Engineering mechanics statics 5th ed solution manual

SOLUTION MANUAL FOR TABLE OF CONTENTS Introduction Engineering and Mechanics Learning Mechanics Fundamental Concepts Units Newtonian Gravitation Vectors Vector Operations and Definitions Scalars and Vectors Rules for Manipulating Vectors Cartesian Components Components in Two Dimensions Components in Three Dimensions Products of Vectors Dot Products Cross Products Mixed Triple Products Forces Types of Forces Equilibrium and Free-Body Diagrams Two-Dimensional Force Systems Three-Dimensional Force Systems Systems of Forces and Moments Two-Dimensional Description of the Moment The Moment Vector Moment of a Force About a Line Couples Equivalent Systems Representing Systems by Equivalent Systems Objects in Equilibrium The Equilibrium Equations Two-Dimensional Applications Statically Indeterminate Objects Three-Dimensional Applications Two-Force and Three-Force Structures in Equilibrium Trusses The Method of Joints The Method of Sections Space Trusses Frames and Machines Centroids and Centers of Mass 316 Centroids Centroids of Areas Centroids of Composite Areas Distributed Loads Centroids of Volumes and Lines The Pappus-Guldinus Theorems Centers of Mass Definition of the Center of Mass Centers of Mass of Objects Centers of Mass of Composite Objects Moments of Inertia Areas Definitions Parallel-Axis Theorems Rotated and Principal Axes Masses Simple Objects Parallel-Axis Theorem Friction Theory of Dry Friction Applications 10 Internal Forces and Moments Beams Axial Force, Shear Force, and Bending Moment Shear Force and Bending Moment Diagrams Relations Between Distributed Load, Shear Force, and Bending Moment Cables Loads Distributed Uniformly Along Straight Lines Loads Distributed Uniformly Along Cables Discrete Loads Liquids and Gasses Pressure and the Center of Pressure Pressure in a Stationary Liquid 11 Virtual Work and Potential Energy Virtual Work Potential Energy   Problem 1.1 The value of is 3.14159265 If C is the circumference of a circle and r is its radius, determine the value of r/C to four significant digits Problem 1.2 The base of natural logarithms is e D 2.718281828 (a) (b) (c) Express e to five significant digits Determine the value of e2 to five significant digits Use the value of e you obtained in part (a) to determine the value of e2 to five significant digits Solution: C D r ) r D D 0.159154943 C To four significant digits we have r D 0.1592 C Solution: The value of e is: e D 2.718281828 (a) To five significant figures e D 2.7183 (b) e2 to five significant figures is e2 D 7.3891 (c) Using the value from part (a) we find e2 D 7.3892 which is not correct in the fifth digit [Part (c) demonstrates the hazard of using rounded-off values in calculations.] Problem 1.3 A machinist drills a circular hole in a panel with a nominal radius r D mm The actual radius of the hole is in the range r D š 0.01 mm (a) To what number of significant digits can you express the radius? (b) To what number of significant digits can you express the area of the hole? Solution: a) The radius is in the range r1 D 4.99 mm to r2 D 5.01 mm These numbers are not equal at the level of three significant digits, but they are equal if they are rounded off to two significant digits Two: r D 5.0 mm b) The area of the hole is in the range from A1 D r1 D 78.226 m2 to A2 D r2 D 78.854 m2 These numbers are equal only if rounded to one significant digit: One: A D 80 mm2 Problem 1.4 The opening in the soccer goal is 24 ft wide and ft high, so its area is 24 ft ð ft D 192 ft2 What is its area in m2 to three significant digits? Solution: A D 192 ft2 1m 3.281 ft D 17.8 m2 A D 17.8 m2 Problem 1.5 The Burj Dubai, scheduled for completion in 2008, will be the world’s tallest building with a height of 705 m The area of its ground footprint will be 8000 m2 Convert its height and footprint area to U.S customary units to three significant digits Solution: h D 705 m A D 8000 m2 3.281 ft 1m D 2.31 ð 103 ft 3.218 ft 1m h D 2.31 ð 103 ft, D 8.61 ð 104 ft2 A D 8.61 ð 104 ft2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 1.6 Suppose that you have just purchased a