Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.1 Two forces are applied as shown to a hook Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: = R 1391 kN, = α 47.8° R = 1391 N 47.8° Copyright © McGraw-Hill Education This is proprietary material solely for authorized instructor use Not authorized for sale or distribution in any manner This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.2 Two forces are applied as shown to a bracket support Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: R = 906 lb, = α 26.6° R = 906 lb 26.6° Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.3 Two structural members B and C are bolted to bracket A Knowing that both members are in tension and that P = 10 kN and Q = 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: α 21.2° R = 20.1 kN, = R = 20.1 kN 21.2° Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.4 Two structural members B and C are bolted to bracket A Knowing that both members are in tension and that P = kips and Q = kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule SOLUTION (a) Parallelogram law: (b) Triangle rule: We measure: = R 8.03 kips, α = 3.8° R = 8.03 kips 3.8° Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.5 The 300-lb force is to be resolved into components along lines a-a′ and b-b′ (a) Determine the angle α by trigonometry knowing that the component along line a-a′ is to be 240 lb (b) What is the corresponding value of the component along b-b′? SOLUTION (a) Using the triangle rule and law of sines: sin b sin 60° = 240 lb 300 lb sin b = 0.69282 = b 43.854° α + b + 60= ° 180° = 180° − 60° − 43.854° α = 76.146° (b) Law of sines: Fbb′ 300 lb = sin 76.146° sin 60° = α 76.1° Fbb′ = 336 lb Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.6 The 300-lb force is to be resolved into components along lines a-a′ and b-b′ (a) Determine the angle α by trigonometry knowing that the component along line b-b′ is to be 120 lb (b) What is the corresponding value of the component along a-a′? SOLUTION Using the triangle rule and law of sines: (a) sin α sin 60° = 120 lb 300 lb sin α = 0.34641 α 20.268° = (b) = α 20.3° α + β + 60= ° 180° β = 180° − 60° − 20.268° = 99.732° Faa′ 300 lb = sin 99.732° sin 60° Faa′ = 341 lb Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.7 A trolley that moves along a horizontal beam is acted upon by two forces as shown (a) Knowing that α = 25°, determine by trigonometry the magnitude of the force P so that the resultant force exerted on the trolley is vertical (b) What is the corresponding magnitude of the resultant? SOLUTION Using the triangle rule and the law of sines: (a) (b) 1600 N P = sin 25° sin 75° P = 3660 N ° 180° 25° + β + 75= β = 180° − 25° − 75° = 80° 1600 N R = sin 25° sin 80° R = 3730 N Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.8 A disabled automobile is pulled by means of two ropes as shown The tension in rope AB is 2.2 kN, and the angle α is 25° Knowing that the resultant of the two forces applied at A is directed along the axis of the automobile, determine by trigonometry (a) the tension in rope AC, (b) the magnitude of the resultant of the two forces applied at A SOLUTION Using the law of sines: TAC R 2.2 kN = = sin 30° sin125° sin 25A TAC = 2.603 kN R = 4.264 kN (a) (b) TAC = 2.60 kN R = 4.26 kN Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.9 Two forces are applied as shown to a hook support Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R SOLUTION Using the triangle rule and law of sines: (a) sin α sin 25° = 50 N 35 N sin α = 0.60374 = α 37.138° (b) = α 37.1° α + β + 25= ° 180° β = 180° − 25° − 37.138° = 117.862° R 35 N = sin117.862° sin 25° R = 73.2 N Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.10 A disabled automobile is pulled by means of two ropes as shown Knowing that the tension in rope AB is kN, determine by trigonometry the tension in rope AC and the value of α so that the resultant force exerted at A is a 4.8kN force directed along the axis of the automobile SOLUTION Using the law of cosines: (3 kN) + (4.8 kN) − 2(3 kN)(4.8 kN) cos 30° TAC = TAC = 2.6643 kN Using the law of sines: sin α sin 30° = kN 2.6643 kN = α 34.3° TAC = 2.66 kN 34.3° Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.26 