Solutions Manual Engineering Mechanics: Statics 2nd Edition Michael E Plesha University of Wisconsin–Madison Gary L Gray The Pennsylvania State University Francesco Costanzo The Pennsylvania State University With the assistance of: Chris Punshon Andrew J Miller Justin High Chris O’Brien Chandan Kumar Joseph Wyne Jonathan Fleischmann Version: May 11, 2012 The McGraw-Hill Companies, Inc Copyright © 2002–2012 Michael E Plesha, Gary L Gray, and Francesco Costanzo This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited Statics 2e Important Information about this Solutions Manual We encourage you to occasionally visit http://www.mhhe.com/pgc2e to obtain the most up-to-date version of this solutions manual Contact the Authors If you find any errors and/or have questions concerning a solution, please not hesitate to contact the authors and editors via email at: plesha@engr.wisc.edu, and stat_solns@email.esm.psu.edu We welcome your input This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited May 11, 2012 Solutions Manual Accuracy of Numbers in Calculations Throughout this solutions manual, we will generally assume that the data given for problems is accurate to significant digits When calculations are performed, all intermediate numerical results are reported to significant digits Final answers are usually reported with or significant digits If you verify the calculations in this solutions manual using the rounded intermediate numerical results that are reported, you should obtain the final answers that are reported to significant digits This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited May 11, 2012 37 Statics 2e Chapter Solutions Problem 2.1 For each vector, write two expressions using polar vector representations, one using a positive value of  and the other a negative value, where  is measured counterclockwise from the right-hand horizontal direction Solution Part (a) rE D 12 in: @ 90ı or rE D 12 in: @ 270ı : (1) FE D 23 N @ 135ı or FE D 23 N @ 225ı : (2) or vE D 15 m=s @ 120ı : (3) Part (b) Part (c) vE D 15 m=s @ 240ı This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited May 11, 2012 38 Solutions Manual Problem 2.2 E and report your result Add the two vectors shown to form a resultant vector R, using polar vector representation Solution Part (a) The vector polygon shown at the right corresponds to the addition of E Note that ˛ is the two position vectors to obtain a resultant position vector R ı ı ı given by ˛ D 180 55 D 125 Knowing this angle, the law of cosines may be used to determine R q R D 101 mm/2 C 183 mm/2 2.101 mm/.183 mm/ cos 125ı D 254:7 mm: (1) R α β 101 mm Next, the law of sines may be used to determine the angle ˇ:  à R 183 mm 183 mm ı D ) ˇ D sin sin 125 D 36:05ı : sin ˛ sin ˇ 254:7 mm 183 mm 55◦ (2) Using these results, we may report the vector RE using polar vector representation as RE D 255 mm @ 36:0ı : (3) Part (b) The vector polygon shown at the right corresponds to the addition of the two E The law of cosines may be used to force vectors to obtain a resultant force vector R determine R q R D 1:23 kip/2 C 1:55 kip/2 2.1:23 kip/.1:55 kip/ cos 45ı D 1:104 kip: (4) Using the law of sines, we find that R 1:23 kip D ı sin 45 sin ˇ ) ˇ D sin  à 1:23 kip ı sin 45 D 51:97ı : 1:104 kip The direction of RE measured from the right-hand horizontal direction is 90ı these results, we may report RE using polar vector representation as 51:97ı D RE D 1:10 kip @ 142ı : This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited (5) 142ı Using (6) May 11, 2012 39 Statics 2e Problem 2.3 E and report your result Add the two vectors shown to form a resultant vector R, using polar vector representation Solution 1.8 m Part (a) The vector polygon shown at the right corresponds to the addition of the two E The law of cosines may be used position vectors to obtain a resultant position vector R to determine R as q R