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Kuldip S. Rattan
Nathan W. Klingbeil
Mathematics for Engineering Applications
Introductory Mathematics for Engineering Applications Kuldip S. Rattan Wright State University
Nathan W. Klingbeil Wright State University
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1 STRAIGHT LINES IN ENGINEERING 1 1.1 Vehicle during Braking 1 1.2 Voltage-Current Relationship in a Resistive Circuit 3 1.3 Force-Displacement in a Preloaded Tension Spring 6 1.4 Further Examples of Lines in Engineering 8 Problems 19
2 QUADRATIC EQUATIONS IN ENGINEERING 32 2.1 A Projectile in a Vertical Plane 32 2.2 Current in a Lamp 36 2.3 Equivalent Resistance 37 2.4 Further Examples of Quadratic Equations in Engineering 38 Problems 50
3 TRIGONOMETRY IN ENGINEERING 60 3.1 Introduction 60 3.2 One-Link Planar Robot 60 3.2.1 3.2.2
Kinematics of One-Link Robot 60 Inverse Kinematics of One-Link Robot 68
3.3 Two-Link Planar Robot 72 3.3.1 3.3.2 3.3.3
Direct Kinematics of Two-Link Robot 73 Inverse Kinematics of Two-Link Robot 75 Further Examples of Two-Link Planar Robot 79
3.4 Further Examples of Trigonometry in Engineering 89 Problems 97
4 TWO-DIMENSIONAL VECTORS IN ENGINEERING 106 4.1 Introduction 106 4.2 Position Vector in Rectangular Form 107 4.3 Position Vector in Polar Form 107 4.4 Vector Addition 110 4.4.1
Examples of Vector Addition in Engineering 111
5 COMPLEX NUMBERS IN ENGINEERING 132 5.1 Introduction 132 5.2 Position of One-Link Robot as a Complex Number 133
5.3 Impedance of R, L, and C as a Complex Number 134 5.3.1 5.3.2 5.3.3
Impedance of a Resistor R 134 Impedance of an Inductor L 134 Impedance of a Capacitor C 135
5.4 Impedance of a Series RLC Circuit 136 5.5 Impedance of R and L Connected in Parallel 137 5.6 Armature Current in a DC Motor 140 5.7 Further Examples of Complex Numbers in Electric Circuits 141
5.8 Complex Conjugate 145 Problems 147
6 SINUSOIDS IN ENGINEERING 157 6.1 One-Link Planar Robot as a Sinusoid 157 6.2 Angular Motion of the One-Link Planar Robot 159 6.2.1
Relations between Frequency and Period 160
6.3 Phase Angle, Phase Shift, and Time Shift 162 6.4 General Form of a Sinusoid 164 6.5 Addition of Sinusoids of the Same Frequency 166 Problems 173
7 SYSTEMS OF EQUATIONS IN ENGINEERING 184 7.1 Introduction 184 7.2 Solution of a Two-Loop Circuit 184
7.3 Tension in Cables 190 7.4 Further Examples of Systems of Equations in Engineering 193
8 DERIVATIVES IN ENGINEERING 218 8.1 Introduction 218 8.1.1
What Is a Derivative? 218
8.2 Maxima and Minima 221 8.3 Applications of Derivatives in Dynamics 225 8.3.1
Position, Velocity, and Acceleration 226
8.4 Applications of Derivatives in Electric Circuits 240 8.4.1 8.4.2
Current and Voltage in an Inductor 243 Current and Voltage in a Capacitor 247
8.5 Applications of Derivatives in Strength of Materials 250 8.5.1
Maximum Stress under Axial Loading 256
8.6 Further Examples of Derivatives in Engineering 261 Problems 266
9 INTEGRALS IN ENGINEERING 278 9.1 Introduction: The Asphalt Problem 278 9.2 Concept of Work 283 9.3 Application of Integrals in Statics 286 9.3.1 9.3.2
Center of Gravity (Centroid) 286 Alternate Definition of the Centroid 293
9.4 Distributed Loads 296 9.4.1
Hydrostatic Pressure on a Retaining Wall 296 Distributed Loads on Beams: Statically Equivalent Loading 298
9.5 Applications of Integrals in Dynamics 302 9.5.1
Graphical Interpretation 309
9.6 Applications of Integrals in Electric Circuits 314 9.6.1
Current, Voltage, and Energy Stored in a Capacitor 314
9.7 Current and Voltage in an Inductor 322 9.8 Further Examples of Integrals in Engineering 327 Problems 334
10 DIFFERENTIAL EQUATIONS IN ENGINEERING 345 10.1 Introduction: The Leaking Bucket 345
10.2 Differential Equations 346 10.3 Solution of Linear DEQ with Constant Coefficients 347 10.4 First-Order Differential Equations 348 10.5 Second-Order Differential Equations 374 10.5.1 Free Vibration of a Spring-Mass System 374 10.5.2 Forced Vibration of a Spring-Mass System 379 10.5.3 Second-Order LC Circuit 386
ANSWERS TO SELECTED PROBLEMS 401 INDEX
This book is intended to provide first-year engineering students with a comprehensive introduction to the application of mathematics in engineering. This includes math topics ranging from precalculus and trigonometry through calculus and differential equations, with all topics set in the context of an engineering application. Specific math topics include linear and quadratic equations, trigonometry, 2-D vectors, complex numbers, sinusoids and harmonic signals, systems of equations and matrices, derivatives, integrals, and differential equations. However, these topics are covered only to the extent that they are actually used in core first- and second-year engineering courses, including physics, statics, dynamics, strength of materials, and electric circuits, with occasional applications from upper-division courses. Additional motivation is provided by a wide range of worked examples and homework problems representing a variety of popular engineering disciplines. While this book provides a comprehensive introduction to both the math topics and their engineering applications, it provides comprehensive coverage of neither. As such, it is not intended to be a replacement for any traditional math or engineering textbook. It is more like an advertisement or movie trailer. Indeed, everything covered in this book will be covered again in either an engineering or mathematics classroom. This gives the instructor an enormous amount of freedom − the freedom to integrate math and physics by immersion. The freedom to leverage student intuition, and to introduce new physical contexts for math without the constraint of prerequisite knowledge. The freedom to let the physics help explain the math and the math to help explain the physics. The freedom to teach math to engineers the way it really ought to be taught − within a context, and for a reason. Ideally, this book would serve as the primary text for a first-year engineering mathematics course, which would replace traditional math prerequisite requirements for core sophomore-level engineering courses. This would allow students to advance through the first two years of their chosen degree programs without first completing the required calculus sequence. Such is the approach adopted by Wright State University and a growing number of institutions across the country, which are now enjoying significant increases not only in engineering student retention but also in engineering student performance in their first required calculus course. Alternatively, this book would make an ideal reference for any freshman engineering program. Its organization is highly compartmentalized, which allows instructors to pick and choose which math topics and engineering applications to cover. Thus, any institution wishing to increase engineering student preparation and
motivation for the required calculus sequence could easily integrate selected topics into an existing freshman engineering course without having to find room in the curriculum for additional credit hours. Finally, this book would provide an outstanding resource for nontraditional students returning to school from the workplace, for students who are undecided or are considering a switch to engineering from another major, for math and science teachers or education majors seeking physical contexts for their students, or for upper-level high school students who are thinking about studying engineering in college. For all of these students, this book represents a onestop shop for how math is really used in engineering.
The authors would like to thank all those who have contributed to the development of this text. This includes their outstanding staff of TA’s, who have not only provided numerous suggestions and revisions, but also played a critical role in the success of the first-year engineering math program at Wright State University. The authors would also like to thank their many colleagues and collaborators who have joined in their nationwide quest to change the way math is taught to engineers. Special thanks goes to Jennifer Serres, Werner Klingbeil and Scott Molitor, who have contributed a variety of worked examples and homework problems from their own engineering disciplines. Thanks also to Josh Deaton, who has provided detailed solutions to all end-of-chapter problems. Finally, the authors would like to thank their wives and families, whose unending patience and support have made this effort possible.
This material is based upon work supported by the National Science Foundation under Grant Numbers EEC-0343214, DUE-0618571, DUE-0622466 and DUE0817332. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of the National Science Foundation.
Straight Lines in Engineering
In this chapter, the applications of straight lines in engineering are introduced. It is assumed that the students are already familiar with this topic from their high school algebra course. This chapter will show, with examples, why this topic is so important for engineers. For example, the velocity of a vehicle while braking, the voltagecurrent relationship in a resistive circuit, and the relationship between force and displacement in a preloaded spring can all be represented by straight lines. In this chapter, the equations of these lines will be obtained using both the slope-intercept and the point-slope forms.
VEHICLE DURING BRAKING The velocity of a vehicle during braking is measured at two distinct points in time, as indicated in Fig. 1.1. t, s 1.5 2.5
Figure 1.1 A vehicle while braking.
The velocity satisfies the equation v(t) = at + vo
where vo is the initial velocity in m/s and a is the acceleration in m/s2 . (a) Find the equation of the line v(t) and determine both the initial velocity vo and the acceleration a. (b) Sketch the graph of the line v(t) and clearly label the initial velocity, the acceleration, and the total stopping time on the graph. The equation of the velocity given by equation (1.1) is in the slope-intercept form y = mx + b, where y = v(t), m = a, x = t, and b = vo . The slope m is given by m=
Chapter 1 Straight Lines in Engineering Therefore, the slope m = a can be calculated using the data in Fig. 1.1 as a=
v2 − v1 5.85 − 9.75 = = −3.9 m/s2 . t2 − t1 2.5 − 1.5
The velocity of the vehicle can now be written in the slope-intercept form as v(t) = −3.9 t + vo . The y-intercept b = vo can be determined using either one of the data points. Using the data point (t, v) = (1.5, 9.75) gives 9.75 = −3.9 (1.5) + vo . Solving for vo gives vo = 15.6 m/s. The y-intercept b = vo can also be determined using the other data point (t, v) = (2.5, 5.85), yielding 5.85 = −3.9 (2.5) + vo . Solving for vo gives vo = 15.6 m/s. The velocity of the vehicle can now be written as v(t) = −3.9 t + 15.6 m/s. The total stopping time (time required to reach v(t) = 0) can be found by equating v(t) = 0, which gives 0 = −3.9 t + 15.6. Solving for t, the stopping time is found to be t = 4.0 s. Figure 1.2 shows the velocity of the vehicle after braking. Note that the stopping time t = 4.0 s and the initial velocity Velocity, m/s Initial velocity, v0 (y -intercept)
Stopping time (x -intercept) 0 0 Figure 1.2 Velocity of the vehicle after braking.
1.2 Voltage-Current Relationship in a Resistive Circuit
vo = 15.6 m/s are the x- and y-intercepts of the line, respectively. Also, note that the slope of the line m = −3.90 m/s2 is the acceleration of the vehicle during braking.
VOLTAGE-CURRENT RELATIONSHIP IN A RESISTIVE CIRCUIT For the resistive circuit shown in Fig. 1.3, the relationship between the applied voltage Vs and the current I flowing through the circuit can be obtained using Kirchhoff’s voltage law (KVL) and Ohm’s law. For a closed-loop in an electric circuit, KVL states that the sum of the voltage rises is equal to the sum of the voltage drops, i.e., Kirchhoff’s voltage law: ⇒
Figure 1.3 Voltage and current in a resistive circuit.
Applying KVL to the circuit of Fig. 1.3 gives Vs = VR + V.
Ohm’s law states that the voltage drop across a resistor VR in volts (V) is equal to the current I in amperes (A) flowing through the resistor multiplied by the resistance R in ohms (Ω), i.e., VR = I R.
Substituting equation (1.3) into equation (1.2) gives a linear relationship between the applied voltage Vs and the current I as Vs = I R + V.
The objective is to find the value of R and V when the current flowing through the circuit is known for two different voltage values given in Fig. 1.3. The voltage-current relationship given by equation (1.4) is the equation of a straight line in the slope-intercept form y = mx + b, where y = Vs , x = I, m = R, and b = V. The slope m is given by Δ y Δ Vs = . Δx ΔI Using the data in Fig. 1.3, the slope R can be found as m=R=
20 − 10 = 10 Ω. 1.1 − 0.1
Chapter 1 Straight Lines in Engineering Therefore, the source voltage can be written in slope-intercept form as Vs = 10 I + b. The y-intercept b = V can be determined using either one of the data points. Using the data point (Vs , I) = (10, 0.1) gives 10 = 10 (0.1) + V. Solving for V gives V = 9 V. The y-intercept V can also be found by finding the equation of the straight line using the point-slope form of the straight line (y − y1 ) = m(x − x1 ) as Vs − 10 = 10(I − 0.1) ⇒ Vs = 10 I − 1.0 + 10. Therefore, the voltage-current relationship is given by Vs = 10 I + 9.
Comparing equations (1.4) and (1.5), the values of R and V are given by R = 10 Ω,
Figure 1.4 shows the graph of the source voltage Vs versus the current I. Note that the slope of the line m = 10 is the resistance R in Ω and the y-intercept b = 9 is the voltage V in volts. The values of R and V can also be determined by switching the interpretation of x and y (the independent and dependent variables). From the voltage-current relationship Vs = I R + V, the current I can be written as a function of Vs as 1 V Vs − . (1.6) R R This is an equation of a straight line y = m x + b, where x is the applied voltage Vs , y is the current I, m = R1 is the slope, and b = − V is the y-intercept. The slope and R I=
Figure 1.4 Voltage-current relationship for the data given in Fig. 1.3.
1.2 Voltage-Current Relationship in a Resistive Circuit
y-intercept can be found from the data given in Fig. 1.3 using the slope-intercept method as Δy ΔI . = m= Δ x Δ Vs Using the data in Fig. 1.3, the slope m can be found as 1.1 − 0.1 = 0.1. 20 − 10 Therefore, the current I can be written in slope-intercept form as m=
I = 0.1 Vs + b. The y-intercept b can be determined using either one of the data points. Using the data point (Vs , I) = (10, 0.1) gives 0.1 = 0.1 (10) + b. Solving for b gives b = −0.9. Therefore, the equation of the straight line can be written in the slope-intercept form as I = 0.1 Vs − 0.9.
Comparing equations (1.6) and (1.7) gives 1 = 0.1 R
and V = −0.9 ⇒ V = 0.9 (10) = 9 V. R Figure 1.5 is the graph of the straight line I = 0.1Vs − 0.9. Note that the y-intercept V 1 is − = − 0.9 A and the slope is = 0.1. R R −
y-intercept 0.9 Figure 1.5 Straight line with I as independent variable for the data given in Fig. 1.3.
Chapter 1 Straight Lines in Engineering
FORCE-DISPLACEMENT IN A PRELOADED TENSION SPRING The force-displacement relationship for a spring with a preload fo is given by f = k y + fo ,
where f is the force in Newtons (N), y is the displacement in meters (m), and k is the spring constant in N/m. y f
Figure 1.6 Force-displacement in a preloaded spring.
The objective is to find the spring constant k and the preload fo , if the values of the force and displacement are as given in Fig. 1.6. Method 1 Treating the displacement y as an independent variable, the forcedisplacement relationship f = k y + fo is the equation of a straight line y = mx + b, where the independent variable x is the displacement y, the dependent variable y is the force f , the slope m is the spring constant k, and the y-intercept is the preload fo . The slope m can be calculated using the data given in Fig. 1.6 as m=
4 5−1 = = 5. 0.9 − 0.1 0.8
The equation of the force-displacement equation in the slope-intercept form can therefore be written as f = 5y + b. The y-intercept b can be found using one of the data points. Using the data point (f , y) = (5, 0.9) gives 5 = 5 (0.9) + b. Solving for b gives b = 0.5 N. Therefore, the equation of the straight line can be written in slope-intercept form as f = 5y + 0.5. Comparing equations (1.8) and (1.9) gives k = 5N∕m, fo = 0.5N.
