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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 3.6 – Equilibrium of Truss Members Connected to a Gusset Plate | Hibbeler 14th Edition

The gusset plate is subjected to the forces of three members. Determine the tension force in member C and its angle θ for equilibrium. The forces are concurrent at point O. Take F=8 kN.

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

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.

Figure 3.1: Engineering Mechanics: Statics Equilibrium of Particle
Figure 3.1

Solution:

Free-body diagram:

Free-Body Diagram for Problem 3.1 of Engineering Mechanics: Statics 14th Edition by Russell C. Hibbeler

Equations of Equilibrium:

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

∑Fx=0F1cos⁡60°+F2sin⁡70°−5cos⁡30°−45(7)=00.5F1+0.9397F2=9.9301(1)\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.

∑Fy=0−F1sin⁡60°+F2cos⁡70°+5sin⁡30°−35(7)=0−0.8660F1+0.3420F2=1.7(2)\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 F1 F_1 and F2 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

F1=1.83 kNF2=9.60 kNF_1=1.83 \: \text{kN}\\ F_2=9.60 \: \text{kN}

Hibbeler Statics 14E P1.1 — Converting mass to weight in newtons

What is the weight in newtons of an object that has a mass of (a) 8 kg, (b) 0.04 kg, and (c) 760 Mg?

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-1

Solution:

Part A: To convert the given mass in kilogram to newton force, we simply need to multiply by the acceleration due to gravity of 9.81 m/s2. We need to take into account that 1 kg m/s2 =1 N 1\:\text{kg m/s}^2\:=1\:\text{N} .

8 kg=8 kg×9.81 m/s2=78.48 N\begin {aligned} 8\:\text{kg} & =8\:\text{kg}\times 9.81\:\text{m/s}^2 \\ &=78.48\:\text{N} \end {aligned}

Part B: Using the same principle from Part A, we have

0.04 kg=0.04 kg×9.81 m/s2=0.3924 N\begin {aligned} 0.04\:\text{kg}&=0.04\:\text{kg}\times 9.81\:\text{m/s}^2\\ &=0.3924\:\text{N} \end {aligned}

Part C: So, we are given 760 Mg (megagram). We know that 1 Mg is equivalent to 1000 kg. Therefore, 760 Mg is equal to 760,000 kg. Therefore, we have

760 000 kg=760 000 kg×9.81 m/s2=7 455 600 N\begin {aligned} 760\:000\:\text{kg}&=760\:000\:\text{kg}\times 9.81\:\text{m/s}^2\\ &=7\:455\:600\:\text{N} \end{aligned}
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Engineering Mechanics: Statics 14th Edition by RC Hibbeler

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General Principles of Statics of Rigid Bodies Fourteenth Edition by RC Hibbeler Cover Photo

Chapter 1: General Principles

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

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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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