D 1:8 m/2 C 2:3 m/2 2.1:8 m/.2:3 m/ cos 65ı D 2:243 m: 65° R 2.3 m (1) The law of sines may be used to determine the angle ˛ as R 2:3 m D ı sin 65 sin ˛ ) ˛ D sin 2:3 m sin 65ı D 68:34ı : 2:243 m (2) Using polar vector representation, the resultant is RE D 2:243 m @ 68:34ı : If desired, this resultant may be stated using a positive angle, where 360ı (3) 68:34ı D 291:7ı , as RE D 2:243 m @ 291:7ı : (4) Part (b) The vector polygon shown at the right corresponds to the addition of the two E The law of cosines may be used to force vectors to obtain a resultant force vector R determine R as q R D kN/2 C 8:2 kN/2 2.6 kN/.8:2 kN/ cos 20ı D 3:282 kN: (5) Noting that ˇ appears to be an obtuse angle (see the Common Pitfall margin note in the text), we will use the law of sines to determine ˛ as  à kN R kN ı D ) ˛ D sin sin 20 D 38:70ı : (6) sin ˛ sin 20ı 3:282 kN This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited May 11, 2012 40 Solutions Manual Angle ˇ is obtained using 20ı C ˛ C ˇ D 180ı ; ˇ D 180ı 20ı (7) 38:70ı D 121:3ı : (8) 180ı (9) Using polar vector representation, the resultant is RE D 3:282 kN @ RE D 3:282 kN @ 121:3ı / 58:70ı : If desired, the resultant may be stated using a positive angle, where 360ı (10) 58:70ı D 301:3ı , as RE D 3:282 kN @ 301:3ı : This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited (11) May 11, 2012 41 Statics 2e Problem 2.4 E and report your result Add the two vectors shown to form a resultant vector R, using polar vector representation Solution 30° Part (a) The vector polygon shown at the right corresponds to the addition of the two E Since the 54 N force is vertical, the angle force vectors to obtain a resultant force vector R ı ˛ may be obtained by inspection as ˛ D 90 C 30ı D 120ı The law of cosines may be used to determine R as q R D 48 N/2 C 54 N/2 2.48 N/.54 N/ cos 120ı D 88:39 N: 48 N R 54 N (1) The law of sines may be used to determine the angle ˇ as 54 N R D sin ˇ sin ˛ ) ˇ D sin 54 N sin 120ı D 31:95ı : 88:39 N (2) Using polar vector representation, the resultant is RE D 88:39 N @ 180ı RE D 88:39 N @ 30ı ˇ/ (3) 118:1ı : If desired, this resultant may be stated using a positive angle, where 360ı (4) 118:1ı D 241:9ı , as RE D 88:39 N @ 241:9ı : Part (b) The vector polygon shown at the right corresponds to the addition E Given the 20° of the two position vectors to obtain a resultant position vector R ı ı 100 mm 20 and 30 angles provided in the problem statement, we determine the angle ı ı ı opposite R to be 70 C30 D 100 The law of cosines may be used to determine R as q R D 100 mm/2 C 80 mm/2 2.100 mm/.80 mm/ cos 100ı D 138:5 mm: This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited (5) R 70° 30° 80 mm 20° (6) May 11, 2012 42 Solutions Manual The law of sines may be used to determine the angle ˛ as R 80 mm D sin ˛ sin 100ı ) ˛ D sin 80 mm sin 100ı D 34:67ı : 138:5 mm (7) Using polar vector representation, the resultant is RE D 138:5 mm @ 34:67ı 20ı (8) RE D 138:5 mm @ 14:67ı : (9) This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc It may be used and/or possessed only by permission of McGraw-Hill, and must be surrendered upon request of McGraw-Hill Any duplication or distribution, either in print or electronic form, without the permission of McGraw-Hill, is prohibited May 11, 2012 ... electronic form, without the permission of McGraw-Hill, is prohibited May 11, 2012 Solutions Manual Accuracy of Numbers in Calculations Throughout this solutions manual, we will generally assume that... deck of a garden tractor is shown where FE1 is horizontal Determine the magnitude of FE2 so that the resultant of these two forces is vertical, and determine the magnitude of this resultant Solution. .. without the permission of McGraw-Hill, is prohibited May 11, 2012 52 Solutions Manual Problem 2.12 Arm OA of a robot is positioned as shown Determine the value for angle ˛ of arm AB so that the