Tag Archives: Engineering Mechanics: Statics

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:

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}

Chapter 3: Equilibrium of a Particle

Coplanar Force Systems

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

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Three-Dimensional Force Systems

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

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 14^th 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 $1\:\text{kg m/s}^2\:=1\:\text{N}$ .

\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

\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

\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 $\textbf{F}_{\text{R}} = \textbf{F}_1 + \textbf{F}_2$ 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 $\textbf{F}_{\text{R}}$

\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.

\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 $\phi$ .

\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, $\text{W}=\text{mg}$ , we can solve for the mass by dividing the weight by the acceleration due to gravity. That is

\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.

\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.

\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 $\textbf{g}=5.30 \ \text{ft/s}^2$ .

\textbf{W}=155\left(\frac{5.30}{32.2}\right)=25.5\:\text{lb}

Part E

\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:

\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

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

Weight of the 12 kg particle

\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 13^th Edition by RC Hibbeler, Problem 1-19}
_{Engineering Mechanics: Statics 14^th Edition by RC Hibbeler, Problem 1-17}

Solution:

\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 13^th Edition by RC Hibbeler, Problem 1-18}
_{Engineering Mechanics: Statics 14^th Edition by RC Hibbeler, Problem 1-15}

Solution:

To prove that F is in Newtons, we have

\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

\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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Engineering Mechanics: Statics 14th Edition by RC Hibbeler Solution Manual

You can purchase the complete solution manual of the Engineering Mechanics: Statics 14th Edition by Russell Hibbeler in this page.

General Principles: Statics of Rigid Bodies 14th Edition by RC Hibbeler

Force System Resultants

Equilibrium of a Rigid Body

Structural Analysis

Internal Forces

Friction

Center of Gravity and Centroid

Moment of Inertia

Virtual Work

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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

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

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Three-Dimensional Force Systems

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

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?

Solution:

Determine the magnitude of the resultant force \textbf{F}_{\text{R}} = \textbf{F}_1 + \textbf{F}_2 and its direction, measured counterclockwise from the positive x axis.

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.

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.

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

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.

Force System Resultants

Equilibrium of a Rigid Body

Structural Analysis

Internal Forces

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

Determine the magnitude of the resultant force $\textbf{F}_{\text{R}} = \textbf{F}_1 + \textbf{F}_2$ and its direction, measured counterclockwise from the positive x axis.