Determine the resultant of the three forces of Problem 2.18 PROBLEM 2.18 Determine the x and y components of each of the forces shown SOLUTION Force x Comp (lb) y Comp (lb) 40 lb +20.00 –34.64 50 lb –38.30 –32.14 60 lb +54.38 +25.36 Rx = +36.08 Ry = −41.42 R Rx i + Ry j = = (+36.08 lb)i + (−41.42 lb) j tan a = Ry Rx 41.42 lb 36.08 lb tan a = 1.14800 = a 48.942° tan a = R= 41.42 lb sin 48.942° R = 54.9 lb 48.9° Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.27 Determine the resultant of the three forces of Problem 2.19 PROBLEM 2.19 Determine the x and y components of each of the forces shown SOLUTION Components of the forces were determined in Problem 2.19: Force x Comp (N) y Comp (N) 80 N +61.3 +51.4 120 N +41.0 +112.8 150 N –122.9 +86.0 Rx = −20.6 Ry = + 250.2 = R Rx i + Ry j (−20.6 N)i + (250.2 N) j = Ry tan a = Rx 250.2 N 20.6 N tan a = 12.1456 a 85.293° = tan a = R= 250.2 N sin 85.293° R = 251 N 85.3° Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.28 For the collar loaded as shown, determine (a) the required value of α if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant SOLUTION Rx = ΣFx = (100 N) cos α + (150 N) cos (α + 30°) − (200 N) cos α Rx = −(100 N) cos α + (150 N) cos (α + 30°) (1) Ry = ΣFy =−(100 N)sin α − (150 N)sin (α + 30°) − (200 N)sin α −(300 N)sin α − (150 N)sin (α + 30°) Ry = (a) (2) For R to be vertical, we must have Rx = We make Rx = in Eq (1): −100 cos α + 150 cos (α + 30°) =0 −100 cos α + 150 (cos α cos 30° − sin α sin 30°) = 29.904 cos α = 75sin α 29.904 75 = 0.39872 tan a = = a 21.738° (b) = α 21.7° Substituting for α in Eq (2): Ry = −300sin 21.738° − 150sin 51.738° = −228.89 N = R |R = y | 228.89 N R = 229 N Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.29 A hoist trolley is subjected to the three forces shown Knowing that α = 40°, determine (a) the required magnitude of the force P if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant SOLUTION Rx = ΣFx= P + (200 lb)sin 40° − (400 lb) cos 40° Rx= P − 177.860 lb (1) = Ry = ΣFy (200 lb) cos 40° + (400 lb)sin 40° Ry = 410.32 lb (a) For R to be vertical, we must have Rx = Set Rx = in Eq (1) 0= P − 177.860 lb P = 177.860 lb (b) (2) P = 177.9 lb Since R is to be vertical: = R R= 410 lb y R = 410 lb Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.30 A hoist trolley is subjected to the three forces shown Knowing that P = 250 lb, determine (a) the required value of α if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant SOLUTION Rx = ΣF = 250 lb + (200 lb)sin α − (400 lb) cos α x Rx = 250 lb + (200 lb)sin α − (400 lb) cos α (1) Ry = = ΣFy (200 lb) cos α + (400 lb)sin α (a) For R to be vertical, we must have Rx = Rx = in Eq (1) Set 0= 250 lb + (200 lb)sin α − (400 lb) cos α (400 lb) cos α (200 lb)sin α + 250 lb = cos α sin α + 1.25 = cos α = sin α + 2.5sin α + 1.5625 4(1 − sin α ) =sin α + 2.5sin α + 1.5625 =5sin α + 2.5sin α − 2.4375 Using the quadratic formula to solve for the roots gives sin α = 0.49162 = α 29.447° or (b) = α 29.4° Since R is to be vertical: R = R= (200 lb) cos 29.447° + (400 lb)sin 29.447° y R = 371 lb Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.31 For the post loaded as shown, determine (a) the required tension in rope AC if the resultant of the three forces exerted at point C is to be horizontal, (b) the corresponding magnitude of the resultant SOLUTION 960 24 Rx = ΣFx = − TAC + (500 N) + (200 N) 1460 25 48 Rx = − TAC + 640 N 73 1100 Ry = ΣFy = − TAC + (500 N) − (200 N) 1460 25 55 Ry = − TAC + 20 N 73 (a) (2) For R to be horizontal, we must have Ry = Set Ry = in Eq (2): − 55 TAC + 20 N = 73 TAC = 26.545 N (b) (1) TAC = 26.5 N Substituting for TAC into Eq (1) gives 48 Rx = − (26.545 N) + 640 N 73 Rx = 622.55 N = R R= 623 N x R = 623 N Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.32 Two cables are tied together at C and are loaded as shown Knowing that α = 30°, determine the tension (a) in cable AC, (b) in cable BC SOLUTION Free-Body Diagram Force Triangle Law of sines: TAC TBC kN = = sin 60° sin 35° sin 85° (a) = TAC kN (sin 60°) sin 85° TAC = 5.22 kN (b) = TBC kN (sin 35°) sin 85° TBC = 3.45 kN Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.33 Two cables are tied together at C and are loaded as shown