1.3 Force-Displacement in a Preloaded Tension Spring
Method 2 Now treating the force f as an independent variable, the forcef 1 displacement relationship f = ky + fo can be written as y = f − o . This relationk k ship is the equation of a straight line y = mx + b, where the independent variable x is the force f , the dependent variable y is the displacement y, the slope m is the 1 reciprocal of the spring constant , and the y-intercept is the negated preload k f divided by the spring constant − o . The slope m can be calculated using the data k given in Fig. 1.6 as 0.9 − 0.1 0.8 = = 0.2. 5−1 4
The equation of the displacement y as a function of force f can therefore be written in slope-intercept form as y = 0.2f + b. The y-intercept b can be found using one of the data points. Using the data point (y, f ) = (0.9, 5) gives 0.9 = 0.2 (5) + b. Solving for b gives b = −0.1. Therefore, the equation of the straight line can be written in the slope-intercept form as y = 0.2f − 0.1. Comparing equation (1.10) with the expression y = 1 = 0.2 k
(1.10) f 1 f − o gives k k
Therefore, the force-displacement relationship for a preloaded spring given in Fig. 1.6 is given by f = 5y + 0.5.
Chapter 1 Straight Lines in Engineering
FURTHER EXAMPLES OF LINES IN ENGINEERING The velocity of a vehicle follows the trajectory shown in Fig. 1.7. The vehicle starts at rest (zero velocity) and reaches a maximum velocity of 10 m/s in 2 s. It then cruises at a constant velocity of 10 m/s for 2 s before coming to rest at 6 s. Write the equation of the function v(t); in other words, write the expression of v(t) for times between 0 and 2 s, between 2 and 4 s, between 4 and 6 s, and greater than 6 s. v(t), m/s 10
Figure 1.7 Velocity profile of a vehicle. Solution
The velocity profile of the vehicle shown in Fig. 1.7 is a piecewise linear function with three different equations. The first linear function is a straight line passing through the origin starting at time 0 sec and ending at time equal to 2 s. The second linear function is a straight line with zero slope (cruise velocity of 10 m/s) starting at 2 s and ending at 4 s. Finally, the third piece of the trajectory is a straight line starting at 4 s and ending at 6 s. The equation of the piecewise linear function can be written as (a) 0 ≤ t ≤ 2: v(t) = mt + b where b = 0 and m =
10 − 0 = 5. Therefore, 2−0 v(t) = 5t m/s.
(b) 2 ≤ t ≤ 4: v = 10 m/s. (c) 4 ≤ t ≤ 6: v(t) = mt + b, 0 − 10 = −5 and the value of b can be calculated using the data 6−4 point (t, v(t)) = (6, 0) as
1.4 Further Examples of Lines in Engineering The value of b can also be calculated using the point-slope formula for the straight line v − v1 = m(t − t1 ), where v1 = 0 and t1 = 6. Thus, v − 0 = −5(t − 6). Therefore, v(t) = −5(t − 6). or v(t) = −5t + 30 m/s. (d) t > 6: v(t) = 0 m/s.
The velocity of a vehicle is given in Fig. 1.8. (a) Determine the equation of v(t) for (i) 0 ≤ t ≤ 3 s (ii) 3 ≤ t ≤ 6 s (iii) 6 ≤ t ≤ 9 s (iv) t ≥ 9 (b) Knowing that the acceleration of the vehicle is the slope of velocity, plot the acceleration of the vehicle. v(t), m/s 24
0 Figure 1.8 Velocity profile of a vehicle.
Chapter 1 Straight Lines in Engineering
(a) The velocity of the vehicle for different intervals can be calculated as (i) 0 ≤ t ≤ 3 s: v(t) = mt + b, where m =
12 − 24 = −4 m/s2 and b = 24 m/s. Therefore, 3−0 v(t) = −4t + 24 m/s.
(ii) 3 ≤ t ≤ 6 s: v(t) = 12 m/s. (iii) 6 ≤ t ≤ 9 s: v(t) = mt + b, 0 − 12 = −4 m/s2 and b can be calculated in slope-intercept 9−6 form using point (t, v(t)) = (9, 0) as
0 = −4(9) + b. Therefore, b = 36 m/s and v(t) = −4t + 36 m/s. (iv) t > 9 s: v(t) = 0 m/s. (b) Since the acceleration of the vehicle is the slope of the velocity in each interval, the acceleration a in m/s2 is given by ⎧−4; ⎪ ⎪ 0; a=⎨ ⎪−4; ⎪ 0; ⎩
0≤t≤3s 3≤t≤6s 6≤t≤9s t>9s
The plot of the acceleration is shown in Fig. 1.9. Acceleration, m/s2
4 Figure 1.9 Acceleration profile of the vehicle in Fig. 1.8.
1.4 Further Examples of Lines in Engineering Example 1-3
In a bolted connector shown in Fig. 1.10, the force in the bolt Fb is related to the external load P as Fb = C P + Fi , where C is the joint constant and Fi is the preload in the bolt. (a) Determine the joint constant C and the preload Fi given the data in Fig. 1.10. (b) Plot the bolt force Fb as a function of the external load P, and label C and Fi on the graph. P
Figure 1.10 External force applied to a bolted connection. Solution
(a) The force-load relationship Fb = CP + Fi is the equation of a straight line, y = mx + b. The slope m is the joint constant C, which can be calculated as C=
ΔFb 600 − 500 100 lb = = = 0.25. ΔP 800 − 400 400 lb
Therefore, Fb (P) = 0.25 P + Fi .
Now, the y-intercept Fi can be calculated by substituting one of the data points into equation (1.11). Substituting the second data point (Fb , P) = (600, 800) gives 600 = 0.25 × 800 + Fi . Solving for Fi yields Fi = 600 − 200 = 400 lb. Therefore, Fb = 0.25 P + 400 is the equation of the straight line, where C = 0.25 and Fi = 400 lb. Note that the joint constant C is dimensionless! (b) The plot of the force Fb in the bolt as a function of the external load P is shown in Fig. 1.11.
Chapter 1 Straight Lines in Engineering Fb (lb) (800, 600)
Plot of the bolt force Fb as a function of the external load P.
For the electric circuit shown in Fig. 1.12, the relationship between the voltage V and the applied current I is given by V = (I + Io )R. Find the values of R and I0 if the voltage across the resistor V is known for the two different values of the current I as shown in Fig. 1.12.
Figure 1.12 Solution
Circuit for Example 1-4.
The voltage-current relationship V = R I + R Io is the equation of a straight line y = mx + b, where the slope m = R can be found from the data given in Fig. 1.12 as R=
ΔV 2.2 − 1.2 1 volt = = = 10 Ω. ΔI 0.2 − 0.1 0.1 amp
Therefore, V = 10(I) + 10 I0 .
The y-intercept b = 10 I0 can be found by substituting the second data point (2.2, 0.2) in equation (1.12) as 2.2 = 100 × 0.2 + 10 I0 . Solving for I0 gives 10 I0 = 2.2 − 2 = 0.2,
1.4 Further Examples of Lines in Engineering
which gives I0 = 0.02 A. Therefore, V = 10 I + 0.2; and R = 10 Ω and I0 = 0.02 A.
The output voltage vo of the operational (Op–Amp) ( amplifier ( ) )circuit shown in 100 100 Fig. 1.13 satisfies the relationship vo = − vin + 1 + vb , where R in R R kΩ is the unknown resistance and vb is the unknown voltage. Fig. 1.13 gives the values of the output voltage for two different values of the input voltage. (a) Determine the value of R and vb . (b) Plot the output voltage vo as a function of the input voltage vin . On the plot, clearly indicate the value of the output voltage when the input voltage is zero (y-intercept) and the value of the input voltage when the output voltage is zero (x-intercept). 100 kΩ R kΩ
Figure 1.13 An Op–Amp circuit as a summing amplifier.
( ) 100 vin + 1 + vb is the equaR 100 tion of a straight line, y = mx + b, where the slope m = − can be found R from the data given in Fig. 1.13 as
(a) The input-output relationship vo =
Δvo −5 − 5 100 −10 = = = = −2. R Δvin 10 − 5 5
Solving for R gives R = 50 Ω. Therefore, ( ( ) ) 100 100 vin + 1 + vb v0 = − 50 50 = −2 vin + 3 vb .
The y-intercept b = 3 vb can be found by substituting the first data point (v0 , vin ) = (5, 5) in equation (1.13) as 5 = −2 × 5 + 3 vb .
Chapter 1 Straight Lines in Engineering Solving for vb yields 3 vb = 5 + 10 = 15, which gives vb = 5 V. Therefore, vo = −2 vin + 15, R = 50 Ω, and vb = 5 V. The x-intercept can be found by substituting vo = 0 in the equation vo = −2 vin + 15 and finding the value of vin as 0 = −2 vin + 15, which gives vin = 7.5 V. Therefore, the x-intercept occurs at Vin = 7.5 V. (b) The plot of the output voltage of the Op–Amp as a function of the input voltage if vb = 5 V is shown in Fig. 1.14. vo, V y−intercept b
Figure 1.14 An Op–Amp circuit as a summing amplifier.
An actuator used in a prosthetic arm (Fig. 1.15) can produce a different amount of force by changing the voltage of the power supply. The force and voltage satisfy the linear relation F = kV, where V is the voltage applied and F is the force produced by the prosthetic arm. The maximum force the arm can produce is F = 44.5 N when supplied with V = 12 volts. (a) Find the force produced by the actuator when supplied with V = 7.3 volts. (b) What voltage is needed to achieve a force of F = 6.0 N? (c) Using the results of parts (a) and (b), sketch the graph of F as a function of voltage V. Use the appropriate scales and clearly label the slope and the results of parts (a) and (b) on your graph.
1.4 Further Examples of Lines in Engineering Solution
(a) The input-output relationship F = k V is the equation of a straight line y = m x, where the slope m = k can be found from the given data as k=
Therefore, the equation of the straight line representing the actuator force F as a function of applied voltage V is given by F = 3.71 V.
Thus, the force produced by the actuator when supplied with 7.3 volts is found by substituting V = 7.3 in equation (1.14) as F = 3.71 × 7.3 = 27.08 N. (b) The voltage needed to achieve a force of 6.0 N can be found by substituting F = 6.0 N in equation (1.14) as 6.0 = 3.71 V 6.0 V= 3.71 = 1.62 volts.
(c) The plot of force F as a function of voltage V can now be drawn as shown in Fig. 1.16. F, N 44.5
6.0 0 Figure 1.16
Plot of the actuator force verses the applied voltage.
Chapter 1 Straight Lines in Engineering
The electrical activity of muscles can be monitored with an electromyogram (EMG). The following root mean square (RMS) value of the amplitude measurements of the EMG signal were taken when a woman was using her hand grip muscles to ensure a lid was tight on a jar. EMG
A, V 0.0005 0.00125
Amplitude measurements of the EMG signal.
The RMS amplitude of the EMG signal satisfies the linear equation A = mF + b
where A is the RMS value of the EMG amplitude in V, F is the applied muscle force in N, and m is the slope. (a) Determine the value of m and b. (b) Plot the RMS amplitude A as a function of the applied muscle force F. (c) Using the equation of the line from part (a), find the RMS value of the amplitude for a muscle force of 200 N.
(a) The input-output relationship A = mF + b is the equation of a straight line y = mx + b, where the slope m can be found from the EMG data given in the table (Fig. 1.17) as m=
V ΔA 0.00125 − 0.0005 0.00075 = = = 4.55 × 10−6 . ΔF 275 − 110 165 N
The y-intercept b can be found by substituting the first data point (A, F) = (0.0005, 110) in equation (1.16) as 0.0005 = 4.55 × 10−6 (110) + b. Solving for b yields b = 5 × 10−7 ≈ 0. Therefore, the equation of the straight line representing the RMS amplitude as a function of applied force is given by A = 4.55 × 10−6 F.
(b) The plot of the RMS amplitude as a function of the applied muscle force is shown in Fig. 1.18.
1.4 Further Examples of Lines in Engineering
Figure 1.18 Plot of the RMS amplitude verses the applied muscle force.
(c) The RMS value of the amplitude for a muscle force of 200 N can be found by substituting F = 200 N in equation (1.17) as A = 4.55 × 10−6 × (200) = 0.91 × 10−3 V.
A civil engineer needs to establish the elevation of the cornerstone for a building located between two benchmarks, B1 and B2, of known elevations as shown in Fig. 1.19. E2 l E1
Sea level Figure 1.19 Elevations along a uniform grade.
The elevation E along the grade satisfies the linear relationship E = m l + E1
where E1 is the elevation of B1, l is the distance from B1 along the grade, and m is the average slope of the grade. (a) Find the equation of the line E and determine the slope m of the grade. (b) Using the equation of the line from part (a), find the elevation of the cornerstone E ∗ if it is located at a distance l = 565 m from B1. (c) Sketch the graph of E as a function of l and clearly indicate both the slope m and elevation E1 of B1.
Chapter 1 Straight Lines in Engineering
(a) The equation of elevation given by equation (1.18) is a straight line in the slopeintercept form y = mx + b, where the slope m can be found from the elevation data given in Fig. 1.19 as m=
ΔE 476.8 − 428.4 = = Δl 1001.2 − 0
48.4 = 0.0483. 1001.2
The y-intercept E1 can be found by substituting the first data point (E, l) = (428.4, 0) in equation (1.18) as 428.4 = 0.0483 × (0) + E1 Solving for E1 yields E1 = 428.4 m. Therefore, the equation of the straight line representing the elevation as a function of distance l is given by E = 0.0483 l + 428.4 m.
(b) The elevation E ∗ of the cornerstone can be found by substituting l = 565 m in equation (1.19) as E ∗ = 0.0483 × (565) + 428.4 = 455.7 m. (c) The plot of the elevation as a function of the length is shown in Fig. 1.20.
Elevation, m 476.8
Figure 1.20 Elevation along a uniform grade.
PROBLEMS 1-1. A constant force F = 2 N is applied to a spring and the displacement x is measured as 0.2 m. If the spring force and displacement satisfy the linear relation F = k x, find the stiffness k of the spring. x k
Figure P1.1 Displacement of a spring in
1-2. The spring force F and displacement x for a close-wound tension spring are measured as shown in Fig. P1.2. The spring force F and displacement x satisfy the linear equation F = k x + Fi , where k is the spring constant and Fi is the preload induced during manufacturing of the spring. (a) Using the given data in Fig. P1.2, find the equation of the line for the spring force F as a function of the displacement x, and determine the values of the spring constant k and preload Fi . (b) Sketch the graph of F as a function of x. Use appropriate axis scales and clearly label the preload Fi , the spring constant k, and both given data points on your graph. F (N) 34.5 57.0
The spring force F and displacement x satisfy the linear equation F = k x + Fi , where k is the spring constant and Fi is the preload induced during manufacturing of the spring. (a) Using the given data, find the equation of the line for the spring force F as a function of the displacement x, and determine the values of the spring constant k and preload Fi . (b) Sketch the graph of F as a function of x and clearly indicate both the spring constant k and preload Fi . F (N) 135 222
Figure P1.3 Close-wound tension spring for
1-4. In a bolted connection shown in Fig. P1.4, the force in the bolt Fb is given in terms of the external load P as Fb = C P + Fi . (a) Given the data in Fig. P1.4, determine the joint constant C and the preload Fi . P
Figure P1.2 Close-wound tension spring for
1-3. The spring force F and displacement x for a close-wound tension spring are measured as shown in Fig. P1.3.
Figure P1.4 Bolted connection for problem P1-4.
Chapter 1 Straight Lines in Engineering (b) Plot the bolt force Fb as a function of the load P and label C and Fi on the graph.
1-5. Repeat problem P1-4 for the data given in Fig. P1.5. P
1-7. The velocity v(t) of a ball thrown upward satisfies the equation v(t) = vo + at, where vo is the initial velocity of the ball in m/s and a is the acceleration in m/s2 . (a) Given the data in Fig. P1.7, find the equation of the line representing the velocity v(t) of the ball, and determine both the initial velocity vo and the acceleration a. (b) Sketch the graph of the line v(t), and clearly indicate both the initial velocity and the acceleration on your graph. Also determine the time at which the velocity is zero. v(t)
Figure P1.5 Bolted connection for problem P1-5.
