Tag Archives: statics

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:

→+ ∑Fx=05 kN+Fsin⁡θ−8 kNcos⁡30°−4 kNcos⁡60°=0Fsin⁡θ=3.9282(1)\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:

+↑∑Fy=08sin⁡30°−4sin⁡60°−Fcos⁡θ=0Fcos⁡θ=0.5359(2)\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)

Fsin⁡θFcos⁡θ=3.92820.5359sin⁡θcos⁡θ=7.3301\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⁡θ=sin⁡θcos⁡θ \tan \theta = \dfrac{\sin \theta}{\cos \theta} :

tan⁡θ=7.3301θ=tan⁡−17.3301θ=82.2°\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

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

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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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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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:

F=G⋅m1m2r2=66.73(10−12)m3/(kg⋅s2)[8 kg(12 kg)(0.8 m)2]=10(10−9) N=10.0 nN\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

W1=8(9.81)=78.5 N\textbf{W}_1=8\left(9.81\right)=78.5\:\text{N}

Weight of the 12 kg particle

W2=12(9.81)=118 N\textbf{W}_2=12\left(9.81\right)=118\:\text{N}
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Expressing the Density of Water in SI Units

Water has a density of 1.94 slug/ft³. What is the density expressed in SI units? Express the answer to three significant figures.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-19
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-17

Solution:

ρw=(1.94 slug1 ft3)(14.59 kg1 slug)(1 ft30.30483 m3)=(1.94 slug1 ft3)(14.59 kg1 slug)(1 ft30.30483 m3)=999.6 kgm3=1.00 Mg/m3\begin{align*} \rho _w & =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\ & =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\ & =999.6\:\frac{\text{kg}}{\text{m}^3}\\ & =1.00\:\text{Mg/m}^3\\ \end{align*}
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Showing How an Equation is Dimensionally Homogeneous

Using the SI system of units, show that Eq. 1–2 is a dimensionally homogeneous equation which gives F in newtons. Determine to three significant figures the gravitational force acting between two spheres that are touching each other. The mass of each sphere is 200 kg and the radius is 300 mm.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-18
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-15

Solution:

To prove that F is in Newtons, we have

F=G⋅m1m2r2=(m3kg⋅s2)(kg⋅kgm2)=kg⋅ms2=N\begin{align*} \text{F} & =\text{G}\cdot \frac{\text{m}_1\text{m}_2}{\text{r}^2}\\ & =\left(\frac{\text{m}^3}{\text{kg}\cdot \text{s}^2}\right)\left(\frac{\text{kg}\cdot \text{kg}}{\text{m}^2}\right)\\ & =\frac{\text{kg}\cdot \text{m}}{\text{s}^2}\\ & =\text{N} \end{align*}

Now, if we substitute the given values into the equation

F=66.73(10−12)[200(200)0.62]=7.41(10−6)N=7.41 μN\begin{align*} \text{F} & = 66.73\left(10^{-12}\right)\left[\frac{200\left(200\right)}{0.6^2}\right]\\ & = 7.41\left(10^{-6}\right) \text{N}\\ & =7.41\ \mu \text{N}\\ \end{align*}
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Evaluation of Expressions to SI Units with Appropriate Prefix

Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) 354 mg (45 km)/(0.0356 kN), (b) (0.00453 Mg)(201 ms), and (c) 435 MN/23.2 mm.