Ferrari F355 coupe and you want to know whether you can use your set of SAE (U.S Customary Units) wrenches to work on it You have wrenches with widths w D 1/4 in, 1/2 in, 3/4 in, and in, and the car has nuts with dimensions n D mm, 10 mm, 15 mm, 20 mm, and 25 mm Defining a wrench to fit if w is no more than 2% larger than n, which of your wrenches can you use? Solution: Convert the metric size n to inches, and compute the percentage difference between the metric sized nut and the SAE wrench The results are: mm inch 25.4 mm D 0.19685 in, 0.19685 0.25 0.19685 D 100 27.0% 10 mm inch 25.4 mm D 0.3937 in, 0.3937 0.5 0.3937 100 D 15 mm inch 25.4 mm D 0.5905 in, 0.5905 0.5 0.5905 100 D C15.3% 20 mm inch 25.4 mm D 0.7874 in, 0.7874 0.75 0.7874 100 D C4.7% 25 mm inch 25.4 mm D 0.9843 in, 0.9843 1.0 0.9843 27.0% n 100 D 1.6% A negative percentage implies that the metric nut is smaller than the SAE wrench; a positive percentage means that the nut is larger then the wrench Thus within the definition of the 2% fit, the in wrench will fit the 25 mm nut The other wrenches cannot be used Problem 1.7 Suppose that the height of Mt Everest is known to be between 29,032 ft and 29,034 ft Based on this information, to how many significant digits can you express the height (a) in feet? (b) in meters? Solution: a) h1 D 29032 ft h2 D 29034 ft The two heights are equal if rounded off to four significant digits The fifth digit is not meaningful Four: h D 29,030 ft b) In meters we have h1 D 29032 ft 1m 3.281 ft D 8848.52 m h2 D 29034 ft 1m 3.281 ft D 8849.13 m These two heights are equal if rounded off to three significant digits The fourth digit is not meaningful Three: h D 8850 m Problem 1.8 The maglev (magnetic levitation) train from Shanghai to the airport at Pudong reaches a speed of 430 km/h Determine its speed (a) in mi/h; (b) ft/s Problem 1.9 In the 2006 Winter Olympics, the men’s 15-km cross-country skiing race was won by Andrus Veerpalu of Estonia in a time of 38 minutes, 1.3 seconds Determine his average speed (the distance traveled divided by the time required) to three significant digits (a) in km/h; (b) in mi/h Solution: a) v D 430 b) v D 430 km h 0.6214 mi km km 1000 m h km v D 392 ft/s Solution: 15 km 1.3 38 C 60 a) vD b) v D 23.7 km/h D 267 mi/h ft 0.3048 m 60 1h mi 1.609 km v D 267 mi/h 1h 3600 s D 23.7 km/h D 14.7 mi/h D 392 ft/s v D 23.7 km/h v D 14.7 mi/h c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 1.10 The Porsche’s engine exerts 229 ft-lb (foot-pounds) of torque at 4600 rpm Determine the value of the torque in N-m (Newton-meters) Solution: Problem 1.11 The kinetic energy of the man in Active Example 1.1 is defined by 12 mv2 , where m is his mass and v is his velocity The man’s mass is 68 kg and he is moving at m/s, so his kinetic energy is 2 2 (68 kg)(6 m/s) D 1224 kg-m /s What is his kinetic energy in U.S Customary units? Solution: T D 229 ft-lb 1m 3.281 ft 1N 0.2248 lb T D 1224 kg-m2 /s2 slug 14.59 kg D 310 N-m ft 0.3048 m Solution: Use Table 1.2 The result is: Problem 1.13 A furlong per fortnight is a facetious unit of velocity, perhaps made up by a student as a satirical comment on the bewildering variety of units engineers must deal with A furlong is 660 ft (1/8 mile) A fortnight is weeks (14 nights) If you walk to class at m/s, what is your speed in furlongs per fortnight to three significant digits? Solution: Solution: 80 mm 120 mm D 20800 mm2 1m A D 20800 mm2 D 0.0208 m2 A D 0.0208 m2 1000 mm A D 200 mm a) b) A D 20800 mm2 in 25.4 mm g D 9.81 v D 12,000 D 903 slug-ft2 /s ft 0.3048 m ft 0.3048 m D 32.185 furlong 660 ft ft s2 3600 s hr D 32.2 24 hr day ft s2 14 day fortnight furlongs fortnight y 40 mm x 120 mm 40 mm 40 mm 200 mm A D 32.2 in Problem 1.15 The cross-sectional area of the C12ð30 American Standard Channel steel beam is A D 8.81 in2 What is its cross-sectional area in mm2 ? m s2 v D m/s D 32.2 in2 T D 903 slug-ft2 /s Problem 1.12 The acceleration due to gravity at sea level in SI units is g D 9.81 m/s2 By converting units, use this value to determine the acceleration due to gravity at sea level in U.S Customary units Problem 1.14 Determine the cross-sectional