1-6. The velocity v(t) of a ball thrown upward satisfies the equation v(t) = vo + at, where vo is the initial velocity of the ball in ft/s and a is the acceleration in ft/s2 . (a) Given the data in Fig. P1.6, find the equation of the line representing the velocity v(t) of the ball, and determine both the initial velocity vo and the acceleration a. (b) Sketch the graph of the line v(t), and clearly indicate both the initial velocity and the acceleration on your graph. Also determine the time at which the velocity is zero. v(t)
Figure P1.7 A ball thrown upward with an
initial velocity vo (t) in problem P1-7.
1-8. A model rocket is fired in the vertical plane. The velocity v(t) is measured as shown in Fig. P1.8. The velocity satisfies the equation v(t) = vo + at, where vo is the initial velocity of the rocket in m/s and a is the acceleration in m/s2 . (a) Given the data in Fig. P1.8, find the equation of the line representing the velocity v(t) of the rocket, and determine both the initial velocity vo and the acceleration a. v(t)
Figure P1.6 A ball thrown upward with an initial
Figure P1.8 A model rocket fired in the vertical
velocity vo (t) in problem P1-6.
plane in problem P1-8.
Problems (b) Sketch the graph of the line v(t) for 0 ≤ t ≤ 8 seconds, and clearly indicate both the initial velocity and the acceleration on your graph. Also determine the time at which the velocity is zero (i.e., the time required to reach the maximum height). 1-9. A model rocket is fired in the vertical plane. The velocity v(t) is measured as shown in Fig. P1.9. The velocity satisfies the equation v(t) = vo + at, where vo is the initial velocity of the rocket in ft/s and a is the acceleration in ft/s2 . (a) Given the data in Fig. P1.9, find the equation of the line representing the velocity v(t) of the rocket, and determine both the initial velocity vo and the acceleration a. (b) Sketch the graph of the line v(t) for 0 ≤ t ≤ 10 seconds, and clearly indicate both the initial velocity and the acceleration on your graph. Also determine the time at which the velocity is zero (i.e., the time required to reach the maximum height).
v(t) (ft/s) 128.8 32.2
(a) Find the equation of the line v(t), and determine both the initial velocity vo and the acceleration a. (b) Sketch the graph of the line v(t), and clearly label the initial velocity, the acceleration, and the total stopping time on the graph. v(t) (m/s) 30 10
Figure P1.10 Velocity of a vehicle during braking
in problem P1-10.
1-11. The velocity of a vehicle is measured at two distinct points in time as shown in Fig. P1.11. The velocity satisfies the relationship v(t) = vo + at, where vo is the initial velocity in ft/s and a is the acceleration in ft/s2 . (a) Find the equation of the line v(t), and determine both the initial velocity vo and the acceleration a. (b) Sketch the graph of the line v(t), and clearly label the initial velocity, the acceleration, and the total stopping time on the graph. v(t) (ft/s) 112.5 37.5
A model rocket fired in the vertical plane Figure P1.11 Velocity of a vehicle during braking in problem P1-9. in problem P1-11.
1-10. The velocity of a vehicle is measured at two distinct points in time as shown in Fig. P1.10. The velocity satisfies the relationship v(t) = vo + at, where vo is the initial velocity in m/s and a is the acceleration in m/s2 .
1-12. The velocity v(t) of a vehicle during braking is given in Fig. P1.12. Determine the equation for v(t) for (a) 0 ≤ t ≤ 2 s (b) 2 ≤ t ≤ 4 s (c) 4 ≤ t ≤ 6 s
Chapter 1 Straight Lines in Engineering (a) 0 ≤ t ≤ 1 s (b) 1 ≤ t ≤ 3 s (c) 3 ≤ t ≤ 4 s a(t), m/s2
Figure P1.14 Acceleration of the robot trajectory.
Figure P1.12 Velocity of a vehicle during braking in
1-13. A linear trajectory is planned for a robot to pick up a part in a manufacturing process. The velocity of the trajectory of one of the joints is shown in Fig. P1.13. Determine the equation of v(t) for (a) 0 ≤ t ≤ 1 s (b) 1 ≤ t ≤ 3 s (c) 3 ≤ t ≤ 4 s v(t), m/s 10
1-15. The temperature distribution in a wellinsulated axial rod varies linearly with respect to distance when the temperature at both ends is held constant as shown in Fig. P1.15. The temperature satisfies the equation of a line T(x) = C1 x + C2 , where C1 and C2 are constants of integration with units of ◦ F/ft and ◦ F, respectively (a) Find the equation of the line T(x), and determine both constants C1 and C2 . (b) Sketch the graph of the line T(x) for 0 ≤ x ≤ 1.5 ft, and clearly label C1 and C2 on your graph. Also, clearly indicate the temperature at the center of the rod (x = 0.75 ft). x
Figure P1.13 Velocity of a robot trajectory.
1-14. The acceleration of the linear trajectory of problem P1-13 is shown in Fig. P1.14. Determine the equation of a(t) for
Figure P1.15 Temperature distribution in a
well-insulated axial rod in problem P1-15.
Problems 1-16. The temperature distribution in a wellinsulated axial rod varies linearly with respect to distance when the temperature at both ends is held constant as shown in Fig. P1.16. The temperature satisfies the equation of a line T(x) = C1 x + C2 , where C1 and C2 are constants of integration with units of ◦ C/m and ◦ C, respectively. (a) Find the equation of the line T(x), and determine both constants C1 and C2 . (b) Sketch the graph of the line T(x) for 0 ≤ x ≤ 0.5 m, and clearly label C1 and C2 on your graph. Also, clearly indicate the temperature at the center of the rod (x = 0.25 m). x
(b) Find the current I if the applied voltage is 10 V. 1-18. A voltage source Vs is used to apply two different voltages (12V and 18V) to the single-loop circuit shown in Fig. P1.18. The values of the measured current are shown in Fig. P1.18. The voltage and current satisfy the linear relation Vs = IR + V, where R is the resistance in ohms, I is the current in amps, and Vs is the voltage in volts. (a) Using the data given in Fig. P1.18, find the equation of the line for Vs as a function of I, and determine the values of R and V. (b) Sketch the graph of Vs as a function of I and clearly indicate the resistance R and voltage V on the graph. I
Vs (volt) 12.0 18.0
Figure P1.16 Temperature distribution in a
well-insulated axial rod in problem P1-16.
1-17. The voltage-current relationship for the circuit shown in Fig. P1.17 is given by Ohm’s law as V = I R, where V is the applied voltage in volts, I is the current in amps, and R is the resistance of the resistor in ohms. (a) Sketch the graph of I as a function of V if the resistance is 5 Ω.
Figure P1.18 Single-loop circuit for problem
1-19. Repeat problem P1-18 for the data shown in Fig. P1.19. I
Figure P1.19 Single-loop circuit for problem Figure P1.17 Resistive circuit for problem P1-17.
Chapter 1 Straight Lines in Engineering
1-20. Repeat problem P1-18 for the data shown in Fig. P1.20.
(b) Sketch the graph of VS as a function of I, and clearly indicate the resistance Rd and the voltage VON on the graph. 1-22. Repeat problem P1-21 for the data given in Fig. P1.22.
I (amp) 24 32 Vs (volt) 2.0 6
Figure P1.20 Single-loop circuit for problem P1-20.
1-21. A linear model of a diode is shown in Fig. P1.21, where Rd is the forward resistance of the diode and VON is the voltage that turns the diode ON. To determine the resistance Rd and voltage VON , two voltage values are applied to the diode and the corresponding currents are measured. The applied voltage VS and the measured current I are given in Fig. P1.21. The applied voltage and the measured current satisfy the linear equation Vs = I Rd + VON . (a) Find the equation of the line for Vs as a function of I and determine the resistance Rd and the voltage VON .
I (amp) 0.035 0.135
Figure P1.22 Linear model of a diode for problem
1-23. The output voltage, vo , of the Op–Amp circuit shown in Fig. ( ) ( P1.23) satisfies v
in − the relationship vo = 1 + 100 R 2 ( ) 100 vb , where R is the unknown resisR tance in kΩ and vb is the unknown voltage in volts. Fig. P1.23 gives the values of the output voltage for two different values of the input voltage.
I (amp) 0.086 0.186
Figure P1.21 Linear model of a diode for problem
Figure P1.23 An Op–Amp circuit as a summing
amplifier for problem P1.23.
Problems (a) Determine the equation of the line for vo as a function of vin and find the values of R and vb . (b) Plot the output voltage vo as a function of the input voltage vin . On the plot, clearly indicate the value of the output voltage when the input voltage is zero (y-intercept) and the value of the input voltage when the output voltage is zero (x-intercept). 1-24. The output voltage, vo , of the Op– Amp circuit shown in Fig.(P1.24 satisfies )
v , the relationship vo = − v2 + 100 R in where R is the unknown resistance in kΩ, vin is the input voltage, and v2 is the unknown voltage. Fig. P1.24 gives the values of the output voltage for two different values of the input voltage vin . (a) Find the equation of the line for vo as a function of vin and determine the values of R and v2 . (b) Plot the output voltage vo as a function of the input voltage vin . Clearly indicate the value of the output voltage when the input voltage is zero (y-intercept) and the value of the input voltage when the output voltage is zero (x-intercept).
Figure P1.24 An Op–Amp circuit for problem
1-25. A DC motor is driving an inertial load JL shown in Fig. P1.25. To maintain a constant speed, two different values of the voltage ea are applied to the motor. The voltage ea and the current ia flowing through the armature winding of the motor satisfy the relationship ea = ia Ra + eb , where Ra is the resistance of the armature winding in ohms and eb is the back-emf in volts. Figure P1.25 gives the values of the current for two different values of the input voltage applied to the armature of the DC motor. (a) Find the equation of the line for ea as a function of ia and determine the values of Ra and eb . (b) Plot the applied voltage ea as a function of the current ia . Clearly indicate the value of the back-emf eb and the winding resistance Ra .
Figure P1.25 Voltage-current data of a DC motor
for problem P1-25.
1-26. Repeat problem P1-25 for the data shown in Fig. P1.26.
Chapter 1 Straight Lines in Engineering i a Ra
the values of the output voltage for two different values of the drain current. Plot the output voltage vo as a function of the input drain current iD . On the plot, clearly indicate the values of RD and VDD . 1-28. Repeat problem P1-27 for the data given in Fig. P1.28.
Figure P1.26 Voltage-current data of a DC motor
in problem P1-26. vi
1-27. In the active region, the output voltage vo of the n-channel enhancementtype MOSFET (NMOS) circuit shown in Fig. P1.27 satisfies the relationship vo = VDD − RD iD , where RD is the unknown drain resistance and VD is the unknown drain voltage. Fig. P1.27 gives
Figure P1.28 NMOS for P1-28.
1-29. An actuator used in a prosthetic arm can produce different amounts of force RD by changing the voltage of the power supply. The force and voltage satisfy the vo linear relation F = kV, where V is the voltage applied and F is the force provi G duced by the prosthetic arm. The maximum force the arm can produce is 30.0 N when supplied with 10 V. S (a) Find the force produced by the actuator when supplied with 6.0 V. (b) What voltage is needed to achieve a force of 5.0 N? (c) Using the results of parts (a) and iD , mA vo , V (b), sketch the graph of F as a func4 8 tion of voltage V. Use the appropri10 2 ate scales and clearly label the slope Figure P1.27 n-channel enhancement-type MOSFET. and the results of parts (a) and (b).
Problems 1-30. The following two measurements of maximum heart rate R (in beats per minute, bpm) were recorded in an exercise physiology laboratory. R, bpm 183 169.5
1-32. A civil engineer needs to establish the elevation of the cornerstone for a building located between two benchmarks, B1 and B2, of known elevations, as shown in Fig. P1.32.
The maximum heart rate R and age A satisfy the linear equation
1-31. The electrical activity of muscles can be monitored with an electromyogram (EMG). The RMS amplitude measurements of the EMG signal when a person is using the hand grip muscle to tighten the lid on a jar is given in the table below: A, V 0.5 E-3 1.25 E-3
The RMS amplitude of the EMG signal satisfies the linear equation R = mA + B, where A is the RMS amplitude in volts, F is the applied muscle force in N, and m is the slope of the line. (a) Using the data provided in the table, find the equation of the line for A(F). (b) Sketch A as a function of F. (c) Using the relationship developed in part (a), find the RMS amplitude for a muscle force of 200 N.
R = mA + B where R is the heart rate in beats per minute and A is the age in years. (a) Using the data provided, find the equation of the line for R. (b) Sketch R as a function of A. (c) Using the relationship developed in part (a), find the maximum heart rate of a 60-year-old person.
Figure P1.32 Elevations along a uniform grade
for exercise P1-32.
The elevation E along the grade satisfies the linear relationship E = m l + E1
where E1 is the elevation of B1, l is the taped distance from B1 along the grade, and m is the rate of change of E with respect to l. (a) Find the equation of the line E and determine the slope m of the linear relationship. (b) Using the equation of the line from part (a), find the elevation of the cornerstone E ∗ if it is located at a distance l = 300 m from B1. (c) Sketch the graph of E as a function of l and clearly indicate both the slope m and elevation E1 of B1. 1-33. A thermocouple is a temperature measurement device, which produces a voltage V proportional to the temperature at the junction of two dissimilar metals. The voltage across a thermocouple is calibrated using the boiling point of water (100◦ C) and the freezing point of Zinc (420◦ C), as shown in Fig. P1.33.
Chapter 1 Straight Lines in Engineering Metal A Voltmeter Measuring junction
Figure P1.33 Thermocouple to measure
temperature in Celsius.
The junction temperature T and the voltage across the thermocouple V satisfy the linear equation T = 𝛼1 V + TR , where 𝛼 is the thermocouple sensitivity in mV/◦ C and TR is the reference temperature in ◦ C. (a) Using the calibration data given in Fig. P1.33, find the equation of the line for the measured temperature T as a function of the voltage V and determine the value of the sensitivity 𝛼 and the reference temperature TR . (b) Sketch the graph of T as a function of V and clearly indicate both the reference temperature TR and the sensitivity 𝛼 on the graph. 1-34. The voltage across a thermocouple is calibrated using the boiling point of water (373◦ K) and the freezing point of Silver (1235◦ K), as shown in Fig. P1.34. Metal A Voltmeter Measuring junction
Figure P1.34 Thermocouple to measure
temperature in Kelvin.
The junction temperature T and the voltage across the thermocouple V satisfy the linear equation T = 𝛼1 V + TR , where 𝛼 is the thermocouple sensitivity in mV/◦ K and TR is the reference temperature in ◦ K. (a) Using the calibration data given in Fig. P1.34, find the equation of the line for the measured temperature T as a function of the voltage V and determine the value of the sensitivity 𝛼 and the reference temperature TR . (b) Sketch the graph of T as a function of V and clearly indicate both the reference temperature TR and the sensitivity 𝛼 on the graph. 1-35. An iron-costantan thermocouple is calibrated by inserting its junction in boiling water (100◦ C) and measuring a voltage V = 5.27 mV, and then inserting the juction in silver chloride at its melting point (455◦ C) and measuring V = 24.88 mV. (a) Find the equation of the line for the measured temperature T as a function of the voltage V and determine the value of the sensitivity 𝛼 and the reference temperature TR . (b) Sketch the graph of T as a function of V and clearly indicate both the reference temperature TR and the sensitivity 𝛼 on the graph. (c) If the thermocouple is mounted in a chemical reactor and the voltage is observed to go from 10.0 mV to 13.6 mV, what is the change in temperature of the reactor? 1-36. Strain is a measure of the deformation of an object. It can be measured using a foil strain gauge shown in Fig. P1.36. The strain being measured (𝜖) and resistance of the sensor satisfy the linear equation R = Ro + Ro S𝜖 𝜖, where Ro is the initial resistance (measured in ohms, Ω) of the sensor with no strain, and S𝜖 is
Problems the gauge factor (a multiplier with NO units). (a) Using the given data, find the equation of the line for the sensor’s resistance R as a function of the strain 𝜖, and determine the values of the gauge factor S𝜖 and initial resistance Ro . (b) Sketch the graph of R as a function of 𝜖, and clearly indicate Ro on your graph.
