Author Archives: Engineering Math

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) 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

\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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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

\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

\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

\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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Skidding on a Curve: Unbanked Curves – Uniform Circular Motion Example Problem

A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is \mu _s=0.60; (b) the pavement is icy and \mu _s=0.25.

Solution:

The forces on the car are gravity mg downward, the normal force FN exerted upward by the road, and a horizontal friction force due to the road. They are shown in the free-body diagram of the car below. The car will follow the curve if the maximum static friction force is greater than the mass times the centripetal acceleration.

Part A

In the vertical direction (y) there is no acceleration. Newton’s second law tells us that the normal force on the car is equal to the weight mg since the road is flat:

\begin{align*}
\sum_{}^{}F_y & =m\ a_c\\
\\
F_N\ -\ mg & =0\\
\\
F_N&=mg\\
\\
F_N &=\left( 1000\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)\\
\\
F_N &=9810\  \text{N}
\end{align*}

In the horizontal direction the only force is friction, and we must compare it to the force needed to produce the centripetal acceleration to see if it is sufficient. The net horizontal force required to keep the car moving in a circle around the curve is

\begin{align*}
\sum_{}^{}F_c & =m\ a_c\\
\\
 & =m\cdot \frac{v^2}{r}\\
\\
& =\left( 1000\ \text{kg} \right)\cdot \frac{\left( 15\ \text{m/s} \right)^2}{50\ \text{m}}\\
\\
&=4500\ \text{N}
\end{align*}

Now we compute the maximum total static friction force (the sum of the friction forces acting on each of the four tires) to see if it can be large enough to provide a safe centripetal acceleration. For (a), \mu _s=0.60, and the maximum friction force attainable is

\begin{align*}
\sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\
\\
&=\left( 0.60 \right)\left( 9810\ \text{N} \right)\\
\\
&=5886\ \text{N}
\end{align*}

Since a force of only 4500 N is needed, and that is, in fact, how much will be exerted by the road as a static friction force, the car can follow the curve.

Part B

The maximum static friction force possible is

\begin{align*}
\sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\
\\
&=\left( 0.25 \right)\left( 9810\ \text{N} \right)\\
\\
&=2452.5\ \text{N}
\end{align*}

The car will skid because the ground cannot exert sufficient force (4500 N is needed) to keep it moving in a curve of radius 50 m at a speed of 54 km/h.