Tag Archives: russell hibbeler

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.

Solving for F_AC 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 F_AB 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.

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 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 14^th 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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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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Statics 3.3 – Solving for the magnitude and direction of a force for equilibrium | Hibbeler 14th Edition

Determine the magnitude and direction θ of F so that the particle is in equilibrium.

Statics 14E Problem 3.3 Forces in Equilibrium with unknown force and its direction

Solution:

Free-body Diagram:

Equilibrium Equation:

Summation of forces in the x-direction:

\begin{aligned} \xrightarrow{+} \: \sum F_x & = 0 & \\ 5 \: \text{kN}+F \sin \theta - 8 \: \text {kN} \cos 30 \degree - 4\: \text{kN} \cos 60 \degree & = 0 & \\ F \sin \theta &= 3.9282 & (1) \end{aligned}

Summation of forces in the y-direction:

\begin{aligned} +\uparrow \sum F_y & = 0 &\\ 8 \sin 30 \degree - 4 \sin 60 \degree - F \cos \theta & =0 &\\ F \cos \theta & = 0.5359 & (2)\\ \end{aligned}

We now have two equations. Divide Eq (1) by (2)

\begin{aligned} \dfrac{F \sin \theta}{F \cos \theta} &= \dfrac{3.9282}{0.5359} \\ \dfrac{ \sin \theta}{ \cos \theta} & = 7.3301 \\ \end{aligned}

We know that $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$ :

\begin{aligned} \tan \theta &=7.3301 \\ \theta & = \tan^{-1}7.3301\\ \textcolor{blue}\theta & \textcolor{blue}{=82.2\degree}\\ \end{aligned}

Substituting this result to equation (1), we have

\begin{aligned} F\sin 82.2 \degree & = 3.9282 \\ \textcolor{blue}F & \textcolor{blue}{=3.96 \ \text{kN}} \end{aligned}

Statics 3.2 – Equilibrium of Truss Members that are Pin Connected | Hibbeler 14th Edition

The members of a truss are pin connected at joint O. Determine the magnitude of F₁ and its angle θ for equilibrium. Set F₂=6 kN.

Figure 3.1: Engineering Mechanics: Statics Equilibrium of Particle — Figure 3.1/3.2

Solution:

Free-body diagram:

Equations of Equilibrium:

The summation of forces in the x-direction:

\begin{aligned} \sum F_x & = 0 &\\ 6 \sin 70 \degree + F_1 \cos \theta - 5 \cos 30 \degree - \dfrac {4}{5} \left(7 \right) & = 0 & \\ F_1 \cos \theta & = 4.2920 & (1) \end{aligned}

The summation of forces in the y-direction:

\begin{aligned} \sum F_y & =0 & \\ 6 \cos 70 \degree+5 \sin 30 \degree - F_1 \sin \theta - \dfrac{3}{5} \left( 7 \right ) & =0 & \\ F_1 \sin \theta &=0.3521 & (2)\\ \end{aligned}

We came up with 2 equations with unknowns $F_1$ and $\theta$ . To solve the equations simultaneously, we can use the method of substitution.

Using equation 1, solve for $F_1$ in terms of $\theta$ .

\begin{aligned} F_1 \cos \theta & = 4.2920 &\\ F_1 & =\dfrac{4.2920}{ \cos \theta } & (3) \\ \end{aligned}

Now, substitute this equation (3) to equation (2).

\begin{aligned} F_1 \sin \theta & = 0.3521 \\ \left ( \dfrac {4.2920}{\cos \theta} \right) \sin \theta & =0.3521 \\ 4.2920 \cdot \dfrac{\sin \theta}{\cos \theta} & = 0.3521 \\ 4.2920 \tan \theta & = 0.3521 \\ \tan \theta & = \dfrac{0.3521}{4.2920} \\ \theta &= \tan ^{-1} \dfrac{0.3521}{4.2920} \\ \theta & = 4.69 \degree \end{aligned}

Substitute the solved value of $\theta$ to equation (3).

\begin{aligned} F_1 & = \dfrac{4.2920}{\cos \theta} \\ F_1 &= \dfrac{4.2920}{\cos 4.69 \degree} \\ F_1 & = 4.31 \text{kN} \end{aligned}

Therefore, the answers to the questions are:

\begin{aligned} F_1= & \:4.31 \: \text {kN} \\ \theta = & \: 4.69 \degree \end{aligned}

Statics 3.1 – Equilibrium of Truss Members in Pin Connection | Hibbeler 14th Edition

The members of a truss are pin connected at joint O. Determine the magnitudes of F₁ and F₂ for equilibrium. Set θ=60.

Solution:

Free-body diagram:

Equations of Equilibrium:

Take the sum of horizontal forces considering forces to the right positive, and equate to zero.

\begin {aligned} \sum{F}_x &= 0 & \\ F_1 \cos{60 \degree}+F_2 \sin{70 \degree}-5\cos{30 \degree}-\dfrac{4}{5}\left(7\right) &= 0 &\\ 0.5F_1+0.9397F_2&=9.9301 &(1)\\ \end {aligned}

Take the sum of vertical forces considering upward forces positive, and equate to zero.

\begin{aligned} \sum F_y&=0 &\\ -F_1\sin60\degree+F_2\cos70\degree+5\sin30\degree-\dfrac{3}{5}\left(7\right)&=0 &\\ -0.8660F_1+0.3420F_2&=1.7 &(2)\\ \end{aligned}

Now, we have two equations with two unknowns $F_1$ and $F_2$ . So, we have a system of two equations. We can solve this using algebra, or we can directly use our calculator with this capability. The answers are

F_1=1.83 \: \text{kN}\\ F_2=9.60 \: \text{kN}

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.

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 .

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.

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.

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

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.

Determine the magnitude and direction θ of F so that the particle is in equilibrium.

Solution:

The members of a truss are pin connected at joint O. Determine the magnitude of F1 and its angle θ for equilibrium. Set F2=6 kN.

Solution:

The members of a truss are pin connected at joint O. Determine the magnitudes of F1 and F2 for equilibrium. Set θ=60.

Solution:

Coplanar Force Systems

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Three-Dimensional Force Systems

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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.

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 $\theta$ .

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.

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.

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.

The members of a truss are pin connected at joint O. Determine the magnitude of F₁ and its angle θ for equilibrium. Set F₂=6 kN.

The members of a truss are pin connected at joint O. Determine the magnitudes of F₁ and F₂ for equilibrium. Set θ=60.