End loops Grid Active grid length
Alignment marks End loops
Backing and encapsulation Solder tabs
Figure P1.36 Foil strain gauge to measure strain in
1-37. Repeat problem P1-36 for the data shown in Fig. P1.37. 1-38. To determine the concentration of a purified protein sample, a graduate student used spectrophotometry to measure the absorbance given in the table below: c (𝜇g/ml) 3.50 8.00
The concentration-absorbance relationship for this protein satisfies a linear equation a = m c + ai , where c is the concentration of a purified protein, a is
End loops Grid Active grid length
Alignment marks End loops
Backing and encapsulation Solder tabs
Figure P1.38 Foil strain gauge to measure strain in
the absorbance of the sample, m is the the rate of change of absorbance a with respect to concentration c, and ai is the y-intercept. (a) Find the equation of the line that describes the concentrationabsorbance relationship for this protein and determine the slope m of the linear relationship. (b) Using the equation of the line from part (a), find the concentration of the sample if this sample had an absorption of 0.486. (c) If the sample is diluted to a concentration of 0.00419 𝜇g/ml, what would you expect the absorption to be? Would this value be accurate? (d) Sketch the graph of absorbance a as a function of concentration c and clearly indicate both the slope m and the y-intercept. 1-39. Repeat problem P1-38 if the absorptionconcentration data for another sample is obtained as c (𝜇g/ml) 4.00 8.50
Chapter 1 Straight Lines in Engineering
1-40. A chemistry student is performing an experiment to determine the temperature-volume behavior of a gas mixture at constant pressure and quantity. Due to technical difficulties, he could only obtain values at two temperatures as shown in the table below: T (◦ C) 50 98
The student knows that the gas volume directly depends on temperature, that is, V(T) = m T + K, where V is the volume in L, T is the temperature in ◦ C, K is the y-intercept in L, and m is the slope of the line in L/◦ C. (a) Find the equation of the line that describes the temperature-volume relationship of the gas mixture and determine the slope m of the linear relationship. (b) Using the equation of the line from part (a), find the temperature of the gas mixture if the volume is 1.15 L. (c) Using the equation of the line from part (a), find the volume of the gas if the temperature is 70◦ C. (d) Sketch the graph of the volumetemperature relationship for the gas mixture from −300◦ C to 100◦ C and clearly indicate both the slope m and the y-intercept. What is the significance of the temperature when V = 0 L? 1-41. To obtain the linear relationship between the Fahrenheit and Celsius temperature scales, the freezing and boiling point of water is used as given in the table below: T (◦ F) 32 212
The relationship between the temperatures in Fahrenheit and Celsius scales satisfies the linear equation T (◦ F) = a T (◦ C) + b. (a) Using the given data, find the equation of the line relating the Fahrenheit and Celsius scales. (b) Sketch the graph of T (◦ F) as a function of T (◦ C), and clearly indicate a and b on your graph. (c) Using the graph obtained in part (b), find the temperature interval in ◦ C if the temperature is between 20◦ F and 80◦ F. 1-42. A thermostat control with dial marking from 0 to 100 is used to regulate the temperature of an oil bath. To calibrate the thermostat, the data for the temperature T (◦ F) versus the dial setting R was obtained as shown in the table below: T (◦ F) 110.0 40.0
The relationship between the temperature T in Fahrenheit and the dial setting R satisfies the linear equation T (◦ F) = a R + b. (a) Using the given data, find the equation of the line relating the temperature to the dial setting. (b) Sketch the graph of T (◦ F) as a function of R, and clearly indicate a and b on your graph. (c) Calculate the thermostat setting needed to obtain a temperature of 320◦ F. 1-43. In a pressure-fed journal bearing, forced cooling is provided by a pressurized lubricant flowing along the axial direction of the shaft (the x-direction) as shown in Fig. P1.43. The lubricant
Figure P1.43 Pressure-fed journal bearing.
pressure satisfies the linear equation p p(x) = s x + ps , l where ps is the supply pressure and l is the length of the bearing. (a) Using the data given in the table (Fig. P1.43), find the equation of the line for the lubricant pressure p(x) and determine the values of the
supply pressure ps and the bearing length l. (b) Calculate the lubricant pressure p(x) if x is 1.0 in. (c) Sketch the graph of the lubricant pressure p(x), and clearly indicate both the supply pressure ps and the bearing length l on the graph.
Quadratic Equations in Engineering In this chapter, the applications of quadratic equations in engineering are introduced. It is assumed that students are familiar with this topic from their high school algebra course. A quadratic equation is a second-order polynomial equation in one variable that occurs in many areas of engineering. For example, the height of a ball thrown in the air can be represented by a quadratic equation. In this chapter, the solution of quadratic equations will be obtained by three methods: factoring, the quadratic formula, and completing the square.
A PROJECTILE IN A VERTICAL PLANE Suppose a ball thrown upward from the ground with an initial velocity of 96 ft/s reaches a height h(t) after time t s as shown in Fig. 2.1. The height is expressed by the quadratic equation h(t) = 96 t − 16 t2 ft. Find the time t in seconds when h(t) = 80 ft.
h(t) = 96 t − 16 t2
Figure 2.1 A ball thrown upward to a height of h(t).
Solution: h(t) = 96 t − 16 t2 = 80 or 16 t2 − 96 t + 80 = 0.
Equation (2.1) is a quadratic equation of the form ax2 + bx + c = 0 and will be solved using three different methods.
2.1 A Projectile in a Vertical Plane Method 1: Factoring
Dividing equation (2.1) by 16 yields t2 − 6 t + 5 = 0.
Equation (2.2) can be factored as (t − 1)(t − 5) = 0. Therefore, t − 1 = 0 or t = 1 s and t − 5 = 0 or t = 5 s. Hence, the ball reaches the height of 80 ft at 1 s and 5 s. Method 2: Quadratic Formula solve for x is given by
If ax2 + bx + c = 0, then the quadratic formula to
√ b2 − 4ac . (2.3) x= 2a Using the quadratic formula in equation (2.3), the quadratic equation (2.2) can be solved as √ 6 ± 36 − 20 t= 2 6±4 . = 2 6+4 6−4 = 1 s and t = = 5 s. Hence, the ball reaches the height of Therefore, t = 2 2 80 ft at 1 s and 5 s. −b ±
First, rewrite the quadratic equation (2.2) as
Method 3: Completing the Square (
−6 (one-half the coefficient of the first-order term) to both 2 sides of equation (2.4) gives ( )2 ( )2 −6 −6 2 t − 6t + = −5 + , 2 2 or
Adding the square of
t2 − 6 t + 9 = −5 + 9.
Equation (2.5) can now be written as
or t − 3 = ±2. Therefore, t = 3 ± 2 or t = 1, 5 s. To check if the answer is correct, substitute t = 1 and t = 5 into equation (2.1). Substituting t = 1 s gives 16 × 1 − 96 × 1 + 80 = 0,
Chapter 2 Quadratic Equations in Engineering which gives 0 = 0. Therefore, t = 1 s is the correct time when the ball reaches a height of 80 ft. Now, substitute t = 5 s, 16 × 52 − 96 × 5 + 80 = 0, which again gives 0 = 0. Therefore, t = 5 s is also the correct time when the ball reaches a height of 80 ft. It can be seen from Fig. 2.2 that the height of the ball is 80 ft at both 1 s and 5 s. The ball is at 80 ft and going up at 1 s, and it is at 80 ft and going down at 5 s. Hence, the maximum height of ball must be halfway between 1 and 5 s, which is 1 + ((5 − 1)∕2) = 3 s. Therefore, the maximum height can be found by substituting t = 3 s in h(t), which is h(3) = 96(3) − 16(3)2 = 144 ft. These three points (height at t = 1, 3, and 5 s) can be used to plot the trajectory of the ball. However, to plot the trajectory accurately, additional data points can be added. The height of the ball at t = 0 is zero since the ball is thrown upward from the ground. To check this, substitute t = 0 in h(t). This gives h(0) = 96(0) − 16(0)2 = 0 ft. The time when the ball hits the ground again can be calculated by equating h(t) = 0. Therefore, 96 t − 16 t2 = 0 6 t − t2 = 0 t (6 − t) = 0. Therefore, t = 0 and 6 − t = 0 or t = 6 s. Since the ball is thrown in the air from the ground (h(t) = 0) at t = 0, it will hit the ground again at t = 6 s. Using these data points, the trajectory of the ball thrown upward with an initial velocity of 96 ft/s is shown in Fig. 2.2. h(t), ft hmax
Figure 2.2 The height of the ball thrown upward with an initial velocity of 96 ft/s.
Suppose now you wish to find the time t in seconds when the height of the ball reaches 144 ft. Setting h(t) = 144 gives h(t) = 96 t − 16 t2 = 144.
2.1 A Projectile in a Vertical Plane
Therefore, 16 t2 − 96 t + 144 = 0 or t2 − 6 t + 9 = 0.
The quadratic equation given in equation (2.6) can also be solved using the three methods as Factoring
Completing the Square t2 − 6t + 9 = 0
t − 6t + 9 = 0 (t − 3)(t − 3) = 0 t−3=0 t=3 s
t2 − 6t = −9 )2 )2 ( −6 −6 2 = −9 + t − 6t + 2 2 2 t − 6t + 9 = 9 − 9 (
t − 3 = ±0 t = 3, 3 t=3 s
Now suppose you wish to find the time t when the height of the ball reaches h(t) = 160 ft. Setting h(t) = 160 gives h(t) = 96t − 16t2 = 160. Therefore, 16t2 − 96t + 160 = 0 or t2 − 6t + 10 = 0. (2.7) The quadratic equation given in equation (2.7) can be solved using the three methods as Factoring
Completing the Square t2 − 6t + 10 = 0
t2 − 6t + 10 = 0 t2 − 6t + 10 = 0 cannot be factored using real integers
36 − 40 √2 6 ± −4 t= 2 √ t = 3 ± −1
t2 − 6t = −10 )2 )2 ( ( −6 −6 t2 − 6t + = −10 + 2 2 t2 − 6t + 9 = −1 (t − 3)2 = −1 √ t − 3 = ± −1 t =3±j
Chapter 2 Quadratic Equations in Engineering √ In the above solution i = j = −1 is the imaginary number, therefore the roots of the quadratic equation are complex. Hence, the ball never reaches the height of 160 ft. The maximum height achieved is 144 ft at 3 s.
CURRENT IN A LAMP A 100 W lamp and a 20 Ω resistor are connected in series to a 120 V power supply as shown in Fig. 2.3. The current I in amperes satisfies a quadratic equation as follows. Using KVL, 120 = VL + VR . LAMP + VL
Figure 2.3 A lamp and a resistor connected to a 120 V supply.
From Ohm’s law, VR = 20 I. Also, since the power is the product of voltage and 100 current, PL = VL I = 100 W, which gives VL = . Therefore, I 100 120 = + 20 I. (2.8) I Multiplying both sides of equation (2.8) by I yields 120 I = 100 + 20 I 2 .
Dividing both sides of equation (2.9) by 20 and rearranging gives I 2 − 6 I + 5 = 0.
The quadratic equation given in equation (2.10) can be solved using the three methods as Factoring
I 2 − 6I + 5 = 0 2
I − 6I + 5 = 0 (I − 1)(I − 5) = 0 I = 1, 5 A
Completing the Square I 2 − 6I + 5 = 0 )2 )2 ( ( −6 −6 I 2 − 6I + = −5 + 2 2 I 2 − 6I + 9 = −5 + 9 (I − 3)2 = 4 I − 3 = ±2 I =3±2 I = 1, 5 A
2.3 Equivalent Resistance
Note that the two solutions correspond to two lamp choices. Case I: For I = 1 A, VL =
100 100 = = 100 V. I 1
Case II: For I = 5 A, VL =
Case I corresponds to a lamp rated at 100 V, and Case II corresponds to a lamp rated at 20 V.
EQUIVALENT RESISTANCE Suppose two resistors are connected in parallel, as shown in Fig. 2.4. If the equivalent R1 R2 = 100 Ω and R1 = 4R2 + 100 Ω, find R1 and R2 . resistance R = R1 + R2
Figure 2.4 Equivalent resistance of two resistors connected in parallel.
The equivalent resistance of two resistors connected in parallel as shown in Fig. 2.4 is given by R1 R2 = 100 Ω. R1 + R2
Substituting R1 = 4R2 + 100 Ω in equation (2.11) gives 4 R22 + 100 R2 (4 R2 + 100)(R2 ) 100 = = . 5 R2 + 100 (4 R2 + 100) + R2
Multiplying both sides of equation (2.12) by 5R2 + 100 yields 100 (5 R2 + 100) = 4 R22 + 100 R2 .
Simplifying equation (2.13) gives 4 R22 − 400 R2 − 10,000 = 0.
Dividing both sides of equation by (2.14) by 4 gives R22 − 100 R2 − 2500 = 0.
Chapter 2 Quadratic Equations in Engineering Equation (2.15) is a quadratic equation in R2 and cannot be factored with whole numbers. Therefore, R2 is solved using the quadratic formula as √ √ 100 ± 10,000 − 4(−2500) 100 ± 2(10,000) = . R2 = 2 2 Therefore,
√ √ 100 ± 100 2 = 50 ± 50 2. R2 = 2
Since R2 canot be negative,
√ R2 = 50 + 50 2 = 120.7 Ω.
Substituting the value of R2 in R1 = 4R2 + 100 Ω yields R1 = 4(120.7) + 100 = 582.8 Ω. Therefore, R1 = 582.8 Ω and R2 = 120.7 Ω.
FURTHER EXAMPLES OF QUADRATIC EQUATIONS IN ENGINEERING A model rocket is fired into the air from the ground with an initial velocity of 98 m/s as shown in Fig. 2.5. The height h(t) satisfies the quadratic equation h(t) = 98 t − 4.9 t2 m.
(a) Find the time when h(t) = 245 m. (b) Find the time it takes the rocket to hit the ground. (c) Use the results of parts (a) and (b) to sketch h(t) and determine the maximum height.
h(t) = 98 t − 4.9 t2
Figure 2.5 A rocket fired vertically in the air.
2.4 Further Examples of Quadratic Equations in Engineering Solution
(a) Substituting h(t) = 245 in equation (2.16), the quadratic equation is given by −4.9 t2 + 98 t − 245 = 0.
Dividing both sides of equation (2.17) by −4.9 gives t2 − 20 t + 50 = 0.
The quadratic equation given in equation (2.18) can be solved using the three methods used in Section 2.1 as Factoring
t2 − 20t + 50 = 0 t2 − 20t + 50 = 0
t2 − 20t + 50 = 0 t2 − 20t = −50
400 − 200 √ 2 t = 10 ± 50
can’t be factored with whole numbers
Completing the Square
t2 − 20t + 100 = −50 + 100
t = 10 ± 7.07 t = 2.93, 17.07 s
(t − 10)2 = 50 √ t − 10 = ± 50 t = 10 ± 7.07 t = 2.93, 17.07 s
(b) Since the rocket hit the ground at h(t) = 0, h(t) = 98 t − 4.9 t2 = 0 4.9 t (20 − t) = 0. Therefore, t = 0 s and t = 20 s. Since the rocket is fired from the ground at t = 0 s, the rocket hits the ground again at t = 20 s. (c) The maximum height should occur halfway between 2.93 and 17.07 s. Therefore, tmax =
2.93 + 17.07 20 = = 10 s. 2 2
Substituting t = 10 s into equation (2.16) yields hmax = 98(10) − 4.9(10)2 = 490 m. The plot of the rocket trajectory is shown in Fig. 2.6. It can be seen from this figure that the rocket is fired from the ground at a height of zero at 0 s, crosses a height of 245 m at 2.93 s, and continues moving up and reaches the maximum height of 490 m at 10 s. At 10 seconds, it starts its downward descent and after crossing the height of 245 m again at 17.07 s, it reaches the ground again at 20 s.
Chapter 2 Quadratic Equations in Engineering h(t), m
490 400 300 245 200 100
Figure 2.6 The height of the rocket fired vertically in the air with an initial velocity of 98 m/s.
