Tag Archives: RC Hibbeler

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

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

FACsin⁡ 45∘=500 Nsin⁡ 75∘FAC=500 N sin⁡45∘sin⁡ 75∘FAC=366.0254 NFAC≈366 N\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.

FABsin⁡ 60∘=500 Nsin⁡ 75∘FAB=500 N sin⁡60∘sin⁡ 75∘FAB=448.2877 NFAB≈448 N\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 θ \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 F \textbf{F} using the cosine law.

F=7002+5002−2(700)(500)cos⁡105∘=959.78 N=960 N\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 θ\theta.

sin⁡(90∘−θ)700=sin⁡105∘959.78sin⁡(90∘−θ)=700sin⁡105∘959.7890∘−θ=sin⁡−1(700sin⁡105∘959.78)θ=90∘−sin⁡−1(700sin⁡105∘959.78)θ=90∘−44.79∘θ=45.2∘\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 θ=60°\theta = 60 \degree and F=450 N \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 FR \textbf{F}_R using the cosine law.

FR=7002+4502−2(700)(450)cos⁡45∘=497.01 N=497 N\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 θ\theta.

sin⁡θ700=sin⁡45∘497.01sin⁡θ=700 sin⁡45∘497.01θ=sin⁡−1(700 sin⁡45∘497.01)This is an ambiguous case θ=84.81∘ or θ=95.19∘\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 θ=95.19∘ \theta = 95.19^{\circ}.

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

ϕ=θ+60∘=95.19∘+60∘=155∘\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

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

From there, we can compute for the mass as

mass=density×volume\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.

V=πr2h=π(0.35 m2)2(2 m)=0.1924 m3\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

mass=density×volume=(2.45×103 kg/m3)×(0.1924 m3)=471.44 kg\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.

Weight=mass×acceleration due to gravity=471.44 kg×9.81 m/s2=4624.78 N\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.

4624.78 N=4624.78 N×1 lb4.4482 N=1039.70 lb=1.04×103 lb=1.04 kip\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 14th Edition by RC Hibbeler, Problem 1-11

Solution:

Part A

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

Part B

8368 N=8.368×103 N=8.368 kN\begin{align*} 8368 \ \text{N} & = 8.368\times 10^3 \ \text{N}\\ & = 8.368 \ \text{kN}\\ \end{align*}

Part C

0.893 kg=0.893×103 g=893 g\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

GN⋅μm=(109 N)(10−6 m)=103 N⋅m=kN⋅m\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

kg/μm=103 g10−6 m=109 gm=Gg/m\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

N/ks2=N(103 s)2=N106 s2=10−6 Ns2=μN/s2\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

kN/μs=103 N10−6 s=109 Ns=GN/s\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

Mg/mm=106 g10−3 m=109 gm=Gg/m\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

mN/μs=10−3 N10−6 s=103 Ns=kN/s\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

μm⋅Mg=(10−6 m)(106 g)=m⋅gThis can also be written as:=mm⋅kg\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 P1.3 — 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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Statics 3.5 – Equilibrium of Truss Members Connected to a Gusset Plate | Hibbeler 14th Edition

The members of a truss are connected to the gusset plate. If the forces are concurrent at point O, determine the magnitudes of F and T for equilibrium. Take θ=90°.

Engineering Mechanics: Statics 14th Edition by RC Hibbeler Problem 3-5 Equilibrium of Truss Members Connected in a Gusset Plate

Solution:

We need to find the angle that force T makes with the positive x-axis first. We call this the angle beta, β. This is depicted in the free-body diagram.

Free-body diagram:

Solving for the values of angles α and β.

tan⁡α=34α=tan⁡−134α=36.8699°\begin{aligned} \tan \alpha & = \dfrac{3}{4} \\ \alpha & = \tan ^{-1} \frac{3}{4} \\ \alpha & = 36.8699 \degree \\ \end{aligned}

Knowing that the sum of angles α and β is 90°, we can solve for the β.

α+β=90°β=90°−αβ=90°−36.8699°β=53.1301°\begin{aligned} \alpha + \beta & = 90\degree \\ \beta & = 90 \degree - \alpha \\ \beta & = 90 \degree - 36.8699 \degree \\ \beta & = 53.1301 \degree \end{aligned}

Equations of Equilibrium:

Summation of forces in the x-direction:

→+∑Fx=0Tcos⁡β−45F=0Tcos⁡53.1301°−45F=0(1)\begin{aligned} \xrightarrow{+} \sum F_x & = 0 \\ T \cos \beta - \frac{4}{5} F & = 0 \\ T \cos 53.1301 \degree - \frac{4}{5} F & = 0 & & \qquad \qquad (1)\\ \end{aligned}

Summation of forces in the y-direction:

+↑∑Fy=09−35F−Tsin⁡β=0Tsin⁡53.1301°+35F=9(2)\begin{aligned} +\uparrow \sum F_y & =0 \\ 9 - \frac{3}{5} F- T \sin \beta & = 0 \\ T \sin 53.1301 \degree + \frac{3}{5}F & = 9 & & \qquad \qquad(2)\\ \end{aligned}

Now, we have two equations with two unknowns. We shall solve the unknowns by solving these equations simultaneously. We can use our calculator, or we can solve this manually using the method of substitution.