area of the beam (a) in m2 ; (b) in in2 T D 310 N-m y A Solution: A D 8.81 in2 25.4 mm in D 5680 mm2 x c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 1.16 A pressure transducer measures a value of 300 lb/in2 Determine the value of the pressure in pascals A pascal (Pa) is one newton per meter squared Solution: Convert the units using Table 1.2 and the definition of the Pascal unit The result: lb in2 300 4.448 N lb 12 in ft N m2 D 2.0683 106 ft 0.3048 m D 2.07 106 Pa Problem 1.17 A horsepower is 550 ft-lb/s A watt is N-m/s Determine how many watts are generated by the engines of the passenger jet if they are producing 7000 horsepower Solution: P D 7000 hp 550 ft-lb/s hp 1m 3.281 ft 1N 0.2248 lb D 5.22 ð 106 W 1m 3.281 ft D 27.4 lb/ft w D 27.4 lb/ft P D 5.22 ð 106 W Problem 1.18 Chapter discusses distributed loads that are expressed in units of force per unit length If the value of a distributed load is 400 N/m, what is its value in lb/ft? Problem 1.19 The moment of inertia of the rectangular area about the x axis is given by the equation I D 13 bh3 The dimensions of the area are b D 200 mm and h D 100 mm Determine the value of I to four significant digits in terms of (a) mm4 ; (b) m4 ; (c) in4 Solution: w D 400 N/m 0.2248 lb 1N Solution: 200 mm 100 mm 3 D 66.7 ð 106 mm4 (a) ID (b) I D 66.7 ð 106 mm4 1m 1000 mm (c) I D 66.7 ð 106 mm4 in 25.4 mm y D 66.7 ð 10 m4 D 160 in4 h x b Problem 1.20 In Example 1.3, instead of Einstein’s equation consider the equation L D mc, where the mass m is in kilograms and the velocity of light c is in meters per second (a) What are the SI units of L? (b) If the value of L in SI units is 12, what is its value in U.S Customary base units? Solution: a) L D mc ) b) L D 12 kg-m/s Units L D kg-m/s 0.0685 slug kg 3.281 ft 1m D 2.70 slug-ft/s L D 2.70 slug-ft/s c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 1.21 The equation D Solution: My I (a) is used in the mechanics of materials to determine normal stresses in beams (a) (b) D (N-m)m My N D D I m4 m D 2000 N-m 0.1 m My D I ð 10 m4 (b) When this equation is expressed in terms of SI base units, M is in newton-meters (N-m), y is in meters (m), and I is in meters to the fourth power (m4 ) What are the SI units of ? If M D 2000 N-m, y D 0.1 m, and I D ð 10 m4 , what is the value of in U.S Customary base units? D 59,700 lb 4.448 N 0.3048 m ft lb ft2 Problem 1.22 The acceleration due to gravity on the Solution: a) The mass does not depend on location The mass in kg is surface of the moon is 1.62 m/s2 (a) What would the 14.59 kg mass of the C-clamp in Active Example 1.4 be on the surface mass D 0.397 kg D 0.397 kg 0.0272 slug slug of the moon? (b) What would the weight of the C-clamp in newtons be on the surface of the moon? b) The weight on the surface of the moon is W D mg D 0.397 kg 1.62 m/s2 D 0.643 N Problem 1.23 The ft ð ft ð ft cube of iron weighs 490 lb at sea level Determine the weight in newtons of a m ð m ð m cube of the same material at sea level Solution: The weight density is D The weight of the m3 cube is: WD VD 490 lb ft3 1m W D 0.643N ft 490 lb ft3 ft 0.3048 m 1N 0.2248 lb ft ft D 77.0 kN Problem 1.24 The area of the Pacific Ocean is 64,186,000 square miles and its average depth is 12,925 ft Assume that the weight per unit volume of ocean water is 64 lb/ft3 Determine the mass of the Pacific Ocean (a) in slugs; (b) in kilograms Problem 1.25 The acceleration due to gravity at sea level is g D 9.81 m/s2 The radius of the earth is 6370 km The universal gravitational constant is G D 6.67 ð 10 11 N-m2 /kg2 Use this information to determine the mass of the earth Solution: The volume of the ocean is V D 64,186,000 mi2 12,925 ft 64 lb/ft3 32.2 ft/s2 (a) mD VD (b) m D 4.60 ð 1019 slugs Solution: Use Eq (1.3) a D 5,280 ft mi D 2.312 ð 1019 ft3 2.312 ð 1019 ft3 D 4.60 ð 1019 slugs 14.59 kg slug D 6.71 ð 1020 kg GmE Solve for the mass, R2 m 9.81 m/s2 6370 km 103 gR2 km mE D D G N-m2 6.67 10 11 kg2 D 5.9679 1024 kg D 5.97 1024 kg c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 1.26 A person weighs 180 lb at sea level The radius of the earth is 3960 mi What force is exerted on the person