The equivalent resistance R of two resistors R1 and R2 connected in parallel as shown in Fig. 2.4 is given by R=
(a) Suppose R2 = 2R1 + 4 Ω and the equivalent resistance R = 8.0 Ω. Substitute these values in equation (2.19) to obtain the following quadratic equation for R1 : 2R21 − 20R1 − 32 = 0. (b) Solve for R1 by each of the following methods: (i) Completing the square. (ii) The quadratic formula. Also, determine the value of R2 corresponding to the only physical solution for R1 . Solution
(a) Substituting R2 = 2R1 + 4 and R = 8.0 in equation (2.19) gives 8.0 =
2 R21 + 4 R1 R1 (2 R1 + 4) . = 3 R1 + 4 R1 + (2 R1 + 4)
Multiplying both sides of equation (2.20) by (3R1 + 4) yields 8.0(3 R1 + 4) = 2 R21 + 4 R1 , or 24.0 R1 + 32.0 = 2 R21 + 4 R1 .
Rearranging terms in equation (2.21) gives 2 R21 − 20 R1 − 32 = 0.
2.4 Further Examples of Quadratic Equations in Engineering
(b) The quadratic equation given in equation (2.22) can be now be solved to find the values of R1 . (i) Method 1: Completing the square: Dividing both sides of equation (2.22) by 2 gives R21 − 10 R1 − 16 = 0.
Taking 16 on the other side of equation (2.23) and adding
both sides yields R21 − 10 R1 + 25 = 16 + 25.
Now, writing both sides of equation (2.24) as squares yields √ (R1 − 5)2 = (± 41)2 = (± 6.4)2 . Therefore, R1 − 5 = ± 6.4, which gives the values of R1 as 5 + 6.4 = 11.4 Ω and 5 − 6.4 = −1.4 Ω. Since the value of R1 cannot be negative, R1 = 11.4 Ω and R2 = 2R1 + 4 = 2(11.4) + 4 = 26.8 Ω. (ii) Method 2: Solving equation (2.22) using the quadratic formula: √ (−20)2 − 4(2)(−32) R1 = 4 √ 20 ± 656 20 ± 25.6 = = = 11.4, −1.4. 4 4 Since R1 cannot be negative, R1 = 11.4 Ω. Substituting R1 = 11.4 Ω in R2 = 2 R1 + 4 gives 20 ±
R2 = 2(11.4) + 4 = 26.8 Ω.
An assembly of springs shown in Fig. 2.7 has an equivalent stiffness k, given by
If k2 = 2k1 + 4 lb/in. and the equivalent stiffness is k = 3.6 lb/in., find k1 and k2 as follows: (a) Substitute the values of k and k2 into equation (2.25) to obtain the following quadratic equation for k1 : 5k21 − 2.8 k1 − 14.4 = 0.
Chapter 2 Quadratic Equations in Engineering (b) Using the method of your choice, solve equation (2.26) and determine the values of both k1 and k2 . k1
Figure 2.7 An assembly of three springs. Solution
(a) Substituting k2 = 2 k1 + 4 and k = 3.6 in equation (2.25) yields 3.6 = k1 +
2 k21 + 4 k1 k1 (2 k1 + 4) . = k1 + 3 k1 + 4 k1 + (2 k1 + 4)
Multiplying both sides of equation (2.27) by (3k1 + 4) gives 3.6(3 k1 + 4) = k1 (3 k1 + 4) + 2 k21 + 4 k1 10.8 k1 + 14.4 = 3 k21 + 4 k1 + 2k21 + 4 k1 10.8 k1 + 14.4 = 5 k21 + 8 k1 .
Rearranging terms in equation (2.28) gives 5 k21 − 2.8 k1 − 14.4 = 0.
(b) The quadratic equation (2.29) can be solved using the quadratic formula as √ 2.8 ± (−2.8)2 − 4(5)(−14.4) k1 = 10 =
= 2.0, −1.44. Since k1 cannot be negative, k1 = 2.0 lb/in. Now, substituting k1 = 2.0 in k2 = 2 k1 + 4 yields k2 = 2(2) + 4 = 8.0. Therefore, k2 = 8.0 lb/in.
2.4 Further Examples of Quadratic Equations in Engineering Example 2-4
A capacitor C and an inductor L are connected in series as shown in Fig. 2.8. The 1 , where 𝜔 is the angular fretotal reactance X in ohms is given by X = 𝜔L − 𝜔C quency in rad/s. (a) Suppose L = 1.0 H and C = 0.25 F. If the total reactance is X = 3.0 Ω, show that the angular frequency 𝜔 satisfies the quadratic equation 𝜔2 − 3𝜔 − 4 = 0. (b) Solve the quadratic equation for 𝜔 by each of the following methods: factoring, completing the square, and the quadratic formula. L
Figure 2.8 Series connection of L and C. Solution
(a) The total reactance of the series combination of L and C shown in Fig. 2.8 is given by X = 𝜔L −
Substituting L = 1.0 H, C = 0.25 F, and X = 3.0 Ω in equation (2.30) yields 3.0 = 𝜔(1) −
Multiplying both sides of equation (2.31) by 𝜔 gives 3𝜔 = 𝜔2 − 4.
Rearranging terms in equation (2.32) yields 𝜔2 − 3𝜔 − 4 = 0.
(b) The quadratic equation (2.33) can be solved by three different methods: factoring, completing the squares, and the quadratic formula. (i) Method 1: Factoring: The quadratic equation (2.33) can be factored as (𝜔 − 4)(𝜔 + 1) = 0, which gives 𝜔 − 4 = 0 or 𝜔 + 1 = 0. Therefore, 𝜔 = 4 rad/s or 𝜔 = −1 rad/s. Since 𝜔 cannot be negative, 𝜔 = 4 rad/s. (ii) Method 2: Completing the squares: The quadratic equation (2.33) can be written as 𝜔2 − 3𝜔 = 4.
Chapter 2 Quadratic Equations in Engineering ( Adding
9 to both sides of equation (2.34) gives 4 ( ) ( ) 9 9 =4+ . 𝜔2 − 3𝜔 + 4 4
Writing both sides of equation (2.35) as a square gives ( )2 ( )2 5 3 = ± . 𝜔− 2 2
Taking the square root of both sides of equation (2.36) yields 𝜔−
Therefore, 3 5 ± , 2 2 3 5 3 5 which gives 𝜔 = + = 4 rad/s or 𝜔 = − = −1 rad/s. Since 𝜔 cannot 2 2 2 2 be negative, 𝜔 = 4 rad/s. (iii) Method 3: Quadratic formula: Solving the quadratic equation (2.33) using the quadratic formula gives √ 3 ± (−3)2 − 4(1)(−4) . (2.37) 𝜔= 2 𝜔=
Equation (2.37) can be written as 𝜔=
which gives 𝜔 = 4, −1. Since 𝜔 cannot be negative, 𝜔 = 4 rad/s.
For the circuit shown in Fig. 2.3, the power P delivered by the voltage source Vs is given by the equation P = I 2 R + I VL . (a) Suppose that P = 96 W, VL = 32 V, and R = 8 Ω. Show that the current I satisfies the quadratic equation I 2 + 4 I − 12 = 0. (b) Solve the quadratic equation for I by each of the following methods: factoring, completing the square, and the quadratic formula.
2.4 Further Examples of Quadratic Equations in Engineering Solution
(a) Substituting P = 96 W, VL = 32 V, and R = 8 Ω into the power delivered P = I 2 R + I VL yields 96 = I 2 (8) + I(32).
Dividing both sides of equation (2.38) by 8 gives 12 = I 2 + 4 I.
Rearranging terms in equation (2.39) yields I 2 + 4 I − 12 = 0.
(b) The quadratic equation given in equation (2.40) can be solved by three different methods: factoring, completing the squares, and the quadratic formula. (i) Method 1: Factoring: The quadratic equation (2.40) can be factored as (I + 6)(I − 2) = 0, which gives I + 6 = 0 or I − 2 = 0. Therefore, I = −6 A or I = 2 A. (ii) Method 2: Completing the square: The quadratic equation (2.40) can be written as I 2 + 4 I = 12. ( )2 4 = 4 to both sides of equation (2.41), Adding 2 I 2 + 4 I + 4 = 12 + 4.
Writing both sides of equation (2.42) as a square yields (I + 2)2 = (± 4)2 .
Taking the square root of both sides of equation (2.43) gives I + 2 = ± 4. Therefore, I = −2 ± 4, which gives I = −2 − 4 = −6 A or I = −2 + 4 = 2 A. (iii) Method 3: Quadratic formula: Solving the quadratic equation (2.40) using the quadratic formula gives √ −4 ± (4)2 − 4(1)(−12) . (2.44) I= 2 Equation (2.44) can be written as √ −4 ± 64 −4 ± 8 = = −2 ± 4, I= 2 2 which gives I = −2 − 4 = −6 A or I = −2 + 4 = 2 A.
Chapter 2 Quadratic Equations in Engineering Case I: For I = −6 A, the power absorbed by the lamp is −6 × 32 = −192 W. Since the power absorbed by the lamp cannot be negative, I = −6 A is not one of the solutions of the quadratic equation given by (2.40). Case II: For I = 2 A, the power absorbed by the lamp is 2 × 32 = 64 W and the power dissipated by the resistor is 96 − 64 = 32 W. The voltage across the resistor VR = 2 × 8 = 16 V and using KVL, Vs = 16 + 32 = 48 V. Therefore, for the applied power of 96 W (source voltage = 48 V), I = 2 A is the solution of the quadratic equation given by (2.40).
A diver jumps off a diving board 1.5 m above the water with an initial vertical velocity of 0.6 m/s as shown in Fig. 2.9. The diver’s height above the water is given by h(t) = −4.905 t2 + 0.6 t + 1.5 m.
vo = 0.6 m/s h(t) = 1.5 1.5 m h(t) = 0
Figure 2.9 Person jumping off a diving board.
(a) Find the time in seconds when the diver hits the water. Use both the quadratic formula and completing the square. (b) Find the maximum height of the diver if it is known to occur at t = 0.0612 s. (c) Use the results of parts (a) and (b) to sketch the height h(t) of the diver as a function of time. Solution
(a) The time when the diver hits the water is found by setting h(t) = 0 in equation (2.45) as −4.905 t2 + 0.6 t + 1.5 = 0
Dividing both sides of equation (2.46) by −4.905 gives t2 − 0.1223 t − 0.3058 = 0.
2.4 Further Examples of Quadratic Equations in Engineering
The quadratic equation given in equation (2.47) can be solved using the quadratic formula and completing the square as outlined below. (i) Method 1: Quadratic formula: Solving the quadratic equation (2.47) using the quadratic formula gives t=
(0.1223)2 − 4(1)(−0.3058) . 2
Equation (2.48) can be written as 0.1223 ± t= 2
0.1223 ± 1.1127 = 0.0612 ± 0.5563, 2
which gives t = 0.0612 − 0.5563 = −0.495 s or t = 0.0612 + 0.5564 = 0.617 s. Since the time cannot be negative, it takes 0.617 s for the diver to hit the water. (ii) Method 2: Completing the square: The quadratic equation (2.47) can be written as t2 − 0.1223 t = 0.3058. ( Adding
)2 = 0.0037 to both sides gives
t2 − 0.1223 t + 0.015 = 0.3058 + 0.0037. Writing both sides of equation (2.50) as a perfect square yields √ (t − 0.0612)2 = (± 0.3095)2 .
Taking the square root of both sides gives t − 0.0612 = ± 0.5563. Therefore, t = 0.0612 ± 0.5563, which gives t = 0.0612 − 0.5563 = −0.495 s or t = 0.0612 + 0.5563 = 0.617 s. Since the time cannot be negative, it takes 0.617 s for the diver to hit the water. (b) The maximum height of the diver is found by substituting t = 0.0612 in equation (2.45) as hmax = h(0.0612) = −4.905(0.0612)2 + 0.6(0.0612) + 1.5 = 1.518 m. (c) Using the results of parts (a) and (b), the diver’s height after jumping from the diving board is plotted as shown in Fig. 2.10.
Chapter 2 Quadratic Equations in Engineering h(t), m
Height of the diver after jumping from the diving board.
Pipeline Through Parabolic Hill: A level pipeline is required to pass through a hill having the parabolic profile y = − 0.004 x2 + 0.3 x.
The origin of the x- and y-coordinates is fixed at elevation zero near the base of the hill, as shown in Fig. 2.11.
Hill profile Tunnel entry A
Pipeline path through a parabolic hill.
(a) Write the quadratic equation for a pipeline elevation of y = 2 m. (b) Solve the quadratic equation found in part (a) to determine the positions of the tunnel entry xA and exit xB using both the quadratic formula and completing the square. (c) Find the length of the tunnel.
2.4 Further Examples of Quadratic Equations in Engineering Solution
(a) Since the height y of the tunnel is 2 m, equation (2.52) can be written as 2 = −0.004 x2 + 0.3 x which gives 0.004 x2 − 0.3 x + 2 = 0 or x2 − 75 x + 500 = 0.
(b) The quadratic equation given in equation (2.53) can be solved using the quadratic formula and completing the square as outlined below. (i) Method 1: Quadratic formula: Solving the quadratic equation (2.53) using the quadratic formula gives √ 75 ± (− 75)2 − 4(1)(500) x= . (2.54) 2 Equation (2.54) can be written as √ 75 ± 3625 75 ± 60.2 x= = = 37.5 ± 30.1, 2 2 which gives x = 37.5 − 30.1 = 7.4 m or x = 37.5 + 30.1 = 67.6 m. Therefore, the tunnel entry position is 7.4 m and the tunnel exit position is 67.6 m. (ii) Method 2: Completing the square: The quadratic equation (2.53) can be written as x2 − 75 x = −500. ( Adding
)2 = 1406.25 to both sides gives x2 − 75 x + 1406.25 = −500 + 1406.25.
Writing both sides of equation (2.56) as a perfect square yields √ (x − 37.5)2 = (± 906.25)2 .
Taking the square root of both sides gives x − 37.5 = ± 30.1. Therefore, x = 37.5 ± 30.1, which gives x = 37.5 − 30.1 = 7.4 m or x = 37.5 + 30.1 = 67.6 m. To check if the answer is correct, substitute x = 7.4 and x = 67.6 into equation (2.53).
Chapter 2 Quadratic Equations in Engineering Substituting x = 7.4 m gives (7.42 ) − 75(7.4) + 500 = 0 55 − 555 + 500 = 0 0=0 Now, substituting x = 67.6 m into equation (2.53) gives (67.62 ) − 75(67.6) + 500 = 0 4570 − 5070 + 500 = 0 0=0 Therefore, xA = 7.4 m and xB = 67.6 m are the correct positions of the tunnel entry and exit, respectively. (c) The tunnel length can be found by subtracting the position xA from xB as Tunnel length = xB − xA = 67.6 − 7.4 = 60.2 m
PROBLEMS 2-1. An analysis of a circuit shown in Fig. P2.1 yields the quadratic equation for the current I as 3I 2 + 6 I = 45, where I is in amps. (a) Rewrite the above equation in the form I 2 + aI + b = 0 , where a and b are constants. (b) Solve the equation in part (a) by each of the following methods: factoring, completing the square, and the quadratic formula. R=3Ω I Vs
Figure P2.2 Resistive
circuit for problem P2-2.
2-3. Repeat problem P2-1 for the circuit shown in Fig. P2.3, which yields the quadratic equation for the current I as 210 = 10 I 2 + 40 I, where I is in amps.
Figure P2.1 Resistive
circuit for problem P2-1.
2-2. Repeat problem P2-1 for the circuit shown in Fig. P2.2, which yields the quadratic equation for the current I as 3I 2 − 6 I = 45, where I is in amps.
Figure P2.3 Resistive
circuit for problem P2-3.
Problems 2-4. The current flowing through the inductor shown in Fig. P2.4 is given by the quadratic equation i(t) = t2 − 8t. Find t when (a) i(t) = 9 A (use the quadratic formula), and (b) i(t) = 84 A (use completing the square).
If the total resistance of the circuit is R = 100 Ω and R2 = 2R1 + 100 Ω, find R1 and R2 as follows: (a) Substitute the values of R and R2 into equation (2.58), and simplify the resulting expression to obtain a single quadratic equation for R1 . (b) Using the method of your choice, solve the quadratic equation for R1 and compute the corresponding value of R2 . R1
v(t) = 2t − 8 Figure P2.4 Current flowing
through an inductor.