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-11
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-18

Solution:

Part A

(354 mg)(45 km)0.0356 kN=[354(10−3) g][45(103) m]0.0356 (103) N=0.447 (103)g⋅mN=0.447 kg⋅m/N\begin{align*} \frac{\left(354\:\text{mg}\right)\left(45\:\text{km}\right)}{0.0356\:\text{kN}} & = \frac{\left[354\left(10^{-3}\right)\:\text{g}\right]\left[45\left(10^3\right)\:\text{m}\right]}{0.0356\:\left(10^3\right)\:\text{N}}\\ & = \frac{0.447\:\left(10^3\right)\text{g}\cdot \text{m}}{\text{N}}\\ & = 0.447\:\text{kg}\cdot \text{m/N} \end{align*}

Part B

(0.00453 Mg)(201 ms)=[4.53(10−3)(103)kg][201 (10−3)s]=0.911 kg⋅s\begin{align*} \left(0.00453\:\text{Mg}\right)\left(201\:\text{ms}\right) & =\left[4.53\left(10^{-3}\right)\left(10^3\right)\text{kg}\right]\left[201\:\left(10^{-3}\right)\text{s}\right]\\ & =0.911\:\text{kg}\cdot \text{s}\\ \end{align*}

Part C

435 MN/23.2 mm=435 (106) N23.2 (10−3) m=18.75(109) Nm=18.8 GN/m\begin{align*} 435\:\text{MN}/23.2\:\text{mm} & =\frac{435\:\left(10^6\right)\:\text{N}}{23.2\:\left(10^{-3}\right)\:\text{m}}\\ & = \frac{18.75\left(10^9\right)\:\text{N}}{\text{m}}\\ & =18.8\:\text{GN/m} \end{align*}
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Converting Measurement from Pounds Per Square Inch to Pascal

The pascal (Pa) is actually a very small unit of pressure. To show this, convert 1 Pa = 1N/m² to lb/ft². Atmospheric pressure at sea level is 14.7 lb/in². How many pascals is this?

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-7
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-16

Solution:

First, we will convert 1 Pa to lb/ft².

1 Pa=1 Nm2(1 lb4.4482 N)(0.30482 m21 ft2)=20.9(10−3) lb/ft2\begin{align*} 1\ \text{Pa} & =\frac{1\:\text{N}}{\text{m}^2}\left(\frac{1\:\text{lb}}{4.4482\:\text{N}}\right)\left(\frac{0.3048^2\:\text{m}^2}{1\:\text{ft}^2}\right)\\ &=20.9\left(10^{-3}\right)\:\text{lb/ft}^2 \end{align*}

Next, we convert 14.7 lb/in2 to Pa

14.7 lb/in2=14.7 lbin2(4.448 N1 lb)(144 in21 ft2)(1 ft20.30482 m2)=101.3(103) N/m2=101.3(103)Pa\begin{align*} 14.7 \ \text{lb/in}^2 & =\frac{14.7\:\text{lb}}{\text{in}^2}\left(\frac{4.448\:\text{N}}{1\:\text{lb}}\right)\left(\frac{144\:\text{in}^2}{1\:\text{ft}^2}\right)\left(\frac{1\:\text{ft}^2}{0.3048^2\:\text{m}^2}\right)\\ & =101.3\left(10^3\right)\:\text{N/m}^2\\ & =101.3\left(10^3\right) \text{Pa}\\ \end{align*}
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Expressing units to correct SI form using appropriate prefix

Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) m/ms, (b) μkm, (c) ks/mg, and (d) km·μN.

Statics of Rigid Bodies 14th Edition by RC Hibbeler, Problem 1-9

Solution:

Part A

\displaystyle \text{m/ms}=\left(\frac{\text{m}}{10^{-3}\:\text{s}}\right)=10^3\:\text{m/s}=\:\text{km/s}

Part B

\displaystyle \mu \text{km}=10^{-6}\cdot 10^3\:\text{m}=10^{-3}\:\text{m}=\text{mm}

Part C

\displaystyle \text{ks/mg}=\frac{10^3\:s}{10^{-6}\:\text{kg}}=10^9\:\frac{\text{s}}{\text{kg}}=\text{Gs/kg}

Part D

\displaystyle \text{km} \cdot \mu \text{N}=10^3\:\text{m}\cdot 10^{-6}\:\text{N}\:=10^{-3}\:\text{m}\cdot \text{N}=\text{mm}\cdot \text{N}