by the gravitational attraction of the earth if he is in a space station in orbit 200 mi above the surface of the earth? Solution: Use Eq (1.5) W D mg RE r WE g D g RE RE C H 3960 3960 C 200 D WE D 180 0.90616 D 163 lb Problem 1.27 The acceleration due to gravity on the surface of the moon is 1.62 m/s2 The moon’s radius is RM D 1738 km (a) What is the weight in newtons on the surface of the moon of an object that has a mass of 10 kg? (b) Using the approach described in Example 1.5, determine the force exerted on the object by the gravity of the moon if the object is located 1738 km above the moon’s surface Solution: Problem 1.28 If an object is near the surface of the earth, the variation of its weight with distance from the center of the earth can often be neglected The acceleration due to gravity at sea level is g D 9.81 m/s2 The radius of the earth is 6370 km The weight of an object at sea level is mg, where m is its mass At what height above the earth does the weight of the object decrease to 0.99 mg? Solution: Use a variation of Eq (1.5) a) W D mgM D 10 kg 1.26 m/s2 D 12.6 N b) Adapting equation 1.4 we have aM D gM W D 12.6 N RM r The force is then F D maM D 10 kg 1.62 m/s2 1738 km 1738 km C 1738 km D 4.05 N F D 4.05 N W D mg RE RE C h D 0.99 mg Solve for the radial height, h D RE p 1 0.99 D 6370 1.0050378 1.0 D 32.09 km D 32,100 m D 32.1 km Problem 1.29 The planet Neptune has an equatorial diameter of 49,532 km and its mass is 1.0247 ð 1026 kg If the planet is modeled as a homogeneous sphere, what is the acceleration due to gravity at its surface? (The universal gravitational constant is G D 6.67 ð 10 11 N-m2 /kg2 ) Solution: Problem 1.30 At a point between the earth and the moon, the magnitude of the force exerted on an object by the earth’s gravity equals the magnitude of the force exerted on the object by the moon’s gravity What is the distance from the center of the earth to that point to three significant digits? The distance from the center of the earth to the center of the moon is 383,000 km, and the radius of the earth is 6370 km The radius of the moon is 1738 km, and the acceleration due to gravity at its surface is 1.62 m/s2 Solution: Let rEp be the distance from the Earth to the point where the gravitational accelerations are the same and let rMp be the distance from the Moon to that point Then, rEp C rMp D rEM D 383,000 km The fact that the gravitational attractions by the Earth and the Moon at this point are equal leads to the equation mN m mN mN D G m ) gN D G rN r rN Note that the radius of Neptune is rN D 12 49,532 km D 24,766 km N-m2 1.0247 ð 1026 kg km Thus gN D 6.67 ð 10 11 2 24766 km 1000 m kg We have: W D G D 11.1 m/s2 gE RE rEp gN D 11.1 m/s2 D gM RM rMp , where rEM D 383,000 km Substituting the correct numerical values leads to the equation 9.81 m s2 6370 km rEp D 1.62 m s2 1738 km rEM rEp , where rEp is the only unknown Solving, we get rEp D 344,770 km D 345,000 km c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 2.1 In Active Example 2.1, suppose that the vectors U and V are reoriented as shown The vector V is vertical The magnitudes are jUj D and jVj D Graphically determine the magnitude of the vector U C 2V 45Њ U V Solution: Draw the vectors accurately and measure the resultant R D jU C 2Vj D 5.7 R D 5.7 Problem 2.2 Suppose that the pylon in Example 2.2 is moved closer to the stadium so that the angle between the forces FAB and FAC is 50° Draw a sketch of the new situation The magnitudes of the forces are jFAB j D 100 kN and jFAC j D 60 kN Graphically determine the magnitude and direction of the sum of the forces exerted on the pylon by the cables Solution: Accurately draw the vectors and measure the magnitude and direction of the resultant jFAB C FAC j D 146 kN ˛ D 32° c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 2.3 The magnitude jFA j D 80 lb and the angle ˛ D 65° The magnitude jFA C FB j D 120 lb Graphically determine the magnitude of FB FB FC ␤ a FA Solution: Accurately draw the vectors and measure the magnitude of FB jFB j D 62 lb Problem 2.4 The magnitudes jFA j D 40 N, jFB j D 50 N, and jFC j D 40 N The angle ˛ D 50° and