Figure P2.6 Series parallel combination
2-5. The voltage across the capacitor shown in Fig. P2.5 is given by the quadratic equation v(t) = t2 − 6t. Find t when (a) v(t) = 16 V (use the quadratic formula), and (b) v(t) = 27 V (use completing the square).
2-7. Repeat problem P2-6 if R = 1600 Ω and R2 = R1 + 500 Ω. 2-8. The energy dissipated by a resistor shown in Fig. P2.8 varies with time t in s according to the quadratic equation W = 3t2 + 6t. Solve for t if (a) W = 3 joules (b) W = 9 joules (c) W = 45 joules
i(t) = (0.2t − 0.6) mA Figure P2.5 Voltage across a
Figure P2.8 Resistive circuit
for problem P2-8.
2-6. In the purely resistive circuit shown in Fig. P2.6, the total resistance R of the circuit is given by R = R1 +
2-9. The equivalent capacitance C of two capacitors connected in series as shown in Fig. P2.9 is given by C=
Chapter 2 Quadratic Equations in Engineering If the total capacitance is C = 120 𝜇F and C2 = C1 + 100 𝜇F, find C1 and C2 as follows: (a) Substitute the values of C and C2 in equation (2.59) and obtain the quadratic equation for C1 . (b) Solve the quadratic equation for C1 obtained in part (a) by each of the following methods: factoring, completing the square, and the quadratic formula. Also, compute the corresponding values of C2 . C1
Figure P2.11 Series combination of two capacitors.
2-12. The equivalent inductance L of two inductors connected in parallel as shown in Fig. P2.12 is given by L=
If the total inductance L = 80 mH and L1 = L2 + 300 mH, find L1 and L2 as follows: (a) Substitute the values of L and L1 in equation (2.61) and obtain the quadratic equation for L2 . (b) Solve the quadratic equation for L2 obtained in part (a) by completing the square, and the quadratic formula. Also, compute the corresponding values of L1 .
Figure P2.9 Series combination of
2-10. Repeat problem P2-9 if C = 75 𝜇F and C2 = C1 + 200 𝜇F. 2-11. The equivalent capacitance C of three capacitors connected in series-parallel as shown in Fig. P2.11 is given by C C C = 25 + 1 2 . C1 + C2
Figure P2.12 Parallel combination of two
If the total capacitance is C = 125 𝜇F and C2 = C1 + 100 𝜇F, find C1 and C2 as follows: (a) Substitute the values of C and C2 in equation (2.60) and obtain the quadratic equation for C1 . (b) Solve the quadratic equation for C1 obtained in part (a) by completing the square and the quadratic formula. Also, compute the corresponding values of C2 .
2-13. Repeat problem P2-12 if L = 150 mH and L1 = L2 + 400 mH. 2-14. The equivalent inductance L of three inductors connected in series-parallel as shown in Fig. P2.14 is given by L = 125 +
(a) Suppose L2 = L1 + 200 mH and that the equivalent inductance is L = 200 mH. Substitute these values in equation (2.62) and obtain
Problems the following quadratic equation: L21 + 50 L1 − 15, 000 = 0.
(b) Solve the quadratic equation (2.63) for L1 by each of the following methods: factoring, completing the square, and the quadratic formula. 125 mH
2-16. The ball shown in Fig. P2.16 is dropped from a height of 1000 meters. The ball falls according to the quadratic equation h(t) = 1000 − 4.905t2 . Find the time t in seconds for the ball to reach a height h(t) of (a) 921.52 m
1000 m h(t) Figure P2.14 Series-parallel combination of
2-15. A model rocket is launched in the vertical plane at time t = 0 s as shown in Fig. P2.15. The height of the rocket (in feet) satisfies the quadratic equation h(t) = 64 t − 16 t2 . (a) Find the value(s) of the time t when h(t) = 48 ft. (b) Find the value(s) of the time t when h(t) = 60 ft. (c) Find the time required for the rocket to hit the ground. (d) Based on your solution to parts (a) through (c), determine the maximum height of the rocket and sketch the height h(t).
Figure P2.16 A ball dropped
from a height of 1000 m.
2-17. At time t = 0, a ball is thrown vertically from the top of the building at a speed of 56 ft/s, as shown in Fig. P2.17. The height of the ball at time t is given by h(t) = 32 + 56t − 16t2 ft. v = 56 ft/s
32 ft h(t) h(t) = 64 t − 16 t2
Figure P2.15 A model rocket for
Figure P2.17 A ball thrown vertically
from the top of a building.
Chapter 2 Quadratic Equations in Engineering (a) Find the value(s) of the time t when h(t) = 32 ft. (b) Find the time required for the ball to hit the ground. (c) Use the results to determine the maximum height, and sketch the height h(t) of the ball.
2-18. Two springs connected in series shown in Fig. P2.18 can be represented by a single equivalent spring. The stiffness of the equivalent spring is given by keq =
where k1 and k2 are the spring constants of the two springs. If keq = 1.2 N/m and k2 = 2k1 − 1 N/m, find k1 and k2 as follows: (a) Substitute the values of keq and k2 in equation (2.64) and obtain the quadratic equation for k1 . (b) Solve the quadratic equation for k1 obtained in part (a) by completing the square, and the quadratic formula. Also, compute the corresponding values of k2 . k1
Figure P2.18 Series combination of two springs.
2-19. The equivalent stiffness of a seriesparallel combination of three springs shown in Fig. P2.19 is given by k k k = k2 + 1 2 . k1 + k2
obtained in part (a) and determine the values of k2 . Also, find the values of k1 . k2
Figure P2.19 Series-parallel combination of three
2-20. An assembly of three springs connected in series as shown in Fig. P2.20 has an equivalent stiffness k given by k=
k1 k2 k3 . k2 k3 + k1 k3 + k1 k2
(a) Suppose k2 = 6 lb/in., k3 = k1 + 8 lb/in., and the equivalent stiffness is k = 2 lb/in. Substitute these values into equation (2.66) to obtain the following quadratic equation: 4 k21 + 8 k1 − 96 = 0.
(b) Solve equation (2.67) for k1 by each of the following methods: (i) factoring, (ii) quadratic formula, and (iii) completing the square. For each case, determine the value of k3 corresponding to the only physical solution for k1 . k1
If k1 = k2 + 2 lb/in and the equivalent stiffness is k = 1.75 lb/in, find k1 and k2 as follows: (a) Substitute the values of k and k1 into equation (2.65) and obtain the quadratic equation for k2 . (b) Using the method of your choice, solve the quadratic equation
Figure P2.20 Series
combination of three springs.
Problems 2-21. Consider a capacitor C and an inductor L connected in parallel, as shown in Fig. P2.21. The total reactance X in ohms is 𝜔L , where 𝜔 is the given by X = 1 − 𝜔2 LC angular frequency in rad/s. (a) Suppose L = 1.0 mH and C = 1 F. If the total reactance is X = 1.0 Ω, show that the angular frequency 𝜔 satisfies the quadratic equation 𝜔2 + 𝜔 − 1000 = 0. (b) Solve the quadratic equation for 𝜔 by the methods of completing the square and the quadratic formula.
(a) Suppose R1 = 100, Ra = Rb = R, and Rc = 100 + R, all measured in ohms. Substitute these values into equation (2.68) to obtain the following quadratic equation for R. R2 − 300 R − 10, 000 = 0 (b) Solve the quadratic equation for R by the methods of completing the square and the quadratic formula. a
Rc Figure P2.24 Delta to Y conversion
Figure P2.21 Parallel connection
2-22. Assume that the total reactance in problem P2-21 is X = −1 Ω. Show that the angular frequency 𝜔 satisfies the quadratic equation 𝜔2 − 𝜔 − 1000 = 0, and find the value of 𝜔 using both the quadratic formula and completing the square. 2-23. If L = 0.5 H, C = 0.005 F, and the total reactance in problem P2-21 is X = Ω, show that the angular frequency − 20 3 𝜔 satisfies the quadratic equation 𝜔2 − 30 𝜔 − 400 = 0. Also, find the value of 𝜔 using both the quadratic formula and completing the square. 2-24. When converting resistances connected in a Δ formation to a Y formation as shown in Fig. P2.24, the resistance R1 is obtained as R1 =
Ra Rb Ra + Rb + Rc
2-25. When converting resistances connected in a Δ formation to a Y formation as shown in Fig. P2.24, the resistance R1 is given by equation (2.68). (a) Suppose R1 = 6, Ra = R, Rb = 3 R, and Rc = 100 + R, all measured in ohms. Substitute these values into equation (2.68) to obtain the following quadratic equation for R. R2 − 10 R − 200 = 0 (b) Solve the quadratic equation for R by the methods of completing the square and the quadratic formula. 2-26. When converting resistances connected in a Δ formation to a Y formation as shown in Fig. P2.24, the resistance R2 is obtained as R2 =
Ra Rc Ra + Rb + Rc
Chapter 2 Quadratic Equations in Engineering (a) Suppose R2 = 12, Ra = R, Rb = 3 R, and Rc = 100 + R, all measured in ohms. Substitute these values into equation (2.69) to obtain the following quadratic equation for R.
R + 40 R − 1200 = 0 (b) Solve the quadratic equation for R by the methods of completing the square and the quadratic formula. 2-27. When converting resistances connected in a Δ formation to a Y formation as shown in Fig. P2.24, the resistance R3 is obtained as R3 =
Rb Rc Ra + Rb + Rc
(a) Suppose R3 = 36, Ra = R, Rb = 3 R, and Rc = 100 + R, all measured in ohms. Substitute these values into equation (2.70) to obtain the following quadratic equation for R. R + 40 R − 1200 = 0 (b) Solve the quadratic equation for R by the methods of completing the square and the quadratic formula. 2-28. The characteristic equation of a series RLC circuit shown in Fig. P2.28 is given as R 1 s+ = 0. L LC
for problem P2-28.
2-29. The characteristic equation of a parallel RLC circuit shown in Fig. P2.29 is given as 1 1 s+ = 0. (2.72) s2 + RC LC If R = 200 Ω, L = 50 mH, and C = 0.2 𝜇F, solve the quadratic equation (2.72) for the values of s by the methods of completing the square and the quadratic formula.
Figure P2.28 Series RLC circuit
(a) If R = 7 Ω, L = 1 H, and C = 0.1 F, solve the quadratic equation (2.71) for the values of s (called the eigenvalues of the system) using the methods of completing the square and the quadratic formula. (b) Repeat part (a) if R = 10 Ω, L = 1 1 F. H, and C = 25
− Figure P2.29 Parallel RLC circuit for
2-30. The characteristic equation of a mass, spring, and damper system shown in Fig. P2.30 is given by m s2 + c s + k = 0.
(a) If m = 1 kg, c = 3 Ns/m, and k = 2 N/m, solve the quadratic equation (2.73) for the values of s using the methods of completing the square and the quadratic formula. (b) Repeat part (a) if m = 1 kg, c = 2 Ns/m, and k = 1 N/m.
Problems x (displacement) k m
c Figure P2.30 Mass, spring, and damper system for
2-31. The perimeter of a rectangle shown in Fig. P2.31 is given by ) ( A +L . (2.74) P=2 L
Area = 36 m2 Perimeter = 30 m L
Figure P2.31 A rectangle of length L
If the perimeter P = 30 m and the area A = W × L = 36 m2 , find the length L and width W as follows: (a) Substitute the values of P and A in equation (2.74) and obtain the quadratic equation for L.
(b) Solve the quadratic equation for L obtained in part (a) by completing the square, and the quadratic formula. Also, compute the corresponding values of W. 2-32. A diver jumps off a diving board 2.0 m above the water with an initial vertical velocity of 0.981 m/s as shown in Fig. P2.32. The height h(t) above the water is given by h(t) = −4.905 t2 + 0.981t + 2.0 m. (a) Find the time in seconds when the diver hits the water. Use both the quadratic formula and completing the square. (b) Find the maximum height of the diver if it is known to occur at t = 0.1 s. (c) Use the results of parts (a) and (b) to sketch the height h(t) of the diver. 2-33. A level pipeline is required to pass through a hill having a parabolic profile y = − 0.005 x2 + 0.35 x.
The origin of the x- and y-coordinates is fixed at elevation zero near the base of the hill, as shown in Fig. P2.33.
vo = 0.981 m/s h(t) = 2.0 2.0 m h(t) = 0
Figure P2.32 Diver jumping off a diving board.
Chapter 2 Quadratic Equations in Engineering
fraction porosity P by
Hill profile Tunnel exit Pipeline path B
E = 304 (1 − 1.9 P + 0.9 P2 ).
Figure P2.33 Pipeline path through a parabolic hill.
(a) Write the quadratic equation for a pipeline elevation of y = 3 m. (b) Solve the quadratic equation found in part (a) to determine the positions of the tunnel entry xA and exit xB using both the quadratic formula and completing the square. (c) Find the length of the tunnel. 2-34. A research group is using a drop test to measure the force of attenuation of a helmet liner they designed to reduce the occurance of brain injuries for soldiers and athletes. The helmet attached to a weight is propelled downward with an initial velocity vi of 3 m/s from an initial height of 30 m. The behavior of the falling helmet is characterized by a quadratic equation h(t) = 30 − 3 t − 4.9 t2 . (a) Write the quadratic equation for time t when the helmet and weight hit the ground, i.e., h(t) = 0 m. (b) Solve the quadratic equation for t obtained in part (a) by completing the square, and the quadratic formula. 2-35. The modulus of elasticity (E) is a measure of a material’s resistance to deformation; the larger the modulus, the stiffer the material. During fabrication of a ceramic material from a powder form, pores were generated that affect the strength of the material. The magnitude of elasticity is related to volume
(a) If a porous sample of silicon nitride has a modulus of elasticity of E = 209 GPa, obtain the quadratic equation for volume fraction porosity P. (b) Solve the quadratic equation found in part (a) using both the methods of completing the square and the quadratic formula. (c) Repeat parts (a) and (b) to find the volume fraction porosity of a silicon nitride sample with a modulus of elasticity of 25 GPa. 2-36. Consider the following reaction having an equilibrium constant of 4.66 × 10−3 at a certain temperature: A(g) + B(g) ⇌ C(g) If 0.300 mol of A and 0.100 mol of B are mixed in a container and allowed to reach equilibrium, the concentrations of A = 0.300 − x and B = 0.100 − x reaction that form the concentration of C = 2x are related to the equilibrium constant by the expression 4.66 × 10−3 =
(2 x)2 . (0.300 − x) (0.100 − x)
(a) Write the quadratic equation for x. (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula. (c) Find the equilibrium concentration A, B, and C. 2-37. Consider the following reaction having an equilibrium constant of 1.00 at 1105◦ K temperature: CO(g) + H2 O(g) ⇌ CO2 (g) + H2 (g) Suppose the feed to a reactor contains 1.000 mol of CO and 2.000 mol
Problems of H2 O(g), and the reaction mixture comes to equilibrium at 1105 K. The concentrations of CO = 1.000 − x and H2 O = 2.000 − x reaction that form the concentration of CO2 = x and H2 = x are related to the equilibrium constant as 2x 1.00 = (1.000 − x) (2.000 − x) (a) Write the quadratic equation for x. (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula. (c) Find the equilibrium concentration of CO, H2 O, CO2 , and H2 . 2-38. An engineering co-op wants to hire an asphalt contractor to widen the truck entrance to the corporate headquarters as shown in Fig. P2.38.
(a) If A = 200 sq. ft., obtain the quadratic equation for b. (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula. (c) Find the dimensions of the width and height. 2-39. Repeat problem P2-38 if the total area of the new asphalt needs to be 300 sq. ft. 2-40. A city wants to hire a contractor to build a walkway around the swimming pool in one of its parks. The dimensions of the walkway along with the dimensions of the pool are shown in Fig. P2.40. The area of the walkway is given by A = (50 + 5 x)(30 + 2 x) − 1500 (a) If A = 4500 sq. ft., obtain the quadratic equation for x. (b) Solve the quadratic equation found in part (a) by completing the square and the quadratic formula.
corporate headquarters driveway. Walkway
The height h is required to be 10 ft. more than the width b, and the total area of the new asphalt is given by 1 A = b(10 + b). 2
Figure P2.38 New asphalt dimensions of the
Figure P2.40 Walkway around the swimming pool.
