Tag Archives: Engineering Mechanics

Torsion Featured Image: Chapter 3 of the book of Andrew Pytel and Ferdinand Singer Strength of Materials 4th Edition

Chapter 3: Torsion

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Simple Strain Featured Image: Chapter 2 of the book of Andrew Pytel and Ferdinand Singer Strength of Materials 4th Edition

Chapter 2: Simple Strain

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Simple Stress Featured Image: Strength of Materials 4th Edition by Andrew Pytel and Ferdinand Singer

Chapter 1: Simple Stress

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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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Hibbeler Statics 14E P2.2 — Resultant of a System of Two Forces

Determine the magnitude of the resultant force FR=F1+F2\textbf{F}_{\text{R}} = \textbf{F}_1 + \textbf{F}_2 and its direction, measured counterclockwise from the positive x axis. 

Engineering Mechanics: Statics figure for Problem 2-3

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 2-1
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-3

SOLUTION:

The parallelogram law of the force system is shown.

Consider the triangle AOB.

Using cosine law to solve for the resultant force FR\textbf{F}_{\text{R}}

FR=(250)2+(375)2−2(250)(375)cos⁡ 75∘=393.2 lb=393 lb\begin{align*} \textbf{F}_\text{R} & =\sqrt{\left(250\right)^2+\left(375\right)^2-2\left(250\right)\left(375\right) \cos\:75^{\circ} }\\ & =393.2 \ \text{lb}\\ & =393\:\text{lb}\\ \end{align*}

The value of angle θ can be solved using sine law. 

393.2sin⁡ (75∘)=250sin⁡ θsin⁡θ=250 sin⁡75°393.2θ=sin⁡−1(250 sin⁡75°393.2)θ=37.89∘\begin{align*} \frac{393.2}{\sin\:\left(75^{\circ} \right)} & = \frac{250}{\sin\:\theta } \\ \sin \theta & = \frac{250 \ \sin75 \degree}{393.2}\\ \theta & =\sin^{-1} \left(\frac{250 \ \sin75 \degree}{393.2}\right)\\ \theta & = 37.89^{\circ}\\ \end{align*}

Solve for the unknown angle ϕ \phi .

ϕ=360∘−45∘+37.89∘=353∘\phi =360^{\circ} -45^{\circ} +37.89^{\circ} =353^{\circ}

The resultant force has a magnitude of 393 lb and is located 353º measured counterclockwise from the positive x-axis.

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Computing the mass and weight of a man on earth and on the moon

If a man weighs 155 lb on earth, specify (a) his mass in slugs, (b) his mass in kilograms, and (c) his weight in newtons. If the man is on the moon, where the acceleration due to gravity is gm=5.30 ft/s², determine (d) his weight in pounds, and (e) his mass in kilograms.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-21
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-20

Solution:

Part A

From the formula, W=mg\text{W}=\text{mg}, we can solve for the mass by dividing the weight by the acceleration due to gravity. That is

m=Wg=155 lb32.2 ft/s2=4.81 slug\begin{align*} \text{m} & = \frac{\text{W}}{\textbf{g}}\\ & = \frac{155\ \text{lb}}{32.2 \ \text{ft/s}^2}\\ & = 4.81 \ \text{slug}\\ \end{align*}

Part B

Convert the slug to kilograms, knowing that 1 slug = 14.59 kg.

m=(15532.2slug)(14.59 kg1 kg)=70.2 kg\begin{align*} \begin{align*} \text{m} & = \left( \frac{155}{32.2} \text{slug}\right)\left( \frac{14.59 \ \text{kg}}{1 \ \text{kg}} \right)\\ & = 70.2 \ \text{kg}\\ \end{align*} \end{align*}

Part C

Convert the 155 lb to newtons using 1 lb = 4.448 N.

