Category Archives: Engineering Mechanics: Statics

Hibbeler Statics 14E P1.5 — Representing a number between 0.1 and 1000 Using an Appropriate Prefix

Represent each of the following as a number between 0.1 and 1000 using an appropriate prefix: (a) 45 320 kN, (b) 568(10⁵) mm, (c) 0.00563 mg.

_{Statics of Rigid Bodies 14^th by RC Hibbeler, Problem 1-5}

Solution:

Part A

\begin{align*} 45 \ 320 \ \text{kN} & = 45.3 \times 10^3 \ \text{kN}\\ & = 45.3\times 10^3\times 10^3 \ \text{N}\\ & = 45.3 \times 10^6 \ \text{N}\\ & = 45.3 \ \text{MN} \end{align*}

Part B

\begin{align*} 568\left( 10^5 \right)\ \text{mm} & =568\left( 10^5 \right)\times 10^{-3} \ \text{m}\\ & = 568\times 10^2 \ \text{m} \\ & = 56.8 \times 10^3 \ \text{m}\\ & = 56.8 \ \text{km} \end{align*}

Part C

\begin{align*} 0.00563 \ \text{mg} & = 5.63\times 10^{-3} \ \text{mg}\\ & = 5.63\times 10^{-3}\times 10^{-3} \ \text{g}\\ & = 5.63\times 10^{-6} \ \text{g}\\ & = 5.63 \ \mu\text{g} \end{align*}

Hibbeler Statics 14E P1.4 — Conversion from English units to Metric Units

Convert: (a) 200 lb·ft to N·m, (b) 350 lb/ft³ to kN/m³, (c) 8 ft/h to mm/s. Express the result to three significant figures. Use an appropriate prefix.

_{Statics of Rigid Bodies 14^th by RC Hibbeler, Problem 1-4}

Solution:

Part A

\begin{align*} 200\ \text{lb}\cdot \text{ft} & =\left( 200\ \text{lb}\cdot \text{ft} \right)\left( \frac{4.4482\ \text{N}}{1\ \text{lb}} \right)\left( \frac{0.3048\ \text{m}}{1\ \text{ft}} \right)\\ &=271\ \text{N}\cdot \text{m} \end{align*}

Part B

\begin{align*} 350 \ \text{lb/ft}^3& = \left( \frac{350\ \text{lb}}{1\ \text{ft}^3} \right)\left( \frac{1\ \text{ft}}{0.3048\ \text{m}} \right)^3\left( \frac{4.4482\ \text{N}}{1\ \text{lb}} \right)\left( \frac{1\ \text{kN}}{1000\ \text{N}} \right)\\ & = 55.0\ \text{kN/m}^3 \end{align*}

Part C

\begin{align*} 8 \ \text{ft/hr}& = \left( \frac{8\ \text{ft}}{1\ \text{hr}} \right)\left( \frac{1\ \text{hr}}{3600\ \text{s}} \right)\left( \frac{0.3048\ \text{m}}{1\ \text{ft}} \right)\left( \frac{1000 \ \text{mm}}{1 \ \text{m}} \right)\\ & = 0.677\ \text{mm/s} \end{align*}

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 14^th Edition by RC Hibbeler, Problem 1-3}

Solution:

Part A

\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

\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

\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*}

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°.

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.

\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.

\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:

\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:

\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.

\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:

\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:

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

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 N_B and N_C 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:

\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:

\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 N_Band N_C on the bearing at its contact points B and C for equilibrium are 163.1759 N and 104.8874 N, respectively.

Statics 3.2 – Equilibrium of Truss Members that are Pin Connected | Hibbeler 14th Edition

The members of a truss are pin connected at joint O. Determine the magnitude of F₁ and its angle θ for equilibrium. Set F₂=6 kN.

Figure 3.1: Engineering Mechanics: Statics Equilibrium of Particle — Figure 3.1/3.2

Solution:

Free-body diagram:

Equations of Equilibrium:

The summation of forces in the x-direction:

\begin{aligned} \sum F_x & = 0 &\\ 6 \sin 70 \degree + F_1 \cos \theta - 5 \cos 30 \degree - \dfrac {4}{5} \left(7 \right) & = 0 & \\ F_1 \cos \theta & = 4.2920 & (1) \end{aligned}

The summation of forces in the y-direction:

\begin{aligned} \sum F_y & =0 & \\ 6 \cos 70 \degree+5 \sin 30 \degree - F_1 \sin \theta - \dfrac{3}{5} \left( 7 \right ) & =0 & \\ F_1 \sin \theta &=0.3521 & (2)\\ \end{aligned}

We came up with 2 equations with unknowns $F_1$ and $\theta$ . To solve the equations simultaneously, we can use the method of substitution.

Using equation 1, solve for $F_1$ in terms of $\theta$ .

\begin{aligned} F_1 \cos \theta & = 4.2920 &\\ F_1 & =\dfrac{4.2920}{ \cos \theta } & (3) \\ \end{aligned}

Now, substitute this equation (3) to equation (2).

\begin{aligned} F_1 \sin \theta & = 0.3521 \\ \left ( \dfrac {4.2920}{\cos \theta} \right) \sin \theta & =0.3521 \\ 4.2920 \cdot \dfrac{\sin \theta}{\cos \theta} & = 0.3521 \\ 4.2920 \tan \theta & = 0.3521 \\ \tan \theta & = \dfrac{0.3521}{4.2920} \\ \theta &= \tan ^{-1} \dfrac{0.3521}{4.2920} \\ \theta & = 4.69 \degree \end{aligned}

Substitute the solved value of $\theta$ to equation (3).

\begin{aligned} F_1 & = \dfrac{4.2920}{\cos \theta} \\ F_1 &= \dfrac{4.2920}{\cos 4.69 \degree} \\ F_1 & = 4.31 \text{kN} \end{aligned}

Therefore, the answers to the questions are:

\begin{aligned} F_1= & \:4.31 \: \text {kN} \\ \theta = & \: 4.69 \degree \end{aligned}

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.

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

$Electrical Transmission Towers Background for Force Vectors Statics of Rigid Bodies$

Represent each of the following as a number between 0.1 and 1000 using an appropriate prefix: (a) 45 320 kN, (b) 568(105) mm, (c) 0.00563 mg.

Convert: (a) 200 lb·ft to N·m, (b) 350 lb/ft3 to kN/m3, (c) 8 ft/h to mm/s. Express the result to three significant figures. Use an appropriate prefix.

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).

Part C

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.

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°.

Solution:

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.

Solution:

The members of a truss are pin connected at joint O. Determine the magnitude of F1 and its angle θ for equilibrium. Set F2=6 kN.

Solution:

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

Represent each of the following as a number between 0.1 and 1000 using an appropriate prefix: (a) 45 320 kN, (b) 568(10⁵) mm, (c) 0.00563 mg.

Convert: (a) 200 lb·ft to N·m, (b) 350 lb/ft³ to kN/m³, (c) 8 ft/h to mm/s. Express the result to three significant figures. Use an appropriate prefix.

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 N_B and N_C on the bearing at its contact points B and C for equilibrium.

The members of a truss are pin connected at joint O. Determine the magnitude of F₁ and its angle θ for equilibrium. Set F₂=6 kN.

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