ \text{N/m}\\ & = \text{kN/m} \end{align*}

Part C

mN(kg⋅μs)=10−3 N10−6 kg⋅s=103 N/(kg⋅s)=kN/(kg⋅s)\begin{align*} \frac{\text{mN}}{\left( \text{kg} \cdot \mu \text{s} \right)} & = \frac{10^{-3} \ \text{N}}{10^{-6} \ \text{kg} \cdot \text{s}}\\ & =10^3 \ \text{N}/\left( \text{kg} \cdot \text{s} \right)\\ & = \text{kN}/\left( \text{kg} \cdot \text{s}\right) \end{align*}
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