Category Archives: Engineering Mathematics Blog

Hibbeler Statics 14E P1.19 — Determine the Weight of the Column with a given Density


A concrete column has a diameter of 350 mm and a length of 2 m. If the density (mass/volume) of concrete is 2.45 Mg/m3, determine the weight of the column in pounds.

Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-19


Solution:

The density of any material is given by the formula

\text{density}=\frac{\text{mass}}{\text{volume}}

From there, we can compute for the mass as

\text{mass}=\text{density} \times \text{volume}

We can solve for mass by multiplying density by volume. The density is already given, and we can compute for the volume of the concrete column by the formula of a volume of a cylinder.

\begin{align*}
\text{V} & = \pi \text{r}^2 \text{h}\\
& =\pi \left( \frac{0.35\ \text{m}}{2} \right)^2 \left( 2 \ \text{m} \right)\\
& =0.1924 \ \text{m}^3
\end{align*}
Concrete Column illustration with diameter of 350 mm or 0.35 m, and a height of 2 m

Therefore, the mass of the concrete column is

\begin{align*}
\text{mass} & =\text{density} \times \text{volume}\\
& = \left( 2.45 \times 10^3 \ \text{kg/m}^3 \right)\times \left( 0.1924 \ \text{m}^3 \right)\\
& =471.44 \ \text{kg}\\
\end{align*}

Now, we can solve for the weight by multiplying the mass by the acceleration due to gravity, g.

\begin{align*}
\text{Weight} & = \text{mass} \times \text{acceleration due to gravity} \\
& = 471.44 \ \text{kg} \times 9.81 \ \text{m/s}^2 \\
& = 4624.78 \ \text{N}
\end{align*}

Finally, we can convert the weight in Newtons to weight in pounds.

\begin{align*}
4624.78\ \text{N} & = 4624.78\ \text{N}\times \frac{1\ \text{lb}}{4.4482\ \text{N}}\\
& = 1039.70\ \text{lb}\\
& = 1.04\times 10^3 \  \text{lb}\\
& = 1.04 \ \text{kip}
\end{align*}

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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

\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

\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

\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:

\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

\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

\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

\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

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

Part B

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

Part C

\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).
\qquad \textbf{a}) \quad \frac{dy}{dx}=xy-3x-2y+6
\qquad \textbf{b})\quad \frac{dy}{dx}=\sin \left( x+y \right)
\qquad \textbf{c}) \quad y\frac{dy}{dx}=e^{x-3y^2}


Solution:

Part A

\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

\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

\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) GNµm, (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

\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

\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

\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

\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

\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

\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

\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

\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

\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

\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

\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

\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

\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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