W=155 lb×4.448 N1 lb=689 N\begin{align*} \textbf{W} & = 155 \ \text{lb}\times \frac{4.448 \ \text{N}}{1 \ \text{lb}}\\ & = 689 \ \text{N}\\ \end{align*}

Part D

Using the same formulas, but now g=5.30 ft/s2\textbf{g}=5.30 \ \text{ft/s}^2.

W=155(5.3032.2)=25.5 lb\textbf{W}=155\left(\frac{5.30}{32.2}\right)=25.5\:\text{lb}

Part E

m=155(14.59 kg32.2)=70.2 kg\textbf{m}=155\left(\frac{14.59\:\text{kg}}{32.2}\right)=70.2\:\text{kg}
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The force of gravity acting between two particles

Two particles have a mass of 8 kg and 12 kg, respectively. If they are 800 mm apart, determine the force of gravity acting between them. Compare this result with the weight of each particle.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-20
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-21

Solution:

The force of gravity acting between them:

F=G⋅m1m2r2=66.73(10−12)m3/(kg⋅s2)[8 kg(12 kg)(0.8 m)2]=10(10−9) N=10.0 nN\begin{align*} \textbf{F} & =\textbf{G}\cdot \frac{\text{m}_1\text{m}_2}{\text{r}^2}\\ & =66.73\left(10^{-12}\right) \text{m}^3/ \left( \text{kg} \cdot \text{s}^2 \right) \left[\frac{8 \ \text{kg} \left(12\ \text{kg}\right)}{\left(0.8\ \text{m} \right)^2}\right]\\ &=10\left(10^{-9}\right)\ \text{N}\\ & =10.0 \ \text{nN}\\ \end{align*}

The weight of the 8 kg particle

W1=8(9.81)=78.5 N\textbf{W}_1=8\left(9.81\right)=78.5\:\text{N}

Weight of the 12 kg particle

W2=12(9.81)=118 N\textbf{W}_2=12\left(9.81\right)=118\:\text{N}
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Expressing the Density of Water in SI Units

Water has a density of 1.94 slug/ft³. What is the density expressed in SI units? Express the answer to three significant figures.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-19
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-17

Solution:

ρw=(1.94 slug1 ft3)(14.59 kg1 slug)(1 ft30.30483 m3)=(1.94 slug1 ft3)(14.59 kg1 slug)(1 ft30.30483 m3)=999.6 kgm3=1.00 Mg/m3\begin{align*} \rho _w & =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\ & =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\ & =999.6\:\frac{\text{kg}}{\text{m}^3}\\ & =1.00\:\text{Mg/m}^3\\ \end{align*}
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Showing How an Equation is Dimensionally Homogeneous

Using the SI system of units, show that Eq. 1–2 is a dimensionally homogeneous equation which gives F in newtons. Determine to three significant figures the gravitational force acting between two spheres that are touching each other. The mass of each sphere is 200 kg and the radius is 300 mm.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-18
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-15

Solution:

To prove that F is in Newtons, we have

F=G⋅m1m2r2=(m3kg⋅s2)(kg⋅kgm2)=kg⋅ms2=N\begin{align*} \text{F} & =\text{G}\cdot \frac{\text{m}_1\text{m}_2}{\text{r}^2}\\ & =\left(\frac{\text{m}^3}{\text{kg}\cdot \text{s}^2}\right)\left(\frac{\text{kg}\cdot \text{kg}}{\text{m}^2}\right)\\ & =\frac{\text{kg}\cdot \text{m}}{\text{s}^2}\\ & =\text{N} \end{align*}

Now, if we substitute the given values into the equation

F=66.73(10−12)[200(200)0.62]=7.41(10−6)N=7.41 μN\begin{align*} \text{F} & = 66.73\left(10^{-12}\right)\left[\frac{200\left(200\right)}{0.6^2}\right]\\ & = 7.41\left(10^{-6}\right) \text{N}\\ & =7.41\ \mu \text{N}\\ \end{align*}
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Problem 1-17| General Principles| Engineering Mechanics: Statics| RC Hibbeler

If an object has a mass of 40 slugs, determine its mass in kilograms.

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