Trigonometry in Engineering
INTRODUCTION In this chapter, the direct (forward) and inverse (reverse) kinematics of one-link and two-link planar robots are considered to explain the trigonometric functions and their identities. Kinematics is the branch of mechanics that studies the motion of an object. The direct or forward kinematics is the static geometric problem of determining the position and orientation of the end-effector (hand) of the robot from the knowledge of the joint displacement. In general, the joint displacement can be linear or rotational (angular). But in this chapter, only rotational motion is considered. Furthermore, it is assumed that the planar robot is wristless (i.e., it has no end-effector or hand) and that only the position but not the orientation of the tip of the robot can be changed. Going in the other direction, the inverse or reverse kinematics is the problem of determining all possible joint variables (angles) that lead to the given Cartesian position and orientation of the end-effector. Since no end-effector is considered in this chapter, the inverse kinematics will determine the joint angle(s) from the Cartesian position of the tip.
ONE-LINK PLANAR ROBOT Consider a one-link planar robot of length l (Fig. 3.1) that is being rotated in the x-y plane by a motor mounted at the center of the table, which is also the location of the robot’s joint. The robot has a position sensor installed at the joint that gives the value of the angle 𝜃 of the robot measured from the positive x-axis. The angle 𝜃 is positive in the counterclockwise direction (0◦ to 180◦ ) and it is negative in the clockwise direction (0◦ to −180◦ ). Therefore, as the joint rotates from 0◦ to 180◦ and 0◦ to −180◦ , the tip of the robot moves on a circle of radius l (the length of the link of the robot) as shown in Fig. 3.2. Note that 180◦ = 𝜋 radians. 3.2.1 Kinematics of One-Link Robot In Fig. 3.2, the point P (tip of the robot) can be represented in rectangular or Cartesian coordinates by a pair (x, y) or in polar coordinates by the pair (l, 𝜃). Assuming
3.2 One-Link Planar Robot
Figure 3.1 One-link planar robot.
that the length of the link l is fixed, a change in the angle 𝜃 of the robot changes the position of the tip of the robot. This is known as the direct or forward kinematics of the robot. The position of the tip of the robot (x, y) in terms of l and 𝜃 can be found using the right-angle triangle OAP in Fig. 3.2 as
Adjacent side x = Hypotenuse l
Opposite side y = Hypotenuse l
y Opposite side Hypotenuse B y
(counterclockwise direction) 0 ≤ θ ≤ 180
x Negative θ (clockwise direction) −180 ≤ θ ≤ 0
Adjacent side Figure 3.2 Circular path of the one-link robot tip.
Chapter 3 Trigonometry in Engineering
Use the one-link robot to find the values of cos(𝜃) and sin(𝜃) for 𝜃 = 0◦ , 90◦ , −90◦ , and 180◦ . Also, find the values of x and y.
By inspection, x = l cos(0◦ ) = l ⇒ cos(0◦ ) = 1 y = l sin(0◦ ) = 0 ⇒ sin(0◦ ) = 0
By inspection, l
x = l cos(90◦ ) = 0 ⇒ cos(90◦ ) = 0 y = l sin(90◦ ) = l ⇒ sin(90◦ ) = 1
By inspection, x = l cos(−90◦ ) = 0 ⇒ cos(−90◦ ) = 0 y = l sin(−90◦ ) = −l ⇒ sin(−90◦ ) = −1
By inspection, x = l cos(180◦ ) = −l ⇒ cos(180◦ ) = −1 ⇒ sin(180◦ ) = 0 y = l sin(180◦ ) = 0
3.2 One-Link Planar Robot Example
Find the position P(x, y) of the robot for 𝜃 = 45◦ , −45◦ , 135◦ , and −135◦ .
l 1 x = l cos(45◦ ) = √ ⇒ cos(45◦ ) = √ 2 2 l 1 y = l sin(45◦ ) = √ ⇒ sin(45◦ ) = √ 2 2
l 1 x = l cos(−45◦ ) = √ ⇒ cos(−45◦ ) = √ 2 2 l 1 ◦ ◦ y = l sin(−45 ) = − √ ⇒ sin(−45 ) = − √ 2 2
1 ⇒ cos(135 ) = − √ 2 2 l 1 y = l sin(135◦ ) = √ ⇒ sin(135◦ ) = √ 2 2 ◦
Case IV: 𝜃 = −135◦ l x = l cos(−135◦ ) = − √
1 ⇒ cos(−135 ) = − √ 2 l y = l sin(−135◦ ) = − √ 2 1 ⇒ sin(−135◦ ) = − √ 2 ◦
Chapter 3 Trigonometry in Engineering Examples 3-1 and 3-2 show that in the first quadrant (0◦ 0 dt dy(t) d2y(t) —— = 0, ——– 0 dt
dy(t) d2y(t) —— = 0, ——– > 0 (minima) dt dt2 dy(t) —— 0 = 0, dt dt2 To test whether the point where the slope (first derivative) of the trajectory of the ball thrown upward is zero is a maximum or a minimum, the student obtains the second derivative of the height as d2 y(t)
d (4.43 − 9.81 t) = −9.81 0. 2 dt2 Therefore, the displacement is maximum at time 𝜋 . tmax = 2𝜔 The maximum displacement can be found by substituting tmax = equation (8.18) as v0 sin(𝜔 tmax ) 𝜔 v 𝜋 = 0 sin 𝜔 2 v0 = 𝜔 v = √ 0 3EI m l3
8.3 Applications of Derivatives in Dynamics
Note: The local maxima or minima of trigonometric functions can also be obtained without derivatives. The plot of the beam displacement given by equation (8.18) is shown in Fig. 8.19. y(t) v0 — ω
Displacement plot to find maximum value.
It can be seen from Fig. 8.19 that the maximum value of the beam displacev v ment is simply the amplitude ymax = 0 = √ 0 and the time where the dis𝜔 3EI placement is maximum is given by 𝜔tmax = or
𝜋 . 2𝜔 (b) The position when the velocity is zero is simply the maximum value √ m l3 . (8.24) ymax = v0 3EI The acceleration found in part (a) is given by a(t) = −v0 𝜔 sin 𝜔t. Therefore, the acceleration when the velocity is zero is given by tmax =
a(tmax ) = −v0 𝜔 sin (𝜔 tmax ) ( ) 𝜋 = −v0 𝜔 sin 2 = −v0 𝜔 √ 3EI . = −v0 m l3
Chapter 8 Derivatives in Engineering Note:
a(t) = −v0 𝜔 sin 𝜔t = −𝜔2
) sin 𝜔t = −𝜔2 y(t). Therefore, the accelera-
tion is maximum when the displacement y(t) is maximum. Since the second derivative of a sinusoid is also a sinusoid of the same frequency (scaled by −𝜔2 ), this is a general result for harmonic motion of any system.
APPLICATIONS OF DERIVATIVES IN ELECTRIC CIRCUITS Derivatives play a very important role in electric circuits. For example, the relationship between voltage and current for both the inductor and the capacitor is a derivative relationship. The relationship between power and energy is also a derivative relationship. Before discussing the applications of derivatives in electric circuits, the relationship between different variables in circuit elements is discussed briefly here. Consider a circuit element as shown in Fig. 8.20, where v(t) is the voltage in volts (V) and i(t) is the current in amperes (A). Note that the current always flows through the circuit element and the voltage is always across the element. i(t)
− Figure 8.20 Voltage and current in a circuit element.
The voltage v(t) is the rate of change of electric potential energy w(t) (in joules (J)) per unit charge q(t) (in coulomb (C)), that is, the voltage is the derivative of the electric potential energy with respect to charge, written as v(t) =
The current i(t) is the rate of change (i.e., derivative) of electric charge per unit time (t in s), written as i(t) =
The power p(t) (in watts (W)) is the rate of change (i.e., derivative) of electric energy per unit time, written as dw(t)
W. dt Note that the power can be written as the product of voltage and current using the chain rule of derivatives: p(t) =
dw(t) dw dq = × , dt dq dt
8.4 Applications of Derivatives in Electric Circuits
The chain rule of derivative is a rule for differentiating composition of functions; for example, if f is a function of g and g is a function of t, then the derivative of composite function f (g(t)) with respect to t can be written as df dg df = × . dt dg dt
For example, the function f (t) = sin(2𝜋t) can be written as f (t) = sin(g(t)), where g(t) = 2𝜋t. By the chain rule of equation (8.26), df df dg = × dt dg dt d d = (sin(g)) × (2𝜋t) dt dt d = cos(g) × (2𝜋t) dt = cos(g) × (2𝜋) = 2𝜋 cos(2𝜋t). The chain rule is also useful in differentiating the power of sinusoidal function such as y1 (t) = sin2 (2𝜋t) or the power of the polynomial function such as y2 (t) = (2t + 10)2 . The derivatives of these functions are obtained as ) dy1 d ( (sin(2𝜋t))2 = dt dt d = 2 (sin(2𝜋t))1 × (sin(2𝜋t)) dt = 2 sin(2𝜋t) × (2𝜋 cos(2𝜋t)) = 4𝜋 sin(2𝜋t) cos(2𝜋t) and ) dy2 d ( (2t + 10)2 = dt dt d = 2 (2t + 10)1 × (2t + 10) dt = 2 (2t + 10) × (2) = 4 (2t + 10) The following example will illustrate some of the derivative relationships discussed above.
Chapter 8 Derivatives in Engineering
For a particular circuit element, the charge is 1 sin 250 𝜋t C 50 and the voltage supplied by the voltage source shown in Fig. 8.21 is q(t) =
v(t) = 100 sin 250 𝜋t V.
Voltage applied to a particular circuit element.
Find the following quantities: (a) The current, i(t). (b) The power, p(t). (c) The maximum power pmax delivered to the circuit element by the voltage source. Solution
(a) Current: The current i(t) can be determined by differentiating the charge q(t) as dq(t) i(t) = dt ( ) d 1 = sin 250 𝜋t dt 50 =
1 (250 𝜋 cos 250 𝜋t) 50
or i(t) = 5 𝜋 cos 250 𝜋t A.
(b) Power: The power p(t) can be determined by multiplying the voltage given in equation (8.28) and the current calculated in equation (8.29) as p(t) = v(t) i(t) = (100 sin 250 𝜋t) (5 𝜋 cos 250 𝜋t) = 500 𝜋(sin 250 𝜋t) (cos 250 𝜋t)
8.4 Applications of Derivatives in Electric Circuits
or p(t) = 250 𝜋sin 500 𝜋t W.
The power p(t) = 250 𝜋 sin 500 𝜋t W in equation (8.30) is obtained by using the double-angle trigonometric identity sin(2 𝜃) = 2 sin 𝜃 cos 𝜃 or sin(500 𝜋t) , which gives p(t) = 250 𝜋 sin 500 𝜋t. (sin 250 𝜋t) (cos 250 𝜋t) = 2 (c) Maximum Power Delivered to the Circuit: As discussed in Section 8.3, the maximum value of a trigonometric function such as p(t) = 250 𝜋(sin 500 𝜋t) can be found without differentiating the function and equating the result to zero. Since −1 ≤ sin 500 𝜋t ≤ 1, the power delivered to the circuit element is maximum when sin 500 𝜋t = 1. Therefore, pmax = 250 𝜋W, which is simply the amplitude of the power.
8.4.1 Current and Voltage in an Inductor The current-voltage relationship for an inductor element (Fig. 8.22) is given by v(t) = L
where v(t) is the voltage across the inductor in V, i(t) is the current flowing through the inductor in A, and L is the inductance of the inductor in henry (H). Note that if the inductance is given in mH (1 mH (millihenry) = 10−3 H), it must be converted to H before using it in equation (8.31).
Circuit element Figure 8.22 Inductor as a circuit element.
For the inductor shown in Fig. 8.22, if L = 100 mH and i(t) = t e−3t A. di(t) . dt (b) Find the value of the current when the voltage is zero. (c) Use the results of (a) and (b) to sketch the current i(t). (a) Find the voltage v(t) = L
Chapter 8 Derivatives in Engineering
(a) The voltage v(t)is determined as v(t) = L
d where dt = dt (t e−3t ). To differentiate the product of two functions (t and −3t e ), the product rule of differentiation (Table 8.3) is used:
d d d (f (t) g(t)) = f (t) (g(t)) + g(t) (f (t)). dt dt dt
Substituting f (t) = t and g(t) = e−3 t in equation (8.32) gives ( ) ( ) ( −3t ) d −3t d d ( −3t ) te e = (t) (e ) + (t) dt dt dt ) ( ) ( = (t) −3 e−3t + (1) e−3t = e−3t (−3 t + 1) .
Therefore, v(t) = 0.1 e−3t (−3 t + 1) V.
(b) To find the current when the voltage is zero, first find the time t when the voltage is zero and then substitute this time in the expression for current. Setting equation (8.34) equal to zero gives 0.1 e−3t (−3 t + 1) = 0. Since e−3t is never zero, it follows that (−3 t + 1) = 0, which gives t = 1∕3 sec. Therefore, the value of the current when the voltage is zero is determined by substituting t = 1∕3 into the current i(t) = t e−3 t , which gives ( ) ( ) ( ) −3 13 1 1 = e i 3 3 =
or i = 0.123 A. (c) Since the voltage is proportional to the derivative of the current, the slope of the current is zero when the voltage is zero. Therefore, the current i(t) is maximum (imax = 0.123 A) at t = 1∕3 s. Also, at t = 0, i(0) = 0 A. Using these values along with the values of the current at t = 1 s (i(1s) = 0.0498 A) and at t = 2 s (i(2s) = 0.00496 A), the approximate sketch of the current can be drawn
8.4 Applications of Derivatives in Electric Circuits
as shown in Fig. 8.23. Note that i = 0.123 A must be a maximum (as opposed to a minimum) value, since it is the only location of zero slope and is greater than the values of i(t) at either t = 0 or t = 2 s. Hence, no second derivative test is required! i(t), A 0.14 imax = 0.123
0.12 0.1 0.08 0.06 0.04 0.02 0 0
Approximate sketch of the current waveform for Example 8-7.
As seen in dynamics, derivatives are frequently used in circuits to sketch functions for which no equations are given, as illustrated in the following example:
For the given input voltage (square wave) shown in Fig. 8.24, plot the current i(t) and the power p(t) if L = 500 mH. Assume i(0) = 0 A and p(0) = 0 W. v(t), V i(t)
−9 Figure 8.24 A square wave voltage applied to an inductor.
Chapter 8 Derivatives in Engineering
di(t) For an inductor, the current-voltage relationship is given by v(t) = L . Since dt the voltage is known, the rate of change of the current is given by (500 × 10−3 )
di(t) = v(t) dt di(t) 1 = v(t) dt 0.5
or di(t) = 2 v(t). dt Therefore, the slope of the current is twice the applied voltage. Since v(t) = ± 9 V (constant in each interval), the current waveform has a constant slope of ± 18 A/s, in other words, the current waveform is a straight line with a constant slope of ± 18 A/s. In the interval 0 ≤ t ≤ 2, the current waveform is a straight line with a slope of 18 A/s that starts at 0 A (i(0) = 0 A). Therefore, the value of the current at t = 2 s is 36 A. In the interval, 2 0; therefore, 𝜎𝜃 has a minimum 2 d𝜃 value at 𝜃 = 90◦ .
Maximum Value of 𝝈𝜽 :
Substituting 𝜃 = 0◦ in equation (8.52) gives 𝜎max = 𝜎 cos2 (0◦ ) = 𝜎.
This means that the largest normal stress during axial loading is simply the applied stress 𝜎!