Determine the tension (a) in cable AC, (b) in cable BC SOLUTION Free-Body Diagram Force Triangle Law of sines: TAC TBC 400 lb = = sin 60° sin 40° sin 80° (a) = TAC 400 lb (sin 60°) sin 80° TAC = 352 lb (b) = TBC 400 lb (sin 40°) sin 80° TBC = 261 lb Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.34 Two cables are tied together at C and are loaded as shown Determine the tension (a) in cable AC, (b) in cable BC SOLUTION Free-Body Diagram Force Triangle W = mg = (200 kg)(9.81 m/s ) = 1962 N Law of sines: TAC TBC 1962 N = = sin 15° sin 105° sin 60° (a) TAC = (1962 N) sin 15° sin 60° TAC = 586 N (b) TBC = (1962 N)sin 105° sin 60° TBC = 2190 N Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.35 Two cables are tied together at C and loaded as shown Determine the tension (a) in cable AC, (b) in cable BC SOLUTION Free Body Diagram at C: ΣFx = 0: − 12 ft 7.5 ft TAC + TBC = 12.5 ft 8.5 ft TBC = 1.08800TAC = ΣFy 0: (a) 3.5 ft ft TAC + TBC= − 396 lb 12 ft 8.5 ft 3.5 ft ft TAC + (1.08800TAC ) − 396 lb = 12.5 ft 8.5 ft (0.28000 + 0.51200)TAC = 396 lb TAC = 500.0 lb (b) TBC = (1.08800)(500.0 lb) TAC = 500 lb TBC = 544 lb Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.36 Two cables are tied together at C and are loaded as shown Knowing that P = 500 N and α = 60°, determine the tension in (a) in cable AC, (b) in cable BC SOLUTION Free-Body Diagram Law of sines: Force Triangle TAC TBC 500 N = = sin 35° sin 75° sin 70° (a) = TAC 500 N sin 35° sin 70° TAC = 305 N (b) = TBC 500 N sin 75° sin 70° TBC = 514 N Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.37 Two forces of magnitude T A = kips and T B = 15 kips are applied as shown to a welded connection Knowing that the connection is in equilibrium, determine the magnitudes of the forces T C and T D SOLUTION Free-Body Diagram = ΣFx 15 kips − kips −= TD cos 40° TD = 9.1379 kips ΣFy = TD sin 40° − TC = (9.1379 kips)sin 40° − TC = TC = 5.8737 kips TC = 5.87 kips TD = 9.14 kips Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.38 Two forces of magnitude T A = kips and T C = kips are applied as shown to a welded connection Knowing that the connection is in equilibrium, determine the magnitudes of the forces T B and T D SOLUTION Free-Body Diagram Σ Fx = TB − kips − TD cos 40° =0 Σ Fy = TD sin 40° − kips = (1) kips sin 40° TD = 14.0015 kips TD = Substituting for T D into Eq (1) gives: TB − kips − (14.0015 kips) cos 40° =0 TB = 16.7258 kips TB = 16.73 kips TD = 14.00 kips Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.39 Two cables are tied together at C and are loaded as shown Knowing that P = 300 N, determine the tension in cables AC and BC SOLUTION Free-Body Diagram = ΣFx − TCA sin 30A + TCB sin 30A − P cos 45° − 200N = For P = 200N we have, −0.5TCA + 0.5TCB + 212.13 − 200 = (1) ΣFy = TCA cos 30° − TCB cos 30A − P sin 45A = 0.86603TCA + 0.86603TCB − 212.13 = (2) Solving equations (1) and (2) simultaneously gives, TCA = 134.6 N TCB = 110.4 N Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.40 Two forces P and Q are applied as shown to an aircraft connection Knowing that the connection is in equilibrium and that P = 500 lb and Q = 650 lb, determine the magnitudes of the forces exerted on the rods A and B SOLUTION Free-Body Diagram Resolving the forces into x- and y-directions: R = P + Q + FA + FB = Substituting components: R= −(500 lb) j + [(650 lb) cos 50°]i − [(650 lb)sin 50°]j + FB i − ( FA cos 50°)i + ( FA sin 50°) j = In the y-direction (one unknown force): −500 lb − (650 lb)sin 50° + FA sin 50° = Thus, FA = 500 lb + (650 lb)sin 50° sin 50° = 1302.70 lb In the x-direction: Thus, FA = 1303 lb (650 lb) cos 50° + FB − FA cos 50° = = FB FA cos 50° − (650 lb) cos 50° = (1302.70 lb) cos 50° − (650 lb) cos 50° = 419.55 lb FB = 420 lb Copyright © McGraw-Hill Education All rights reserved No reproduction or distribution without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ ... https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/ PROBLEM 2.40 Two forces P and Q are applied as shown... without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/... without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Statics and Mechanics of Materials 2nd Edition by Beer Full file at https://TestbankDirect.eu/