ˇ D 80° Graphically determine the magnitude of FA C FB C FC FB FC ␤ a FA Solution: Accurately draw the vectors and measure the magnitude of FA C FB C FC R D jFA C FB C FC j D 83 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 11.39 Each homogenous bar is of mass m and length L The spring is unstretched when ˛ D If mg D kL, determine the value of ˛ in the range < ˛ < 90° for which the system is in equilibrium k α α Solution: The potential energy of the spring is The non zero position of equilibrium is, when W D kL, 1 D Reduce: V D 12 ks2 cos ˛ cos ˛ D 12 , Noting that from which s D 2L cos ˛ , cos ˛ D 12 , V D 2kL2 cos ˛ ˛ D cos The potential energy of the bars is Vbars D 1 D 60° WL WL cos ˛ C WL cos ˛ C cos ˛ D 2WL cos ˛, 2 where the datum point is the lower pin joint The total energy is V D 2kL2 cos ˛ C 2WL cos ˛ The equilibrium condition is dV D d˛ 2kL W cos ˛ sin ˛ D Problem 11.40 Determine whether the equilibrium position found in Problem 11.39 is stable or unstable Solution: Use the solution to Problem 11.39 The condition for an equilibrium point is dV D d˛ 2kL W cos ˛ sin ˛ D The condition for stability is d2 V d˛2 ˛D60° D 2kL sin2 ˛ W C 2kL W cos2 ˛ cos ˛ ˛D60° D 1.5 > 0, so the position is stable 892 c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 11.41 The pinned bars are held in place by the linear spring Each bar has weight W and length L The spring is unstretched when ˛ D 90° Determine the value of ˛ in the range < ˛ < 90° for which the system is in equilibrium (See Example 11.5.) L a a k Solution: The potential energy is VD2 W L sin ˛ C k 2L cos ˛ 2 For equilibrium we must have ˛ D sin Solving we find dV D WL cos ˛ d˛ 4kL2 cos ˛ sin ˛ D W 4kL Problem 11.42 Determine whether the equilibrium position found in Problem 11.41 is stable or unstable (See Example 11.5.) Solution: See 11.41 d2 V D d˛2 D WL sin ˛ 4kL2 cos2 ˛ C 4kL2 sin2 ˛ WL sin ˛ C 8kL2 sin2 4kL2 We need to evaluate this expression at the equilibrium position (given in problem 11.41) d2 V D d˛2 WL W 4kL C 8kL2 W 4kL 4kL2 D W2 4k 4kL2 > For stability we therefore need W > 4kL Assuming that an equilibrium position exists (Problem 11.41) then this condition cannot be met We conclude that the equilibrium position is unstable c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 893 Problem 11.43 The bar weighs 15 lb The spring is unstretched when ˛ D The bar is in equilibrium when ˛ D 30° Determine the spring constant k Solution: From the cosine law, the length of the spring is d2 D 22 C 42 p from which d D k 16 cos ˛, cos ˛ ft α The spring extension is Dd 2D2 p cos ˛ The potential energy of the spring is VD ft k2 The potential energy of the bar is Vbar D W cos ˛ The total potential energy is Vtot D k2 C W cos ˛ Noting d sin ˛ D p , d˛ cos ˛ the equilibrium condition is p k sin ˛ cos ˛ dV p D d˛ cos ˛ W sin ˛ D Solve for the spring constant: kD p W cos ˛ p cos ˛ Substitute values: p cos ˛ D 1.2393, kD 15 1.2393 0.2393 D 9.71 lb/ft 894 c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 11.44 Determine whether the equilibrium positions of the bar in Problem 11.43 are stable or unstable Solution: Use the solution to Problem 11.43 The condition for equilibrium is p k sin ˛ cos ˛ dV p D d˛ cos ˛ For brevity write L D dV D d˛ 8k L L p W sin ˛ D cos ˛ The derivative is W sin ˛ D 8k W 8k L sin ˛, from which the second derivative is d2 V 16k D sin2 ˛ C 8k d˛2 L W 8k L cos ˛ The condition for stability is determined from d2 V d˛2 ˛D˛i For ˛ D 0, d2 V d˛2 D WD 15 < 0, ˛D0 so that the equilibrium position is unstable For ˛ D 30° , d2 V d˛2 ˛D30° D 20.4 > 0, so that the equilibrium point is stable c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 895 Problem 11.45 (a) Determine the couple exerted on the beam at A (b) Determine the vertical force exerted on the beam at A 100 N 200 N-m A 30° 2m Solution: (a) Perform a virtual rotation about A: υU D MA υ 200υ C 100 sin 30° υ D 0, from which MA 200 C 100 sin 30° υ D 0, from which MA D 200 (b) 50 D 100 N m Perform a virtual translation of the bar in the y-direction: υU D Ay υy C 100 sin 30° υy D 0, from which Ay C 50 υy D 0, or Ay D 50 N Problem 11.46 The