Chapter 8 Derivatives in Engineering Value of 𝜽 where 𝝉𝜽 is maximum: (a) First, find the derivative of 𝝉𝜽 with respect to 𝜽: The derivative of 𝜏𝜃 given by equation (8.53) is given by d𝜏𝜃 d = (𝜎 sin 𝜃 cos 𝜃) d𝜃 d𝜃 d =𝜎 (sin 𝜃 cos 𝜃) d𝜃 ( ) sin 2 𝜃 d =𝜎 d𝜃 2 𝜎 = (2 cos 2 𝜃) 2
(b) Next, equate the derivative in equation (8.55) to zero and solve the resulting equation 𝜎 cos 2 𝜃 = 0 for the value of 𝜃 (between 0 and 90◦ ) where 𝜏𝜃 is maximum: cos 2 𝜃 = 0
Therefore, 𝜏𝜃 has a local maximum or minimum at 𝜃 = 45◦ . To find whether 𝜏𝜃 has a maximum or minimum at 𝜃 = 45◦ , the second derivative test is performed. The second derivative of 𝜏𝜃 is given by d2 𝜏𝜃 d𝜃 2
d (𝜎 cos 2 𝜃) d𝜃 = 𝜎 (−sin 2 𝜃)(2) =
= −2 𝜎 sin 2 𝜃 For 0 ≤ 𝜃 ≤ 90◦
sin 2 𝜃 > 0, therefore,
d𝜃 2 0. Since the second derivative is negative, 𝜏𝜃 has a maximum value at 𝜃 = 45◦ .
6 s: v(t) = 0 m/s (b) 0 ≤ t ≤ 2 s: quadratic with increasing slope/concave up
P8-15 (a) 0 ≤ t ≤ 2 s: a(t) = −4 m/s2 2 8 s: v(t) = 12 m/s (b) 0 ≤ t ≤ 4 s: quadratic with decreasing slope/concave down Slope = 0 at t = 4 s 1 x(4) = 0 + × 4 × 12 = 24 m 2 4 5 ms: q(t) = 2 mC
4 16 s: v(t) = 0 m/s
ANSWERS TO SELECTED PROBLEMS P9-21
v(0) = 0 V, v(1) = −10 V and v(2) = 0 V
0 ≤ t ≤ 2 s: quadratic with increasing slope/concave up x(t) = 5 t2 , x(2) = 20 m
0 ≤ t ≤ 2 s: quadratic with increasing slope/concave up
2 4 s Since i(t) is constant over each interval, v(t) is linear between each interval
0 ≤ t ≤ 3 s: v(t) = 10 t m/s 3 12 s: v(t) = 30 m/s
0 ≤ t ≤ 0.04 s: xunbelted (t) = 0.509 sin(12.5 𝜋 t) m
xunbelted (0.04) − xunbelted (0.04) = 0.407 m p̂ x
(b) w(t) = 7.5 + 2.5 cos(20 t)− 10 cos(10 t) mJ
P9-29 (a) vo (t) = sin(100 t) − 100 t V
(b) w(t) = 7.5 + 2.5 cos(20 t)− 10 cos(10 t) mJ ) ( P9-31 (a) vo (t) = 10 1 − e−10 t + 10 V ( ) (b) w(t) = 0.01 1 − e−20 t + 0.005 J
htran (t) = C e− A t Q
)× A2 𝜔2 + k2 (k sin𝜔 t − A 𝜔 cos 𝜔 t) k Q (A 𝜔 e− A t 2 +k +k sin𝜔 t − A 𝜔 cos 𝜔 t)
ANSWERS TO SELECTED PROBLEMS R
vtran (t) = C e− RC t V
(c) i(t) = 50 e− L t mA
v(t) = 10 e− RC t V
(b) iss (t) = 0 A − 12
(c) i(t) = 50 e−10,000 t mA
P10-17 (a) itran (t) = C e−10,000 t A, 𝜏 = 0.1 ms
P10-19 (a) vtran (t) = C e−100 t V (b) vss (t) = 10 V (c) v(t) = 10(1 − e−100 t ) V
vtran (t) = C e− 20 t V
v(t) = 10 − 5 e− 20 t V
vtran (t) = A e− RC t V, 𝜏 = RC s
P10-11 (a) vtran (t) = C e−2.5 t V, 𝜏 = 0.4 s
P10-13 (a) vtran (t) = C e−100 t V
(b) iss (t) = 0.5 A (c) i(t) = −1.5 e−50 t + 0.5 A
(c) vo (t) = −13.33 e−5 t +13.33 cos(10 t)+ 6.67 sin(10 t) V
(b) vss (t) = 10 V ) ( (c) v(t) = 10 1 − e−100 t V
(d) 𝜏 = 0.01 s v(0.01) = 6.321 V v(0.02) = 8.647 V v(0.04) = 9.817 V v(∞) = 10 V
P10-21 (a) itran (t) = C e−50 t A, 𝜏 = 0.02 s
(b) vo,ss (t) = 13.33 cos(10 t) + 6.67 sin(10 t) V
(c) v(t) = 10 sin(0.01t) V
P10-23 (a) vo,tran (t) = C e−5 t V, 𝜏 = 0.2 s
(b) vss (t) = 10 sin(0.01t) V
P10-15 (a) itran (t) = C e− L t A, 𝜏 =
(e) 𝜏 = 0.01 s v(0.01) = 6.32 V v(0.02) = 8.65 V v(0.04) = 9.82 V
P10-25 (a) Ptran (t) = A e− RC t mmHg (b) Pss (t) = 60 R mmHg 1
(c) P(t) = 240 − 234 e− 8 t mmHg √ P10-27 (a) ytran (t) = C3 cos( 10 √ t)+ C4 sin( 10 t) m
ANSWERS TO SELECTED PROBLEMS (b) yss (t) = 0 m
√ (c) y(t) = 0.2 cos( 10 t) m P10-29 (a) ytran (t) = C3 cos(4 t)+ C4 sin(4 t) m (b) yss (t) = 3∕16 m (c) y(t) = 0.028 cos(4 t − 1.107)+ 3∕16 m √ P10-31 (a) xtran (t) = C3 cos( 40 √ t)+ C4 sin( 40 t) m √ 𝜔n = 40 rad/s or f = 1.01 Hz
v(t) = 10 (1 − cos(500 t)) V
vtran (t) = C3 cos(250 t)+ C4 sin(250 t) V 𝜔n = 250 rad/s, f = 125∕𝜋 Hz
vss (t) = 5.1 sin(245 t) V
v(t) = 5 sin(250 t)+ 5.1 sin(245 t) V (√ ) 3k 𝜃tran (t) = C3 cos t + m (√ ) 3k t rad C4 sin m √ 3k 𝜔n = rad/s m F cos(𝜔 t) rad 𝜃ss (t) = 1 kl − 3 m l 𝜔2 ( ) 𝜋 3k 𝜃(t) = cos t rad 18 m
P10-33 (a) ytran (t) = C3 cos(0.61 t)+ C4 sin(0.61 t) m 𝜔n = 0.61 rad/s (b) yss (t) = 0.5 cos(0.5 t) m (c) ymin = −0.5 m at t = 2 𝜋 s (√ ) √ m k P10-35 (a) x(t) = Vo cos t m k m (c) T, F, T, T, F P10-37 itran (t) = C3 cos(100 t)+ C4 sin(100 t) A iss (t) = 0 A i(t) = 0.1 sin(100 t) A
vtran (t) = C3 cos(500 t)+ C4 sin(500 t) V 𝜔n = 500 rad/s, f = 250∕𝜋 Hz
(b) xss (t) = 0.245 m √ (c) x(t) = −0.245 cos( 40 t)+ √ √ 0.7 h sin( 40 t) + 0.245 m
Acceleration, 221, 226–240 Acceleration due to gravity, 221 Addition of sinusoids of same frequency, 166–173, See also under Sinusoids Addition of two complex numbers, 137 Angular frequency, sinusoids, 161 Angular motion of one-link planar robot, 159–162 frequency and period, relations between, 160–162 Antiderivative, 280 Armature current in DC motor, 140–141 Asphalt problem, 278–283 Average velocity, 219 Bending moment, 250 Capacitor, 314–321 current passing through, 247–250, 314–321 energy stored in, 314–321 impedance of, 135 voltage across, 247–250, 314–321 Center of Gravity (Centroid), 286–293 alternate definition of, 293–296 determination using vertical rectangles, 292 evaluation using horizontal elemental areas, 89 first moment of area, 294 of triangular section, 86 Chain rule of derivative, 241 Complementary solution, 347 Complex conjugate, 145–147 Complex numbers, 132–156 addition of two complex numbers, 137 armature current in a DC motor, 140–141 complex conjugate, 145–147 division of complex number in polar form, 139–140 examples, 141–145 exponential form, 136 impedance of a series RLC circuit, 136–137 impedance of R, L, and C as a, 134–135
multiplication of complex number in polar form, 139 polar form, 136 position of one-link robot as, 133–134 subtraction of two complex numbers, 137 Cos(𝜃), 61 Cramer’s Rule, 188–189, 192 Current, 240 flowing in inductor, 322–326 passing through a capacitor, 314–321 Definite integral, 280 Deflection, 251 Derivatives, 218–277 applications in dynamics, 225–240 applications in electric circuits, 240–250 applications in strength of materials, 250–260, See also Material strength chain rule of, 241 definition, 218–221 examples of, 261–266 maxima and minima, 221–225 Differential equations (DEQ), 345–400 first-order differential equations, 348–374 leaking bucket, 345–346 linear DEQ, solution with constant coefficients, 347–348 second-order differential equations, 374–390, See also separate entry Direct kinematics of two-link robot, 73–74 Distributed loads, 251, 296–302 on beams, statically equivalent loading, 297–302 hydrostatic pressure on a retaining wall, 296–297 Division of complex number in polar form, 139–140 Dynamics derivatives applications in, 225–240 integrals applications in, 302–314 Elbow-up solution for 𝜃1 , 77–79 Electric circuits, 314–321
INDEX Electric circuits (continued ) current and voltage in a capacitor, 314–321 derivatives applications in, 240–250 integrals applications in, 314–321 Energy stored in a capacitor, 314–321 Equivalent resistance, 37–38 Euler’s formula, 134 Exponential form, complex numbers, 136 Factoring, 33 First-order differential equations, 348–374 Force-displacement relationship, 6–7 Free-body diagram (FBD), 117, 119 Graphical method solution, for two-loop circuit, 185–186 Homogeneous solution, 347 Hydrostatic pressure on a retaining wall, 296–297 Impedance of capacitor, 135 of inductor, 134–135, 137–140 of resistor, 134, 137–140 of a series RLC circuit, 136–137 Indefinite integrals, 281 Inductor as a circuit element, 243 current and voltage in, 243–247, 322–326 impedance of, 134–135 Integrals, 278–344, See also Distributed loads Asphalt problem, 278–283 concept of work, 283–286, See also Work definite integral, 280 in dynamics, 302–314 in electric circuits, 314–321 examples, 327–333 indefinite integrals, 281 inductor, current and voltage in, 322–326 inverse operations, 281 in statics, 286–296, See also Statics Inverse kinematics of one-link robot, 68–72 Inverse kinematics of two-link robot, 75–79 Inverse operations, 281 Kirchhoff’s voltage law (KVL), 3, 184 Linear DEQ, solution with constant coefficients, 347–348 steady-state solution, 348 transient solution, 347 Linear frequency, sinusoids, 161 Low-pass filter, 373
Material strength, derivatives applications in, 250–260 bending moment, 250 deflection, 251 distributed load, 251 maximum stress under axial loading, 256–260 maximum value, 259 minimum value, 259 moment, 255 shear force, 251, 255 slope, 252 Matrix algebra method solution tension in cables, 191–192 for two-loop circuit, 186–188 Maxima, 221–225 Maximum stress under axial loading, 256–260 inclined section of the rectangular bar, 256 normal and shear stresses acting on the inclined cross section, 258 rectangular bar under axial loading, 256 Method of undetermined coefficients, 348 Minima, 221–225 Moment, 255, 298 Multiplication of complex number in polar form, 139 Newton’s first law, 117 Newton’s second law, 375 Ohm’s law, 3, 184 One-link planar robot angular motion of, 159–162 as a sinusoid, 157–159 Particular solution, 348 Period, one-link planar robot, 161 Phase angle, 162–164 Phase shift, 162–164 Polar form complex numbers, 136 position vector in, 107–110 Position, 226–240 Position of one-link robot as a complex number, 133–134 Position vector in polar form, 107–110 in rectangular form, 107 Power, 240 Quadratic equations, 32–59 current in a lamp, 36–37 equivalent resistance, 37–38 examples, 38–50
INDEX pipeline through parabolic hill, 48–50 projectile in a vertical plane, 32–36 resistors in parallel, 37 Quadratic formula, 33 Rectangular form, position vector in, 107 Reference angle, 65 Relative velocity, 114 Repeated roots, 347 Resistive circuit, voltage-current relationship in, 3–5 Resistor, impedance of, 134 Resistors in parallel, 37 Resonance, 383 Second-order differential equations, 374–390 forced vibration of a spring-mass system, 379–386 free vibration of a spring-mass system, 374–379 second-order LC circuit, 386–390 Second-order LC circuit, 386–390 Shear force, 251, 255 Sin(𝜃), 61 Sinusoids, 157–183 addition of sinusoids of same frequency, 166–173 amplitude of, 160 general form of, 164–166 one-link planar robot as, 157–159 phase angle, 162–164 phase shift, 162–164 sine function, two cycles of, 159 time shift, 162–164 Slope, 252 Slope-intercept method, 5 Speed at impact, 219 Spring-mass system forced vibration of, 379–386 free vibration of, 374–379 Static equilibrium, 117, 119 Statically equivalent loading, 297–302 Statics, integrals application in, 286–296 Center of Gravity (Centroid), 286–293, See also individual entry Steady-state solution, 348 Straight lines, 1–31 examples, 8–18 force-displacement relationship, 6–7
slope-intercept method, 5 vehicle during braking, 1–3 voltage-current relationship in resistive circuit, 3–5 Substitution method solution tension in cables, 190–191 for two-loop circuit, 185 Subtraction of two complex numbers, 137 Systems of equations, 184–217 examples, 193–205 solution of a two-loop circuit, 184–189, See also Tension in cables Tension in cables, 190–192 Cramer’s Rule, 192 matrix algebra method, 191–192 substitution method, 190–191 Time constant, 351, 357 Time shift, 162–164 Transient solution, 347 Triangular section, centroid of, 86 Trigonometry, 60–105 examples, 89–96 one-link planar robot, 60–72 reference angle, 65 sine and cosine functions, values of, 64 trigonometric functions in four quadrants, 64 two-link planar robot, 72–89 Two-dimensional vectors, 106–131 free-body diagram (FBD), 117, 119 position vector relative velocity, 114 static equilibrium, 117, 119 vector addition, 110–123 Two-loop circuit, solution of, 184–189 Vector addition, 110–123 Velocity, 226–240 Voltage, 240 in capacitor, 314–321 in inductor, 322–326 Voltage-current relationship in resistive circuit, 3–5 Work, 283–286 as area under a constant force curve, 284 as area under a variable force curve, 284
Useful Mathematical Relations Algebra and Geometry Arithmetic Operations a(b + c) = a c + = b d a+c = b
ab + ac ad + bc bd a c + b b
a b = a×d c b c d ad = bc
Exponents and Radicals ( )n xn x = n y y √ 1∕n x = nx √ n xm xn = xm+n xm∕n = xm m ( )m √ x m−n n = x = x xn √ √ √ (xm )n = = xm n n xy = n x n y 1 √ √ n x−n = n x x x n = √ n n n n y (x y) = x y y Factoring Special Polynomials x2 − y2 = (x + y) (x − y) x3 + y3 = (x + y) (x2 − xy + y2 ) x3 − y3 = (x − y) (x2 + xy + y2 ) Quadratic Formula
Slope of line through points P1 (x1 , y1 ) and P2 (x2 , y2 ): y − y1 m= 2 x2 − x1 Point-slope equation of line through point P1 (x1 , y1 ) with slope m: y − y1 = m (x − x1 ) Point-slope equation of line through point P2 (x2 , y2 ) with slope m: y − y2 = m (x − x2 ) Distance Formula Distance between points P1 (x1 , y1 ) and P2 (x2 , y2 ): √ d = (x2 − x1 )2 + (y2 − y1 )2 Area of a Triangle: 1 bh 2 1 = a b sin(𝜃) 2
If a x2 + b x + c = 0. then √ −b ± b2 − 4 a c x= 2a Lines
Equation, Area and Circumference of Circle:
Equation of line with slope m and yintercept b:
(x − k)2 + (y − k)2 = r2 A = 𝜋 r2 C = 2𝜋 r