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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.

Problem 1-10| General Principles| Engineering Mechanics: Statics| RC Hibbeler

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (0.631 Mm)/(8.60kg)², (b) (35 mm)² (48 kg)³.

Solution:

a) \left(0.631\:Mm\right)/\left(8.60\:kg\right)^2=\left(\frac{0.631\left(10^6\right)m}{\left(8.60\right)^2kg^2}\right)=\frac{8532\:m}{kg^2}=8.53\left(10^3\right)m/kg^2=8.53\:km/kg

b) \left(35\:mm\right)^2\left(48\:kg\right)^3=\left[35\left(10^{-3}\right)m\right]^2\left(48\:kg\right)^3=135\:m^2\cdot kg^3

Expressing units in the correct SI form using an appropriate prefix

Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) kN/μs, (b) Mg/mN, and (c) MN/(kg•ms).

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-2

Solution:

Part A

KN/μs=(10)3 N(10)−6 s=(10)9 N/s=GN/s\begin{align*} \text{KN}/\mu\text{s} & = \frac{\left( 10 \right)^3\ \text{N}}{\left( 10 \right)^{-6}\ \text{s}} \\ & =\left( 10 \right)^9 \ \text{N/s}\\ & = \text{GN/s} \end{align*}

Part B

Mg/mN=(106)g(10−3)N=109 g/N=Gg/N\begin{align*} \text{Mg/mN} & =\frac{\left(10^6\right)\text{g}}{\left(10^{-3}\right)\text{N}}\\ & = 10^9\:\text{g/N}\\ & =\text{Gg/N} \end{align*}

Part C

MN/(kg⋅ms)=106 Nkg⋅(10−3)s=109 Nkg⋅s=GN/(kg⋅s)\begin{align*} \text{MN}/\left(\text{kg}\cdot \text{ms}\right) & =\frac{10^6\:\text{N}}{\text{kg}\cdot \left(10^{-3}\right)\text{s}}\\ & =10^9\:\frac{\text{N}}{\text{kg}\cdot \text{s}}\\ & =\text{GN}/\left(\text{kg}\cdot \text{s}\right) \end{align*}
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Rounding off to 3 significant figures

Round off the following numbers to three significant figures: (a) 58 342 m, (b) 68.534 s, (c) 2553 N, and (d) 7555 kg.

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-6

Solution:

Part A

58 342 m=58.342×103 m=58.342 km=58.3 km\begin{align*} 58 \ 342 \ \text{m} & = 58.342\times 10^{3} \ \text{m}\\ & = 58.342 \ \text{km}\\ & = 58.3 \ \text{km} \end{align*}

Part B

68.534 s=68.5 s\begin{align*} 68.534 \ \text{s} & = 68.5 \ \text{s} \end{align*}

Part C

2553 N=2.553 kN=2.55 kN\begin{align*} 2553 \ \text{N} & = 2.553 \ \text{kN}\\ & = 2.55 \ \text{kN} \end{align*}

Part D

7555 kg=7.555×103 kg=7.555×103×103 g=7.555×106 g=7.555 Mg=7.56 Mg\begin{align*} 7555 \ \text{kg} & = 7.555\times 10^3 \ \text{kg}\\ & = 7.555 \times 10^3 \times 10^3 \ \text{g}\\ & = 7.555 \times 10^6 \ \text{g} \\ & = 7.555 \ \text{Mg}\\ & = 7.56 \ \text{Mg} \end{align*}

College Physics by Openstax Chapter 3 Problem 1

Find the following for path A in Figure 3.52:
(a) The total distance traveled, and
(b) The magnitude and direction of the displacement from start to finish.

Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side
Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side

Solution:

Part A

The total distance traveled is 

d=(3×120 m)+(1×120 m)d=480 m  (Answer)\begin{align*} \text{d} & =\left(3\times 120\ \text{m}\right)+\left(1\times 120\:\text{m}\right) \\ \text{d} & =480\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

Part B

The magnitude of the displacement is 

s =(sx)2 +(sy)2s =(1×120 m)2+(3×120 m)2s =379 m  (Answer)\begin{align*} \text{s }& =\sqrt{\left( s_x \right)^{2\:}+\left( s_y \right)^2} \\ \text{s }& = \sqrt{\left(1\times 120\:\text{m}\right)^2+\left(3\times 120\:\text{m}\right)^2} \\ \text{s }& = 379\ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

The direction is

θ=tan⁡−1(sxsy)θ=tan⁡−1(1×120 m3×120 m)θ=71.6∘, E of N  (Answer)\begin{align*} \theta & = \tan^{-1}\left(\frac{s_x}{s_y}\right) \\ \theta & = \tan^{-1}\left(\frac{1\times 120\:\text{m}}{3\times 120 \ \text{m}}\right) \\ \theta & =71.6^{\circ} ,\:\text{E of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
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