Using equation (1), solve for T in terms of F.

Tcos⁡53.1301°−45F=0Tcos⁡53.1301°=45FT=45Fcos⁡53.1301°(3)\begin{aligned} T \cos 53.1301\degree-\frac{4}{5} F & = 0 \\ T \cos 53.1301\degree & = \frac{4}{5} F \\ T & = \dfrac{\frac{4}{5}F}{\cos 53.1301\degree} \qquad \qquad (3)\\ \end{aligned}

Now, substitute this equation (3) to equation (2) to solve for F:

Tsin⁡53.1301°+35F=9(45Fcos⁡53.1301°)sin⁡53.1301°+35F=945F(sin⁡53.1301°cos⁡53.1301°)+35F=945Ftan⁡53.1301°+35F=9F(45⋅tan⁡53.1301°+35)=9F=945⋅tan⁡53.1301°+35F=5.4 kN\begin{aligned} T \sin 53.1301 \degree + \frac{3}{5}F & = 9 \\ \left(\dfrac{\frac{4}{5}F}{\cos 53.1301\degree} \right) \sin 53.1301 \degree + \frac{3}{5}F & = 9 \\ \frac{4}{5}F \left( \dfrac{\sin 53.1301\degree}{\cos 53.1301\degree}\right)+ \frac{3}{5}F & = 9 \\ \frac{4}{5}F \tan 53.1301\degree+\frac{3}{5}F &=9 \\ F\left( \frac{4}{5} \cdot \tan 53.1301\degree +\frac{3}{5}\right) & = 9\\ F & = \dfrac{9}{\frac{4}{5} \cdot \tan 53.1301\degree +\frac{3}{5}} \\ F & = 5.4 \ \text{kN} \\ \end{aligned}

Substitute the value of F to equation (3) to solve for T:

T=45Fcos⁡53.1301°T=45⋅(5.4 kN)cos⁡53.1301°T=7.2 kN\begin{aligned} T & = \dfrac{\frac{4}{5}F}{\cos 53.1301\degree} \\ T & = \dfrac{\frac{4}{5} \cdot \left( 5.4 \ \text{kN}\right)}{\cos 53.1301\degree} \\ T & = 7.2 \ \text{kN} \end{aligned}

Therefore, F=5.4 kN F = 5.4 \ \text{kN} and T=7.2 kN T= 7.2 \ \text{kN} .

Statics 3.4 – Normal Reactions in a Bearing | Hibbeler 14th Edition

The bearing consists of rollers, symmetrically confined within the housing. The bottom one is subjected to a 125-N force at its contact A due to the load on the shaft. Determine the normal reactions NB and NC on the bearing at its contact points B and C for equilibrium.

Engineering Mechanics: Statics 14th Edition Problem 3-4 - The normal reactions on a bearing.

Solution:

Free-body diagram of the roller:

Free-body diagram of Problem 3.4 - Engineering Mechanics Statics 14th Edition by Russell C. Hibbeler| Normal Forces in a Bearing

Equations of Equilibrium:

Note that if we take the sum of forces in the x-direction, there are two unknown forces involve, but if we take the sum of forces in the y-direction, there is only one unknown force involve.

Summation of forces in the y-direction:

+↑∑Fy=0125−NCcos⁡40°=0NC=125cos⁡40°NC=163.1759 N\begin{aligned} +\uparrow \sum F_y & =0& & & & & \\ 125- N_C \cos 40 \degree &=0 & & & & &\\ N_C &=\dfrac{125}{\cos 40 \degree} & & & & & \\ N_C & =163.1759 \ \text{N} \\ \end{aligned}

Summation of forces in the x-direction:

→+∑Fx=0NB−163.1759 sin⁡40°=0NB=163.1759sin⁡40°NB=104.8874 N\begin{aligned} \xrightarrow{+} \sum F_x & =0 \\ N_B - 163.1759\ \sin 40 \degree &=0 \\ N_B &=163.1759 \sin 40\degree \\ N_B & = 104.8874 \ \text{N} \end{aligned}

Therefore, the normal reactions NB and NC on the bearing at its contact points B and C for equilibrium are 163.1759 N and 104.8874 N, respectively.