structure is subjected to a 20 kNm couple Determine the horizontal reaction at C y B 20 kN-m 2m Solution: The interior angle at B is 100° Denote the axial force in BC by RBC Do a virtual rotation of member AB about A: υU D 20 C 2RBC sin 100° υ D 0, A 40° 40° C x from which RBC D 20 D 10.15 kN cos 10° The horizontal component of the axial force is Rx D RBC cos 40° D 7.778 D 7.8 kN (directed parallel to the negative x axis.) 896 c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 11.47 The “rack and pinion” mechanism is used to exert a vertical force on a sample at A for a stamping operation If a force F D 30 lb is exerted on the handle, use the principle of virtual work to determine the force exerted on the sample Solution: Perform a virtual rotation of the handle The virtual work is υU D 8Fυ C Aυx D 0, from which AD 8F υ υx The angular rotation is related to the vertical translation by 2υ D from which υ D υx υx, , and A D 4F D 120 lb in in A F Problem 11.48 If you were assigned to calculate the force exerted on the bolt by the pliers when the grips are subjected to forces F as shown in Fig a, you could carefully measure the dimensions, draw free-body diagrams, and use the equilibrium equations But another approach would be to measure the change in the distance between the jaws when the distance between the handles is changed by a small amount If your measurements indicate that the distance d in Fig b decreases by mm when D is decreased mm, what is the approximate value of the force exerted on the bolt by each jaw when the forces F are applied? Solution: Let L be the distance between the points of application of the forces F and the point of application of the gripping force at the jaw The ratio of the motions indicates that the “effective” axis of rotation of a jaw is located 89 L from the point of application of F Perform a virtual rotation about this axis: F F (a) d D (b) υU D LFυ Lfυ D 0, from which 8F f υ D 0, or f D 8F This result is approximate because some work is done by the mechanism as the handle is closed In addition, the closure of the handles produces a translation of one handle relative to the other in the direction required to close the jaws, and this translation does work since it is associated with moment about the effective axis; hence not all of the virtual work due to a virtual rotation is included in the above expression c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 897 Problem 11.49 The system is in equilibrium The total weight of the suspended load and assembly A is 300 lb (a) By using equilibrium, determine the force F (b) Using the result of (a) and the principle of virtual work, determine the distance the suspended load rises if the cable is pulled downward ft at B B Solution: (a) Isolate the assembly A The sum of the forces: Fy D W F 3F D 0, A where F is the tension in the cable, from which FD (b) W D 100 lb Perform a virtual translation of the assembly A in the vertical direction The virtual work: υU D Wυy C Fυx D 0, from which W υx D D υy F The ratio of translations of the assembly A and the point B is D 3, from which yA D 13 ft yA Problem 11.50 The system is in equilibrium By drawing free-body diagrams and using equilibrium equations, determine the couple M (b) Using the result of (a) and the principle of virtual work, determine the angle through which pulley B rotates if pulley A rotates through an angle ˛ 200 mm 200 N-m M (a) Solution: The pulleys are frictionless and the belts not slip Denote the left pulley by A and the right pulley by B Denote the upper and lower tensions in the belts at pulley A by T3 , T4 , at B by T , T2 For pulley A: (1) T3 T4 0.1 D 200 N m, For pulley B (2) M D T1 T2 0.2 For the center pulley, (3) T1 T2 0.1 D T3 T4 0.2 Combine and solve: M D 200 D 800 N m (b) Perform a virtual rotation of the pulley A The virtual work of the system is υU D M1 υ˛ Mυ D 0, from which (a) 100 mm 100 A mm B 200 mm 200 T Nm 100 A mm T 100 mm T3 T1 200 mm T1 M B T4 T2 T2 200 mm M1 200 υ D D D , υ˛ M 800 from which  D 898 ˛ c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 11.51 The mechanism is in equilibrium Neglect friction