Tag Archives: free statics and dynamics books

Hibbeler Statics 14E P2.4 – Components of a Force Along Two Non-Perpendicular Axes

The vertical force \textbf{F} acts downward at A on the two-membered frame. Determine the magnitudes of the two components of \textbf{F} directed along the axes of AB and AC. Set \textbf{F} = 500 N.

Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-4

Solution:

Draw the components of the force using the parallelogram law. Then the triangulation rule.

Parallelogram Law
Triangulation Rule

Solving for FAC using sine law.

\begin{align*}
\frac{\text{F}_\text{AC}}{\sin \ 45^\circ } & =\frac{500 \ \text{N}}{\sin \ 75^\circ }\\
\text{F}_\text{AC} & = \frac{500 \ \text{N} \ \sin45^\circ }{\sin\ 75^\circ }\\
\text{F}_\text{AC} & =366.0254 \ \text{N}\\
\text{F}_\text{AC} &\approx 366 \ \text{N}
\end{align*}

Solve for FAB using sine law.

\begin{align*}
\frac{\text{F}_\text{AB}}{\sin \ 60^\circ } & =\frac{500 \ \text{N}}{\sin \ 75^\circ }\\
\text{F}_\text{AB} & = \frac{500 \ \text{N} \ \sin60^\circ }{\sin\ 75^\circ }\\
\text{F}_\text{AB} & =448.2877 \ \text{N}\\
\text{F}_\text{AB} &\approx 448 \ \text{N}
\end{align*}
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Hibbeler Statics 14E P2.2 — Solving for an Unknown Force Given the Magnitude and Direction of a Resultant and Another Force

If the magnitude of the resultant force is to be 500 N, directed along the positive y-axis, determine the magnitude of force F and its direction \theta .

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

Solution:

The parallelogram law and the triangulation rule are shown in the figures below.

Engineering Mechanics by RC Hibbeler Problem 2.2 Parallelogram Law
Parallelogram Law
Engineering Mechanics by RC Hibbeler Problem 2.2 Triangulation Rule
Triangulation Rule

Considering the figure of the triangulation rule, we can solve for the magnitude of \textbf{F} using the cosine law.

\begin{align*}
\textbf{F} & = \sqrt{700^2+500^2-2\left( 700 \right)\left( 500 \right)\cos105^{\circ}}\\
& = 959.78 \  \text{N}\\
& = 960 \  \text{N}\\
\end{align*}

Then we use the sine law to solve for the angle \theta. 

\begin{align*}
\frac{\sin \left(90^{\circ}-\theta \right)}{700} & = \frac{\sin 105^{\circ}}{959.78}\\
\sin \left(90^{\circ}-\theta \right) & =\frac{700 \sin 105^{\circ }}{959.78}\\
90^{\circ}-\theta & = \sin^{-1} \left( \frac{700 \sin 105^{\circ }}{959.78} \right)\\
\theta & = 90^\circ-\sin^{-1} \left( \frac{700 \sin 105^{\circ }}{959.78} \right) \\
\theta & =  90^\circ-44.79^\circ\\
\theta & =  45.2^\circ\\
\end{align*}
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Hibbeler Statics 14E P2.1 — Solving for the Magnitude and Direction of the Resultant of Two Coplanar-Concurrent Forces

If \theta = 60 \degree and \textbf{F} = 450 \ \text{N}, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

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

Solution:

The parallelogram law and the triangulation rule are shown in the figures below.

(a) Parallelogram Law
(b) Triangulation Rule

Considering figure (b), we can solve for the magnitude of \textbf{F}_R using the cosine law.

\begin{align*}
\textbf{F}_R & = \sqrt{700^2+450^2-2\left( 700 \right)\left( 450 \right)\cos45^{\circ}}\\
& = 497.01 \ \text{N}\\
& = 497 \ \text{N}
\end{align*}

Then we use the sine law to solve for the interior angle \theta. 

\begin{align*}
\frac{\sin \theta}{700} & = \frac{\sin 45^{\circ}}{497.01}\\
\sin \theta & =\frac{700\ \sin 45^{\circ }}{497.01}\\
\theta & = \sin^{-1} \left( \frac{700\ \sin 45^{\circ }}{497.01} \right)\\
& \text{This is an ambiguous case }\\
\theta & = 84.81^\circ \  or \  \theta =95.19^\circ \\
\end{align*}

In here, the correct angle measurement is \theta = 95.19^{\circ}.

Thus, the direction angle \phi of \textbf{F}_R measured counterclockwise from the positive x-axis, is 

\begin{align*}
\phi & = \theta +60^\circ \\
& = 95.19^\circ +60^\circ \\
& = 155^\circ 
\end{align*}
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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 P2.2 — Resultant of a System of Two Forces

Determine the magnitude of the resultant force \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 \textbf{F}_{\text{R}}

\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. 

\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 \phi .

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

\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. 

\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. 

\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 \textbf{g}=5.30 \ \text{ft/s}^2.

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

Part E

\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:

\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

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

Weight of the 12 kg particle

\textbf{W}_2=12\left(9.81\right)=118\:\text{N}
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