between the horizontal bar and the collar Determine M in terms of F, ˛, and L 2L L α M F Solution: Perform a virtual rotation about the left pin support M D FL sin ˛ C p υx MDF υ˛ x xL sin ˛ L cos ˛ cos ˛ x2 sin2 ˛, an alternate form of the solution cos ˛ M D FL sin ˛ C 4L2 cos2 ˛ C From the identity, cos2 ˛ D is Using the results of the solution to Problem 11.14, MDF Substitute into the expression for the moment, and reduce: Fυx D 0, from which υU D Mυ˛ sin2 ˛ C L2 D From the dimensions given and the cosine law, 2Lx cos ˛, from which x2 2xL cos ˛ 3L2 D 0, which has the solution x D L cos ˛ š p p L cos2 ˛ C 3L2 D L cos ˛ š cos2 ˛ C Since a negative value of x has no meaning here, x D L cos ˛ C p cos2 ˛ C Problem 11.52 In an injection casting machine, a couple M applied to arm AB exerts a force on the injection piston at C Given that the horizontal component of the force exerted at C is kN, use the principle of virtual work to determine M B 350 mm 300 mm 45° A M Solution: Perform a virtual rotation of the crank The virtual work C The distance AC is is υU D Mυ C Fυx D 0, 0.35 x D , sin ˛ sin 45° from which from which x D 0.490 m From the solution to Problem 11.14, MD F 0.3 0.49 sin 45° υx D D 0.374 υ 0.49 0.3 cos 45° υx υ Denote the interior angle ACB by ˇ, and the interior angle ABC by ˛ From the sine law, The moment is M D 0.374F D 0.374 D 1.5 kN m 0.3 0.35 D , sin 45° sin ˇ from which ˇ D sin and ˛ D 180° 0.3 sin 45° 0.35 45° D 37.3° , 37.3° D 97.69° c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 899 Problem 11.53 Show that if bar AB is subjected to a clockwise virtual rotation υ˛, bar CD undergoes a counterclockwise virtual rotation b/a υ˛ B F 400 mm A C kN-m a Solution: The coordinates of pts B and C are α b cos ˇ, Cx D a C b b sin ˇ A We know that Bx 600 mm B By D 400 cos ˛, Cx b y Bx D 400 sin ˛, Cy D D x β C C Cy By D constant The derivative of this equation with respect to ˛ is Cx D2 aCb C2 dBx d˛ dCx d˛ Bx b sin ˇ b cos ˇ C Cy 400 sin ˛ 400 cos ˛ By b sin ˇ b cos ˇ dCy d˛ dˇ d˛ dBy d˛ 400 cos ˛ dˇ C 400 sin ˛ d˛ D At ˛ D 0, ˇ D 0, this equation is 2a 400 C 400 b dˇ d˛ D 0, from which we obtain υˇ D a υ˛ b Problem 11.54 The system in Problem 11.53 is in equilibrium, a D 800 mm, and b D 400 mm Use the principle of virtual work to determine the force F Solution: See the solution of Problem 11.53 When bar AB undergoes a clockwise virtual rotation υ˛, the virtual work is υU D 6υ˛ D 6υ˛ F 0.6υˇ a F 0.6 υ˛ D 0, b so FD 900 6b D kN 0.6a c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 11.55 Show that if bar AB is subjected to a clockwise virtual rotation υ˛, bar CD undergoes a clockwise virtual rotation [ad/ ac C bc bd ]υ˛ C B c d 24 N-m A M a Solution: Denote the interior acute angle formed by BC with the horizontal by ˇ, the obtuse interior angle at C by , and the interior acute angle at D by Perform a virtual translation υX parallel to the bar BC (Note: This is often a useful step where cranks- and connecting-rod-like mechanisms are involved.) Then D b Substitute: υ D υ˛ d CD a BC c b c d a ð C ð BC CD CD BC D ad ac C bc bd υX cos ˇ D dυ˛, and υX sin D CDυÂ, from which d cos ˇ υ D υ˛ CD sin Noting that sin D sin 90° D sin ˇ C and cos ˇ D a , BC sin c , CD D sin ˇ D cos D ˇ C 90° D sin ˇ cos C sin cos ˇ, c d , BC b CD Problem 11.56 The system in Problem 11.55 is in equilibrium, a D 300 mm, b D 350 mm, c D 350 mm, and d D 200 mm Use the principle of virtual work to determine the couple M Solution: Perform a virtual rotation of the crank at A The virtual work is υU D from which M1 υ˛ C Mυ D 0, M1 υ D υ˛ M From the solution Problem 11.55, υ ad D υ˛ ac C bc bd , from which MD M1 ac C bc ad bd D 24 2.625 D 63 N m c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 901 Problem 11.57 The mass of the bar is 10 kg, and it is m in length Neglect the masses of the two collars The spring is unstretched when the bar is vertical (˛ D 0), and the spring constant is k D 100 N/m Determine the values of ˛ at which the bar is in equilibrium k α Solution: The potential energy of the spring is Substitute numerical values: V D 12 ks2 cos ˛ D 0.5095, Noting that s D L V D 12 kL cos ˛ , then or ˛ D cos 0.5095 D 59.369 D 59.4° cos ˛ The potential energy of the bar is Vbar D WL cos ˛, where the datum point is the lower pin joint From which Vtot D kL 2 cos ˛ C WL cos ˛ The condition for equilibrium is dV D d˛ 2kL W cos ˛ sin ˛ D The equilibrium points are ˛ D 0, and the value of ˛ determined by 2kL W cos ˛ D 0, from which cos ˛ D W 2kL Problem 11.58 Determine whether the equilibrium positions of the bar in Problem 11.57 are stable or unstable Solution: Use the solution to Problem 11.57 The equilibrium For ˛ D 0, condition is dV D d˛ 2kL W cos ˛ d2 V d˛2 sin ˛ D D < 0, ˛D0 so the equilibrium point is unstable For ˛ D 59.4° , The stability condition is determined by d2 V d˛2 2kL W D ˛D˛i sin ˛ cos2 d2 V d˛2 ˛ ˛D59.4° D 1.51 > 0, so the equilibrium point is stable C 902 2kL W cos ˛ ˛D˛i c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 11.59 The spring is unstretched when ˛ D 90° Determine the value of ˛ in the range < ˛ < 90° for which the system is in equilibrium 1_ L m 1_ L 1_ L α k Solution: Choose a coordinate system such that the equilibrium position of the spring occurs at x D and at y D L The potential energy is V D mgy C kx2 Noting that y D L sin ˛ and x D L cos ˛, then V D mgL sin ˛ C kL2 cos2 ˛ The equilibrium condition is dV D D mg d˛ kL sin ˛ cos ˛ D This has two solutions: cos ˛ D 0, and sin ˛ D mg kL In the interval < ˛ < 90° , only one solution exists, ˛ D sin mg kL Problem 11.60 Determine whether the equilibrium position found in Problem 11.59 is stable or unstable Solution: Use the solution to Problem 11.59 The stability condition is d2 V D d˛2 mg sin ˛ C kL sin2 ˛ cos2 ˛ D kL sin2 ˛ < The system is unstable c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 903 Problem 11.61 The hydraulic cylinder C exerts a horizontal force at A, raising the weight W Determine the magnitude of the force the hydraulic cylinder must exert to support the weight in terms of W and ˛ W α A C α b Solution: The distance x D 2.5b, so υx D 2.5υb y Notice that y C b2 D constant Taking the derivative of this equation with respect to y, W 2y C 2b y D W x x The virtual work is Wυy F α b υb y υU D 1– b db D 0, dy we obtain υy D b Fυx b υb y F 2.5υb D 0, so FD 904 W b WD 2.5y 2.5 tan ˛ c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Problem 11.62 The homogenous composite object consists of a hemisphere and a cone It is at rest on the plane surface Show that this equilibrium position is p stable only if h < R h R Solution: Use the same strategy used to solve Problem 11.31: a point mass at the composite mass center, suspended (supported) at the radial center Choose a coordinate system with the y axis along the axis of the cone, positive upward, the x axis parallel to the floor, and the origin at the point of contact with the floor The mass of the cone is mcyl D where yD R2 h, is the mass density The center of mass of the cone is h C R The mass of the hemisphere is R3 The center of mass of the hemisphere is located at y D 5R The location of the mass centroid of the composite is R2 yD h h CR C h R2 C 3 R3 5R D h R3 h CR C R 12 h C R 3 For a small angular rotation  the length of the equivalent pendulum is R y The potential energy due to this rotation is V D R y W cos  The point of equilibrium is dV D R d y W sin  D 0, from which  D is a point of equilibrium The stability is determined by d2 V d D[ R y W cos Â]ÂD0 D R y W, ÂD0 from which, if R > y, the system is stable; if R < y, the system is unstable Reduce: R Rh C R2 yD h2 12 Rh h C R 3 R2 R 12 D h C h2 12 , 2R from which, (since the denominator is always positive) if R2 h2 > 0, 12 the system is stable Thus R2 > h2 , or, taking the positive square root: p 3R > h c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 905 ... speed (the distance traveled divided by the time required) to three significant digits (a) in km/h; (b) in mi/h Solution: a) v D 430 b) v D 430 km h 0.6214 mi km km 1000 m h km v D 392 ft/s Solution: ... Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in... in in2 T D 310 N-m y A Solution: A D 8.81 in2 25.4 mm in D 5680 mm2 x c 2008 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright