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    PROBLEM 2.1

    Two forces are applied to an eye bolt fastened to a beam. Determine

    graphically the magnitude and direction of their resultant using (a) the

    parallelogram law, (b) the triangle rule.

    SOLUTION

    (a)

    (b)

    We measure: 8.4 kNR =

    19 =

    8.4 kN=R 19

    1

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    PROBLEM 2.2

    The cable stays AB and AD help support pole AC. Knowing that the

    tension is 500 N in AB and 160 N in AD, determine graphically the

    magnitude and direction of the resultant of the forces exerted by the stays

    atA using (a) the parallelogram law, (b) the triangle rule.

    SOLUTION

    We measure: 51.3 , 59 = =

    (a)

    (b)

    We measure: 575 N, 67= = R

    575 N=R 67

    2

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    PROBLEM 2.3

    Two forces P and Q are applied as shown at point A of a hook support.

    Knowing that P = 15 lb and Q = 25 lb, determine graphically themagnitude and direction of their resultant using (a) the parallelogram law,

    (b) the triangle rule.

    SOLUTION

    (a)

    (b)

    We measure: 37 lb, 76= = R

    37 lb=R 76

    3

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    PROBLEM 2.4

    Two forces P and Q are applied as shown at point A of a hook support.

    Knowing that P = 45 lb and Q = 15 lb, determine graphically themagnitude and direction of their resultant using (a) the parallelogram law,

    (b) the triangle rule.

    SOLUTION

    (a)

    (b)

    We measure: 61.5 lb, 86.5= = R

    61.5 lb=R 86.5

    4

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    PROBLEM 2.5

    Two control rods are attached at A to lever AB. Using trigonometry and

    knowing that the force in the left-hand rod is F1 = 120 N, determine(a) the required force F2 in the right-hand rod if the resultant R of the

    forces exerted by the rods on the lever is to be vertical, (b) the

    corresponding magnitude ofR.

    SOLUTION

    Graphically, by the triangle law

    We measure: 2 108 NF

    77 NR

    By trigonometry: Law of Sines

    2 120

    sin sin 38 sin

    F R

    = =

    90 28 62 , 180 62 38 80 = = = =

    Then:

    2 120 N

    sin 62 sin 38 sin80

    F R= =

    or (a) 2 107.6 NF =

    (b) 75.0 NR =

    5

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    PROBLEM 2.6

    Two control rods are attached at A to lever AB. Using trigonometry and

    knowing that the force in the right-hand rod is F2 = 80 N, determine(a) the required force F1 in the left-hand rod if the resultant R of the

    forces exerted by the rods on the lever is to be vertical, (b) the

    corresponding magnitude ofR.

    SOLUTION

    Using the Law of Sines

    1 80

    sin sin 38 sin

    F R

    = =

    90 10 80 , 180 80 38 62 = = = =

    Then:

    1 80 N

    sin80 sin 38 sin 62

    F R= =

    or (a) 1 89.2 NF =

    (b) 55.8 NR =

    6

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    PROBLEM 2.7

    The 50-lb force is to be resolved into components along lines -a a and

    - .b b (a) Using trigonometry, determine the angle knowing that thecomponent along -a a is 35 lb. (b) What is the corresponding value ofthe component along - ?b b

    SOLUTION

    Using the triangle rule and the Law of Sines

    (a)sin sin 40

    35 lb 50 lb

    =

    sin 0.44995 =

    26.74 =

    Then: 40 180 + + =

    113.3 =

    (b) Using the Law of Sines:

    50 lb

    sin sin 40

    bbF

    =

    71.5 lbbbF =

    7

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    PROBLEM 2.8

    The 50-lb force is to be resolved into components along lines -a a and

    - .b b (a) Using trigonometry, determine the angle knowing that thecomponent along -b b is 30 lb. (b) What is the corresponding value ofthe component along - ?a a

    SOLUTION

    Using the triangle rule and the Law of Sines

    (a)sin sin 40

    30 lb 50 lb

    =

    sin 0.3857 =

    22.7 =

    (b) 40 180 + + =

    117.31 =

    50 lb

    sin sin 40

    aaF

    =

    sin50 lb

    sin 40

    =

    aaF

    69.1lbaaF =

    8

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    PROBLEM 2.9

    To steady a sign as it is being lowered, two cables are attached to the sign

    at A. Using trigonometry and knowing that = 25, determine (a) therequired magnitude of the force P if the resultant R of the two forces

    applied atA is to be vertical, (b) the corresponding magnitude ofR.

    SOLUTION

    Using the triangle rule and the Law of Sines

    Have: ( )180 35 25 = +

    120=

    Then:360 N

    sin 35 sin120 sin 25

    P R= =

    or (a) 489 NP =

    (b) 738 NR =

    9

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    PROBLEM 2.10

    To steady a sign as it is being lowered, two cables are attached to the sign

    atA. Using trigonometry and knowing that the magnitude ofP is 300 N,

    determine (a) the required angle if the resultant R of the two forces

    applied atA is to be vertical,(b) the corresponding magnitude ofR.

    SOLUTION

    Using the triangle rule and the Law of Sines

    (a) Have:360 N 300 N

    sin sin 35=

    sin 0.68829 =

    43.5 =

    (b) ( )180 35 43.5 = +

    101.5=

    Then:300 N

    sin101.5 sin 35

    R=

    or 513 NR =

    10

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    PROBLEM 2.11

    Two forces are applied as shown to a hook support. Using trigonometry

    and knowing that the magnitude ofP is 14 lb, determine (a) the required

    angle if the resultant R of the two forces applied to the support is to be

    horizontal, (b) the corresponding magnitude ofR.

    SOLUTION

    Using the triangle rule and the Law of Sines

    (a) Have:20 lb 14 lb

    sin sin 30=

    sin 0.71428 =

    45.6 =

    (b) ( )180 30 45.6 = +

    104.4=

    Then:14 lb

    sin104.4 sin 30

    R=

    27.1 lbR =

    11

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    PROBLEM 2.12

    For the hook support of Problem 2.3, using trigonometry and knowing

    that the magnitude ofP is 25 lb, determine (a) the required magnitude of

    the force Q if the resultant R of the two forces applied at A is to be

    vertical, (b) the corresponding magnitude ofR.

    Problem 2.3: Two forces P and Q are applied as shown at point A of ahook support. Knowing that P = 15 lb and Q = 25 lb, determinegraphically the magnitude and direction of their resultant using (a) the

    parallelogram law, (b) the triangle rule.

    SOLUTION

    Using the triangle rule and the Law of Sines

    (a) Have:25 lb

    sin15 sin 30

    Q=

    12.94 lbQ =

    (b) ( )180 15 30 = +

    135=

    Thus:25 lb

    sin135 sin 30

    R=

    sin13525 lb 35.36 lb

    sin30R

    = =

    35.4 lbR =

    12

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    PROBLEM 2.13

    For the hook support of Problem 2.11, determine, using trigonometry,

    (a) the magnitude and direction of the smallest force P for which the

    resultant R of the two forces applied to the support is horizontal,

    (b) the corresponding magnitude ofR.

    Problem 2.11: Two forces are applied as shown to a hook support. Usingtrigonometry and knowing that the magnitude of P is 14 lb, determine

    (a) the required angle if the resultant R of the two forces applied to the

    support is to be horizontal, (b) the corresponding magnitude ofR.

    SOLUTION

    (a) The smallest force Pwill be perpendicular to R, that is, vertical

    ( )20 lb sin30P =

    10 lb= 10 lb=P

    (b) ( )20 lb cos30R =

    17.32 lb= 17.32 lbR =

    13

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    PROBLEM 2.14

    As shown in Figure P2.9, two cables are attached to a sign at A to steady

    the sign as it is being lowered. Using trigonometry, determine (a) the

    magnitude and direction of the smallest force P for which the resultant R

    of the two forces applied atA is vertical, (b) the corresponding magnitude

    ofR.

    SOLUTION

    We observe that force P is minimum when is 90 , that is, P is horizontal

    Then: (a) ( )360 N sin35P =

    or 206 N=P

    And: (b) ( )360 N cos35R =

    or 295 NR =

    14

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    PROBLEM 2.15

    For the hook support of Problem 2.11, determine, using trigonometry, the

    magnitude and direction of the resultant of the two forces applied to the

    support knowing that P = 10 lb and = 40.

    Problem2.11: Two forces are applied as shown to a hook support. Using

    trigonometry and knowing that the magnitude of P is 14 lb, determine(a) the required angle if the resultant R of the two forces applied to the

    support is to be horizontal, (b) the corresponding magnitude ofR.

    SOLUTION

    Using the force triangle and the Law of Cosines

    ( ) ( ) ( )( )2 22 10 lb 20 lb 2 10 lb 20 lb cos110R = +

    ( ) 2100 400 400 0.342 lb = +

    2636.8 lb=

    25.23 lbR =

    Using now the Law of Sines

    10 lb 25.23 lb

    sin sin110=

    10 lbsin sin110

    25.23 lb

    =

    0.3724=

    So: 21.87 =

    Angle of inclination ofR, is then such that:

    30 + =

    8.13 =

    Hence: 25.2 lb=R 8.13

    15

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    PROBLEM 2.16

    Solve Problem 2.1 using trigonometry

    Problem 2.1: Two forces are applied to an eye bolt fastened to a beam.

    Determine graphically the magnitude and direction of their resultant

    using (a) the parallelogram law, (b) the triangle rule.

    SOLUTION

    Using the force triangle, the Law of Cosines and the Law of Sines

    We have: ( )180 50 25 = +

    105=

    Then: ( ) ( ) ( )( )2 22 4.5 kN 6 kN 2 4.5 kN 6 kN cos105R = +

    270.226 kN=

    or 8.3801 kNR =

    Now:8.3801 kN 6 kN

    sin105 sin=

    6 kNsin sin105

    8.3801 kN

    =

    0.6916=

    43.756 =

    8.38 kN=R 18.76

    16

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    PROBLEM 2.17

    Solve Problem 2.2 using trigonometry

    Problem 2.2: The cable staysAB andAD help support poleAC. Knowing

    that the tension is 500 N in AB and 160 N in AD, determine graphically

    the magnitude and direction of the resultant of the forces exerted by the

    stays atA using (a) the parallelogram law, (b) the triangle rule.

    SOLUTION

    From the geometry of the problem:

    1 2tan 38.662.5

    = =

    1 1.5tan 30.962.5

    = =

    Now: ( )180 38.66 30.96 110.38 = + =

    And, using the Law of Cosines:

    ( ) ( ) ( )( )2 22 500 N 160 N 2 500 N 160 N cos110.38R = +

    2331319 N=

    575.6 NR =

    Using the Law of Sines:

    160 N 575.6 N

    sin sin110.38=

    160 Nsin sin110.38

    575.6 N

    =

    0.2606=

    15.1 =

    ( )90 66.44 = + =

    576 N=R 66.4

    17

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    PROBLEM 2.18

    Solve Problem 2.3 using trigonometry

    Problem 2.3: Two forces P and Q are applied as shown at point A of a

    hook support. Knowing that P = 15 lb and Q = 25 lb, determinegraphically the magnitude and direction of their resultant using (a) theparallelogram law, (b) the triangle rule.

    SOLUTION

    Using the force triangle and the Laws of Cosines and Sines

    We have:

    ( )180 15 30 = +

    135=

    Then: ( ) ( ) ( ) ( )2 22 15 lb 25 lb 2 15 lb 25 lb cos135R = +

    21380.3 lb=

    or 37.15 lbR =

    and

    25 lb 37.15 lb

    sin sin135=

    25 lbsin sin135

    37.15 lb

    =

    0.4758=

    28.41 =

    Then: 75 180 + + =

    76.59 =

    37.2 lb=R 76.6

    18

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    PROBLEM 2.19

    Two structural members A and B are bolted to a bracket as shown.

    Knowing that both members are in compression and that the force is

    30 kN in member A and 20 kN in member B, determine, usingtrigonometry, the magnitude and direction of the resultant of the forces

    applied to the bracket by membersA andB.

    SOLUTION

    Using the force triangle and the Laws of Cosines and Sines

    We have: ( )180 45 25 110 = + =

    Then: ( ) ( ) ( )( )2 22

    30 kN 20 kN 2 30 kN 20 kN cos110R = +

    21710.4 kN=

    41.357 kNR =

    and

    20 kN 41.357 kN

    sin sin110=

    20 kNsin sin110

    41.357 kN

    =

    0.4544=

    27.028 =

    Hence: 45 72.028 = + =

    41.4 kN=R 72.0

    19

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    PROBLEM 2.20

    Two structural members A and B are bolted to a bracket as shown.

    Knowing that both members are in compression and that the force is

    20 kN in member A and 30 kN in member B, determine, usingtrigonometry, the magnitude and direction of the resultant of the forces

    applied to the bracket by membersA andB.

    SOLUTION

    Using the force triangle and the Laws of Cosines and Sines

    We have: ( )180 45 25 110 = + =

    Then: ( ) ( ) ( )( )2 22

    30 kN 20 kN 2 30 kN 20 kN cos110R = +

    21710.4 kN=

    41.357 kNR =

    and

    30 kN 41.357 kN

    sin sin110=

    30 kNsin sin110

    41.357 kN

    =

    0.6816=

    42.97 =

    Finally: 45 87.97 = + =

    41.4 kN=R 88.0

    20

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    PROBLEM 2.21

    Determine thex andy components of each of the forces shown.

    SOLUTION

    20 kN Force:

    ( )20 kN cos40 ,xF = + 15.32 kNxF =

    ( )20 kN sin 40 ,yF = + 12.86 kNyF =

    30 kN Force:

    ( )30 kN cos70 ,xF = 10.26 kNxF =

    ( )30 kN sin 70 ,yF = + 28.2 kNyF =

    42 kN Force:

    ( )42 kN cos20 ,

    xF = 39.5 kN

    xF =

    ( )42 kN sin 20 ,yF = + 14.36 kNyF =

    21

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    PROBLEM 2.22

    Determine thex andy components of each of the forces shown.

    SOLUTION

    40 lb Force:

    ( )40 lb sin 50 ,xF = 30.6 lbxF =

    ( )40 lb cos50 ,yF = 25.7 lbyF =

    60 lb Force:

    ( )60 lb cos60 ,xF = + 30.0 lbxF =

    ( )60 lb sin 60 ,yF = 52.0 lbyF =

    80 lb Force:

    ( )80 lb cos25 ,xF = + 72.5 lbxF =

    ( )80 lb sin 25 ,yF = + 33.8 lbyF =

    22

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    PROBLEM 2.23

    Determine thex andy components of each of the forces shown.

    SOLUTION

    We compute the following distances:

    ( ) ( )

    ( ) ( )

    ( ) ( )

    2 2

    2 2

    2 2

    48 90 102 in.

    56 90 106 in.

    80 60 100 in.

    OA

    OB

    OC

    = + =

    = + =

    = + =

    Then:

    204 lb Force:

    ( )48

    102 lb ,102xF = 48.0 lbxF =

    ( )90

    102 lb ,102

    yF = + 90.0 lbyF =

    212 lb Force:

    ( )56

    212 lb ,106

    xF = + 112.0 lbxF =

    ( )90

    212 lb ,106

    yF = + 180.0 lbyF =

    400 lb Force:

    ( )80

    400 lb ,100

    xF = 320 lbxF =

    ( )60

    400 lb ,100

    yF = 240 lbyF =

    23

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    PROBLEM 2.24

    Determine thex andy components of each of the forces shown.

    SOLUTION

    We compute the following distances:

    ( ) ( )2 2

    70 240 250 mmOA = + =

    ( ) ( )2 2

    210 200 290 mmOB = + =

    ( ) ( )2 2

    120 225 255 mmOC = + =

    500 N Force:

    70500 N250

    xF =

    140.0 NxF =

    240500 N

    250yF

    = +

    480 NyF =

    435 N Force:

    210435 N

    290xF

    = +

    315 NxF =

    200435 N

    290yF

    = +

    300 NyF =

    510 N Force:

    120510 N

    255xF

    = +

    240 NxF =

    225510 N

    255yF

    =

    450 NyF =

    24

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    PROBLEM 2.25

    While emptying a wheelbarrow, a gardener exerts on each handle AB a

    force P directed along line CD. Knowing that P must have a 135-N

    horizontal component, determine (a) the magnitude of the force P, (b) its

    vertical component.

    SOLUTION

    (a)cos40

    xPP =

    135 N

    cos40=

    or 176.2 NP =

    (b) tan 40 sin 40y xP P P= =

    ( )135 N tan40=

    or 113.3 NyP =

    25

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    PROBLEM 2.26

    Member BD exerts on member ABC a force P directed along line BD.

    Knowing that P must have a 960-N vertical component, determine (a) the

    magnitude of the force P, (b) its horizontal component.

    SOLUTION

    (a)sin35

    yPP =

    960 N

    sin35=

    or 1674 NP =

    (b)tan35

    yx

    PP =

    960 N

    tan35=

    or 1371NxP =

    26

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    PROBLEM 2.27

    Member CB of the vise shown exerts on blockB a force P directed along

    line CB. Knowing that P must have a 260-lb horizontal component,

    determine (a) the magnitude of the force P, (b) its vertical component.

    SOLUTION

    We note:

    CB exerts force P onB along CB, and the horizontal component ofP is 260 lb.xP =

    Then:

    (a) sin50xP P=

    sin50xPP =

    260 lb

    sin50=

    339.4 lb= 339 lbP =

    (b) tan50x yP P=

    tan50

    xy

    PP =

    260 lb

    tan50=

    218.2 lb= 218 lby =P

    27

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    PROBLEM 2.28

    Activator rod AB exerts on crankBCD a force P directed along line AB.

    Knowing that P must have a 25-lb component perpendicular to armBCof

    the crank, determine (a) the magnitude of the force P, (b) its component

    along line BC.

    SOLUTION

    Using the x andy axes shown.

    (a) 25 lbyP =

    Then:sin75

    yPP =

    25 lb

    sin75=

    or 25.9 lbP =

    (b)tan75

    yx

    PP =

    25 lb

    tan 75=

    or 6.70 lbxP =

    28

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    PROBLEM 2.29

    The guy wire BD exerts on the telephone pole AC a force P directed

    along BD. Knowing that P has a 450-N component along line AC,

    determine (a) the magnitude of the force P, (b) its component in a

    direction perpendicular to AC.

    SOLUTION

    Note that the force exerted by BD on the pole is directed along BD, and the component of P along ACis 450 N.

    Then:

    (a)450 N

    549.3 Ncos35

    P = =

    549 NP =

    (b) ( )450 N tan35xP =

    315.1 N=

    315 NxP =

    29

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    PROBLEM 2.30

    The guy wire BD exerts on the telephone pole AC a force P directed

    along BD. Knowing that P has a 200-N perpendicular to the pole AC,

    determine (a) the magnitude of the force P, (b) its component along

    line AC.

    SOLUTION

    (a)sin38

    xPP =

    200 N

    sin38=

    324.8 N= or 325 NP =

    (b)tan38

    xy

    PP =

    200 N

    tan38=

    255.98 N=

    or 256 NyP =

    30

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    PROBLEM 2.31

    Determine the resultant of the three forces of Problem 2.24.

    Problem 2.24: Determine thex andy components of each of the forces

    shown.

    SOLUTION

    From Problem 2.24:

    ( ) ( )500 140 N 480 N= +F i j

    ( ) ( )425 315 N 300 N= +F i j

    ( ) ( )510 240 N 450 N= F i j

    ( ) ( )415 N 330 N= = +R F i j

    Then:

    1 330tan 38.5415

    = =

    ( ) ( )2 2

    415 N 330 N 530.2 NR = + =

    Thus: 530 N=

    R 38.5

    31

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    PROBLEM 2.32

    Determine the resultant of the three forces of Problem 2.21.

    Problem 2.21: Determine thex andy components of each of the forces

    shown.

    SOLUTION

    From Problem 2.21:

    ( ) ( )20 15.32 kN 12.86 kN= +F i j

    ( ) ( )30 10.26 kN 28.2 kN= +F i j

    ( ) ( )42 39.5 kN 14.36 kN= +F i j

    ( ) ( )34.44 kN 55.42 kN= = +R F i j

    Then:

    1 55.42tan 58.134.44

    = =

    ( ) ( )2 2

    55.42 kN 34.44 N 65.2 kNR = + =

    65.2 kNR = 58.2

    32

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    PROBLEM 2.33

    Determine the resultant of the three forces of Problem 2.22.

    Problem 2.22: Determine the x andy components of each of the forces

    shown.

    SOLUTION

    The components of the forces were determined in 2.23.

    x yR R= +R i j

    ( ) ( )71.9 lb 43.86 lb= i j

    43.86tan

    71.9 =

    31.38 =

    ( ) ( )2 2

    71.9 lb 43.86 lbR = +

    84.23 lb=

    84.2 lb=R 31.4

    Force comp. (lb)x comp. (lb)y

    40 lb 30.6 25.7

    60 lb 30 51.96

    80 lb 72.5 33.8

    71.9xR = 43.86yR =

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    PROBLEM 2.34

    Determine the resultant of the three forces of Problem 2.23.

    Problem 2.23: Determine thex andy components of each of the forces

    shown.

    SOLUTION

    The components of the forces were

    determined in Problem 2.23.

    ( ) ( )204 48.0 lb 90.0 lb= +F i j

    ( ) ( )212 112.0 lb 180.0 lb= +F i j

    ( ) ( )400 320 lb 240 lb= F i j

    Thus

    x y= +R R R

    ( ) ( )256 lb 30.0 lb= +R i j

    Now:

    30.0tan

    256 =

    1 30.0tan 6.68256

    = =

    and

    ( ) ( )2 2

    256 lb 30.0 lbR = +

    257.75 lb=

    258 lb=R 6.68

    34

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    PROBLEM 2.35

    Knowing that 35 , = determine the resultant of the three forces

    shown.

    SOLUTION

    300-N Force:

    ( )300 N cos 20 281.9 NxF = =

    ( )300 N sin 20 102.6 NyF = =

    400-N Force:

    ( )400 N cos 55 229.4 NxF = =

    ( )400 N sin 55 327.7 NyF = =

    600-N Force:

    ( )600 N cos 35 491.5 NxF = =

    ( )600 N sin 35 344.1 NyF = =

    and

    1002.8 Nx xR F= =

    86.2 Ny yR F= =

    ( ) ( )2 2

    1002.8 N 86.2 N 1006.5 NR = + =

    Further:

    86.2

    tan 1002.8 =

    1 86.2tan 4.911002.8

    = =

    1007 N=R 4.91

    35

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    PROBLEM 2.36

    Knowing that 65 , = determine the resultant of the three forces

    shown.

    SOLUTION

    300-N Force:

    ( )300 N cos 20 281.9 NxF = =

    ( )300 N sin 20 102.6 NyF = =

    400-N Force:

    ( )400 N cos85 34.9 NxF = =

    ( )400 N sin 85 398.5 NyF = =

    600-N Force:

    ( )600 N cos 5 597.7 NxF = =

    ( )600 N sin 5 52.3 NyF = =

    and

    914.5 Nx xR F= =

    448.8 Ny yR F= =

    ( ) ( )2 2

    914.5 N 448.8 N 1018.7 NR = + =

    Further:

    448.8tan

    914.5 =

    1 448.8tan 26.1914.5

    = =

    1019 N=R 26.1

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    PROBLEM 2.37

    Knowing that the tension in cableBCis 145 lb, determine the resultant of

    the three forces exerted at pointB of beamAB.

    SOLUTION

    Cable BC Force:

    ( )84

    145 lb 105 lb116x

    F = =

    ( )80

    145 lb 100 lb116

    yF = =

    100-lb Force:

    ( )3

    100 lb 60 lb5

    xF = =

    ( )4

    100 lb 80 lb5

    yF = =

    156-lb Force:

    ( )12

    156 lb 144 lb13

    xF = =

    ( )5

    156 lb 60 lb13

    yF = =

    and

    21 lb, 40 lbx x y yR F R F= = = =

    ( ) ( )2 2

    21 lb 40 lb 45.177 lbR = + =

    Further:

    40tan

    21 =

    1 40tan 62.321

    = =

    Thus: 45.2 lb=R 62.3

    37

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    PROBLEM 2.38

    Knowing that 50 , = determine the resultant of the three forces

    shown.

    SOLUTION

    The resultant forceR has thex- andy-components:

    ( ) ( ) ( )140 lb cos50 60 lb cos85 160 lb cos50x xR F= = +

    7.6264 lbxR =

    and

    ( ) ( ) ( )140 lb sin 50 60 lb sin85 160 lb sin 50y yR F= = + +

    289.59 lbyR =

    Further:

    290tan

    7.6 =

    1 290tan 88.57.6

    = =

    Thus: 290 lb=R 88.5

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    PROBLEM 2.39

    Determine (a) the required value of if the resultant of the three forcesshown is to be vertical, (b) the corresponding magnitude of the resultant.

    SOLUTION

    For an arbitrary angle , we have:

    ( ) ( ) ( ) ( )140 lb cos 60 lb cos 35 160 lb cosx xR F = = + +

    (a) So, forR to be vertical:

    ( ) ( ) ( ) ( )140 lb cos 60 lb cos 35 160 lb cos 0x xR F = = + + =

    Expanding,

    ( )cos 3 cos cos35 sin sin 35 0 + =

    Then:

    13

    cos35

    tan sin35

    =

    or

    11 3

    cos35tan 40.265

    sin35

    = = 40.3 =

    (b) Now:

    ( ) ( ) ( )140 lb sin 40.265 60 lb sin 75.265 160 lb sin 40.265y yR R F= = = + +

    252 lbR R= =

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    PROBLEM 2.40

    For the beam of Problem 2.37, determine (a) the required tension in cableBCif the resultant of the three forces exerted at point B is to be vertical,

    (b) the corresponding magnitude of the resultant.

    Problem 2.37: Knowing that the tension in cableBCis 145 lb, determinethe resultant of the three forces exerted at pointB of beamAB.

    SOLUTION

    We have:

    ( ) ( )84 12 3156 lb 100 lb116 13 5

    x x BCR F T= = +

    or 0.724 84 lbx BCR T= +

    and

    ( ) ( )80 5 4

    156 lb 100 lb116 13 5

    y y BCR F T= =

    0.6897 140 lby BCR T=

    (a) So, forR to be vertical,

    0.724 84 lb 0x BCR T= + =

    116.0 lbBCT = (b) Using

    116.0 lbBCT =

    ( )0.6897 116.0 lb 140 lb 60 lbyR R= = =

    60.0 lbR R= =

    40

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    PROBLEM 2.41

    Boom AB is held in the position shown by three cables. Knowing that the

    tensions in cables AC and AD are 4 kN and 5.2 kN, respectively,

    determine (a) the tension in cable AE if the resultant of the tensions

    exerted at point A of the boom must be directed along AB,

    (b) the corresponding magnitude of the resultant.

    SOLUTION

    Choosex-axis along bar AB.

    Then

    (a) Require

    ( ) ( )0: 4 kN cos 25 5.2 kN sin 35 sin 65 0y y AER F T= = + =

    or 7.2909 kNAET =

    7.29 kNAET =

    (b) xR F=

    ( ) ( ) ( )4 kN sin 25 5.2 kN cos35 7.2909 kN cos 65=

    9.03 kN=

    9.03 kNR =

    41

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    PROBLEM 2.42

    For the block of Problems 2.35 and 2.36, determine (a) the required value

    of of the resultant of the three forces shown is to be parallel to the

    incline, (b) the corresponding magnitude of the resultant.

    Problem 2.35: Knowing that 35 , = determine the resultant of the

    three forces shown.

    Problem 2.36: Knowing that 65 , = determine the resultant of the

    three forces shown.

    SOLUTION

    Selecting the x axis along ,aa we write

    ( ) ( )300 N 400 N cos 600 N sinx xR F = = + + (1)

    ( ) ( )400 N sin 600 N cosy yR F = = (2)

    (a) Setting 0yR = in Equation (2):

    Thus600

    tan 1.5400

    = =

    56.3 =

    (b) Substituting for in Equation (1):

    ( ) ( )300 N 400 N cos56.3 600 N sin 56.3xR = + +

    1021.1 NxR =

    1021 NxR R= =

    42

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    PROBLEM 2.43

    Two cables are tied together at Cand are loaded as shown. Determine the

    tension (a) in cable AC, (b) in cable BC.

    SOLUTION

    Free-Body Diagram

    From the geometry, we calculate the distances:

    ( ) ( )2 2

    16 in. 12 in. 20 in.AC = + =

    ( ) ( )2 2

    20 in. 21 in. 29 in.BC = + =

    Then, from the Free Body Diagram of point C:

    16 210: 0

    20 29x AC BCF T T = + =

    or29 4

    21 5BC ACT T=

    and12 20

    0: 600 lb 020 29

    y AC BCF T T = + =

    or12 20 29 4

    600 lb 020 29 21 5

    AC ACT T

    + =

    Hence: 440.56 lbACT =

    (a) 441 lbACT =

    (b) 487 lbBCT =

    43

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    PROBLEM 2.44

    Knowing that 25 , = determine the tension (a) in cable AC, (b) in

    ropeBC.

    SOLUTION

    Free-Body Diagram Force Triangle

    Law of Sines:

    5 kN

    sin115 sin 5 sin 60

    AC BCT T= =

    (a)5 kN

    sin115 5.23 kNsin60

    ACT = =

    5.23 kNACT =

    (b)5 kN

    sin 5 0.503 kNsin60

    BCT = =

    0.503 kNBCT =

    44

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    PROBLEM 2.45

    Knowing that 50 = and that boom AC exerts on pin C a force

    directed long line AC, determine (a) the magnitude of that force, (b) the

    tension in cableBC.

    SOLUTION

    Free-Body Diagram Force Triangle

    Law of Sines:

    400 lb

    sin 25 sin 60 sin 95

    AC BCF T= =

    (a)400 lb

    sin 25 169.69 lbsin95

    ACF = =

    169.7 lbACF =

    (b)400

    sin 60 347.73 lbsin95

    BCT = =

    348 lbBCT =

    45

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    PROBLEM 2.46

    Two cables are tied together at Cand are loaded as shown. Knowing that

    30 , = determine the tension (a) in cableAC, (b) in cableBC.

    SOLUTION

    Free-Body Diagram Force Triangle

    Law of Sines:

    2943 N

    sin 60 sin 55 sin 65

    AC BCT T= =

    (a)2943 N

    sin 60 2812.19 Nsin 65

    ACT = =

    2.81 kNACT =

    (b)2943 N

    sin 55 2659.98 Nsin65

    BCT = =

    2.66 kNBCT =

    46

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    PROBLEM 2.47

    A chairlift has been stopped in the position shown. Knowing that each

    chair weighs 300 N and that the skier in chairEweighs 890 N, determine

    that weight of the skier in chair F.

    SOLUTION

    Free-Body Diagram Point B

    Force Triangle

    Free-Body Diagram Point C

    Force Triangle

    In the free-body diagram of pointB, the geometry gives:

    1 9.9tan 30.5116.8

    AB

    = =

    1 12tan 22.6128.8

    BC

    = =

    Thus, in the force triangle, by the Law of Sines:

    1190 N

    sin 59.49 sin 7.87

    BCT=

    7468.6 NBC

    T =

    In the free-body diagram of point C(with Wthe sum of weights of chairand skier) the geometry gives:

    1 1.32tan 10.397.2

    CD

    = =

    Hence, in the force triangle, by the Law of Sines:

    7468.6 N

    sin12.23 sin100.39

    W=

    1608.5 NW =

    Finally, the skier weight 1608.5 N 300 N 1308.5 N= =

    skier weight 1309 N=

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    PROBLEM 2.48

    A chairlift has been stopped in the position shown. Knowing that each

    chair weighs 300 N and that the skier in chair Fweighs 800 N, determine

    the weight of the skier in chairE.

    SOLUTION

    Free-Body Diagram Point F

    Force Triangle

    Free-Body Diagram Point E

    Force Triangle

    In the free-body diagram of point F, the geometry gives:

    1 12tan 22.6228.8

    EF

    = =

    1 1.32tan 10.397.2

    DF

    = =

    Thus, in the force triangle, by the Law of Sines:

    1100 N

    sin100.39 sin12.23

    EFT=

    5107.5 NBCT =

    In the free-body diagram of point E(with Wthe sum of weights of chairand skier) the geometry gives:

    1 9.9tan 30.5116.8

    AE

    = =

    Hence, in the force triangle, by the Law of Sines:

    5107.5 N

    sin 7.89 sin 59.49

    W=

    813.8 NW =

    Finally, the skier weight 813.8 N 300 N 513.8 N= =

    skier weight 514 N=

    48

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    PROBLEM 2.49

    Four wooden members are joined with metal plate connectors and are in

    equilibrium under the action of the four fences shown. Knowing thatFA = 510 lb and FB = 480 lb, determine the magnitudes of the other two

    forces.

    SOLUTION

    Free-Body Diagram

    Resolving the forces intox andy components:

    ( ) ( )0: 510 lb sin15 480 lb cos15 0x CF F = + =

    or 332 lbCF =

    ( ) ( )0: 510 lb cos15 480 lb sin15 0y DF F = + =

    or 368 lbDF =

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    PROBLEM 2.50

    Four wooden members are joined with metal plate connectors and are in

    equilibrium under the action of the four fences shown. Knowing thatFA = 420 lb and FC = 540 lb, determine the magnitudes of the other two

    forces.

    SOLUTION

    Resolving the forces intox andy components:

    ( ) ( )0: cos15 540 lb 420 lb cos15 0 or 671.6 lbx B BF F F = + + = =

    672 lbBF =

    ( ) ( )0: 420 lb cos15 671.6 lb sin15 0y DF F = + =

    or 232 lbDF =

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    PROBLEM 2.51

    Two forces P and Q are applied as shown to an aircraft connection.

    Knowing that the connection is in equilibrium and the P = 400 lb andQ = 520 lb, determine the magnitudes of the forces exerted on the rodsA andB.

    SOLUTION

    Free-Body Diagram Resolving the forces intox andy directions:

    0A B= + + + =R P Q F F

    Substituting components:

    ( ) ( ) ( )400 lb 520 lb cos55 520 lb sin 55 = + R j i j

    ( ) ( )cos55 sin 55 0B A AF F F+ + =i i j

    In they-direction (one unknown force)

    ( )400 lb 520 lb sin 55 sin 55 0AF + =

    Thus,

    ( )400 lb 520 lb sin 551008.3 lb

    sin55AF

    + = =

    1008 lbAF =

    In thex-direction:

    ( )520 lb cos55 cos55 0B AF F + =

    Thus,

    ( )cos55 520 lb cos55B AF F=

    ( ) ( )1008.3 lb cos 55 520 lb cos 55=

    280.08 lb=

    280 lbBF =

    51

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    PROBLEM 2.52

    Two forces P and Q are applied as shown to an aircraft connection.

    Knowing that the connection is in equilibrium and that the magnitudes ofthe forces exerted on rods A and B are FA = 600 lb and FB = 320 lb,determine the magnitudes ofP and Q.

    SOLUTION

    Free-Body Diagram Resolving the forces intox andy directions:

    0A B= + + + =R P Q F F

    Substituting components:

    ( ) ( ) ( )320 lb 600 lb cos55 600 lb sin 55 = + R i i j

    ( ) ( )cos55 sin 55 0P Q Q+ + =i i j

    In thex-direction (one unknown force)

    ( )320 lb 600 lb cos55 cos55 0Q + =

    Thus,

    ( )320 lb 600 lb cos5542.09 lb

    cos55Q

    + = =

    42.1lbQ =

    In they-direction:

    ( )600 lb sin 55 sin 55 0P Q =

    Thus,

    ( )600 lb sin 55 sin 55 457.01 lbP Q= =

    457 lbP =

    52

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    PROBLEM 2.53

    Two cables tied together at C are loaded as shown. Knowing that

    W = 840 N, determine the tension (a) in cableAC, (b) in cableBC.

    SOLUTION

    Free-Body Diagram From geometry:

    The sides of the triangle with hypotenuse CB are in the ratio 8:15:17.

    The sides of the triangle with hypotenuse CA are in the ratio 3:4:5.

    Thus:

    ( )3 15 15

    0: 680 N 05 17 17

    x CA CBF T T = + =

    or

    1 5200 N

    5 17CA CBT T + = (1)

    and

    ( )4 8 8

    0: 680 N 840 N 05 17 17y CA CBF T T = + =

    or

    1 2290 N

    5 17CA CBT T+ = (2)

    Solving Equations (1) and (2) simultaneously:

    (a) 750 NCAT =

    (b) 1190 NCBT =

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    PROBLEM 2.54

    Two cables tied together at Care loaded as shown. Determine the range

    of values of W for which the tension will not exceed 1050 N in eithercable.

    SOLUTION

    Free-Body Diagram From geometry:

    The sides of the triangle with hypotenuse CB are in the ratio 8:15:17.

    The sides of the triangle with hypotenuse CA are in the ratio 3:4:5.

    Thus:

    ( )3 15 15

    0: 680 N 05 17 17

    x CA CBF T T = + =

    or

    1 5200 N

    5 17CA CBT T + = (1)

    and

    ( )4 8 8

    0: 680 N 05 17 17

    y CA CBF T T W = + =

    or

    1 2 180 N

    5 17 4+ = +CA CBT T W (2)

    Then, from Equations (1) and (2)

    17680 N

    28

    25

    28

    CB

    CA

    T W

    T W

    = +

    =

    Now, with 1050 NT

    25: 1050 N 28

    CA CAT T W= =

    or 1176 NW =

    and

    17: 1050 N 680 N

    28CB CBT T W= = +

    or 609 NW = 0 609 N W

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    PROBLEM 2.55

    The cabin of an aerial tramway is suspended from a set of wheels that can

    roll freely on the support cable ACB and is being pulled at a constant

    speed by cable DE. Knowing that 40 = and = 35, that thecombined weight of the cabin, its support system, and its passengers is

    24.8 kN, and assuming the tension in cable DF to be negligible,determine the tension (a) in the support cable ACB, (b) in the traction

    cableDE.

    SOLUTION

    Note: In Problems 2.55 and 2.56 the cabin is considered as a particle. Ifconsidered as a rigid body (Chapter 4) it would be found that its center of

    gravity should be located to the left of the centerline for the line CD to be

    vertical.

    Now

    ( )0: cos35 cos40 cos 40 0x ACB DEF T T = =

    or

    0.0531 0.766 0ACB DET T = (1)

    and

    ( )0: sin 40 sin 35 sin 40 24.8 kN 0y ACB DEF T T = + =

    or

    0.0692 0.643 24.8 kNACB DET T+ = (2)

    From (1)

    14.426ACB DET T=

    Then, from (2)

    ( )0.0692 14.426 0.643 24.8 kNDE DET T+ =

    and

    (b) 15.1 kNDET =

    (a) 218 kNACBT =

    55

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    PROBLEM 2.56

    The cabin of an aerial tramway is suspended from a set of wheels that can

    roll freely on the support cable ACB and is being pulled at a constant

    speed by cableDE. Knowing that 42 = and = 32, that the tension

    in cable DE is 20 kN, and assuming the tension in cable DF to be

    negligible, determine (a) the combined weight of the cabin, its supportsystem, and its passengers, (b) the tension in the support cableACB.

    SOLUTION

    Free-Body Diagram

    First, consider the sum of forces in thex-direction because there is only one unknown force:

    ( ) ( )0: cos32 cos42 20 kN cos 42 0x ACBF T = =

    or

    0.1049 14.863 kNACBT =

    (b) 141.7 kNACBT =

    Now

    ( ) ( )0: sin 42 sin 32 20 kN sin 42 0

    y ACBF T W = + =

    or

    ( )( ) ( )( )141.7 kN 0.1392 20 kN 0.6691 0W+ =

    (a) 33.1 kNW =

    56

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    PROBLEM 2.57

    A block of weight W is suspended from a 500-mm long cord and two

    springs of which the unstretched lengths are 450 mm. Knowing that theconstants of the springs are kAB = 1500 N/m and kAD = 500 N/m,

    determine (a) the tension in the cord, (b) the weight of the block.

    SOLUTION

    Free-Body Diagram At A First note from geometry:

    The sides of the triangle with hypotenuseAD are in the ratio 8:15:17.

    The sides of the triangle with hypotenuseAB are in the ratio 3:4:5.

    The sides of the triangle with hypotenuseACare in the ratio 7:24:25.

    Then:

    ( )AB AB AB oF k L L=

    and

    ( ) ( )2 2

    0.44 m 0.33 m 0.55 mABL = + =

    So:

    ( )1500 N/m 0.55 m 0.45 mABF =

    150 N=

    Similarly,

    ( )AD AD AD oF k L L=

    Then:

    ( ) ( )2 2

    0.66 m 0.32 m 0.68 mADL = + =

    ( )1500 N/m 0.68 m 0.45 mADF =

    115 N=

    (a)

    ( ) ( )4 7 15

    0: 150 N 115 N 05 25 17

    x ACF T = + =

    or

    66.18 NACT = 66.2 NACT =

    57

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    PROBLEM 2.57 CONTINUED

    (b) and

    ( ) ( ) ( )3 24 8

    0: 150 N 66.18 N 115 N 05 25 17

    yF W = + + =

    or 208 N=W

    58

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    PROBLEM 2.58

    A load of weight 400 N is suspended from a spring and two cords which

    are attached to blocks of weights 3Wand Was shown. Knowing that theconstant of the spring is 800 N/m, determine (a) the value ofW, (b) the

    unstretched length of the spring.

    SOLUTION

    Free-Body Diagram At AFirst note from geometry:

    The sides of the triangle with hypotenuseAD are in the ratio 12:35:37.

    The sides of the triangle with hypotenuseACare in the ratio 3:4:5.

    The sides of the triangle with hypotenuseAB are also in the ratio

    12:35:37.

    Then:

    ( ) ( )4 35 12

    0: 3 05 37 37

    x sF W W F = + + =

    or

    4.4833sF W=

    and

    ( ) ( )3 12 35

    0: 3 400 N 05 37 37

    y sF W W F = + + =

    Then:

    ( ) ( ) ( )3 12 35

    3 4.4833 400 N 05 37 37

    W W W+ + =

    or

    62.841NW =

    and

    281.74 NsF =

    or

    (a) 62.8 NW =

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    PROBLEM 2.58 CONTINUED

    (b) Have spring force

    ( )s AB oF k L L=

    Where

    ( )AB AB AB oF k L L=

    and

    ( ) ( )2 2

    0.360 m 1.050 m 1.110 mABL = + =

    So:

    ( )0281.74 N 800 N/m 1.110 mL=

    or 0 758 mmL =

    60

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    PROBLEM 2.59

    For the cables and loading of Problem 2.46, determine (a) the value of

    for which the tension in cable BC is as small as possible, (b) the

    corresponding value of the tension.

    SOLUTION

    The smallest BCT is when BCT is perpendicular to the direction of ACT

    Free-Body Diagram At C Force Triangle

    (a) 55.0 =

    (b) ( )2943 N sin55BCT =

    2410.8 N=

    2.41kNBCT =

    61

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    PROBLEM 2.60

    Knowing that portionsACandBCof cableACB must be equal, determine

    the shortest length of cable which can be used to support the load shown

    if the tension in the cable is not to exceed 725 N.

    SOLUTION

    Free-Body Diagram: C

    ( )For 725 NT = 0: 2 1000 N 0y yF T = =

    500 NyT =

    2 2 2x yT T T+ =

    ( ) ( )2 22 500 N 725 NxT + =

    525 NxT =

    By similar triangles:

    1.5 m

    725 525

    BC=

    2.07 m =BC

    ( )2 4.14 mL BC= =

    4.14 mL =

    62

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    PROBLEM 2.61

    Two cables tied together at C are loaded as shown. Knowing that the

    maximum allowable tension in each cable is 200 lb, determine (a) the

    magnitude of the largest force P which may be applied at C, (b) the

    corresponding value of.

    SOLUTION

    Free-Body Diagram: C Force Tr iangle

    Force triangle is isoceles with

    2 180 85 =

    47.5 =

    (a) ( )2 200 lb cos 47.5 270 lbP = =

    Since 0,P > the solution is correct. 270 lbP =

    (b) 180 55 47.5 77.5 = = 77.5 =

    63

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    PROBLEM 2.62

    Two cables tied together at C are loaded as shown. Knowing that the

    maximum allowable tension is 300 lb in cable ACand 150 lb in cable BC,

    determine (a) the magnitude of the largest force P which may be applied

    at C, (b) the corresponding value of.

    SOLUTION

    Fr ee-Body Diagram: C Force Tr iangle

    (a) Law of Cosines:

    ( ) ( ) ( )( )2 22 300 lb 150 lb 2 300 lb 150 lb cos85P = +

    323.5 lbP =

    Since 300 lb,P > our solution is correct. 324 lbP =

    (b) Law of Sines:

    sin sin85

    300 323.5

    =

    sin 0.9238 =

    or 67.49 =

    180 55 67.49 57.5 = =

    57.5 =

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    PROBLEM 2.63

    For the structure and loading of Problem 2.45, determine (a) the value of

    for which the tension in cable BC is as small as possible, (b) the

    corresponding value of the tension.

    SOLUTION

    BCT must be perpendicular to ACF to be as small as possible.

    Fr ee-Body Diagram: C Force Tr iangle is

    a r ight tr iangle

    (a) We observe: 55 = 55 =

    (b) ( )400 lb sin 60BCT =

    or 346.4 lbBCT = 346 lbBCT =

    65

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    PROBLEM 2.64

    Boom AB is supported by cable BC and a hinge at A. Knowing that the

    boom exerts on pin B a force directed along the boom and that the tension

    in rope BD is 70 lb, determine (a) the value of for which the tension in

    cable BC is as small as possible, (b) the corresponding value of the

    tension.

    SOLUTION

    Free-Body Diagram: B (a) Have: 0BD AB BC+ + =T F T

    where magnitude and direction of BDT are known, and the direction

    of ABF is known.

    Then, in a force triangle:

    By observation, BCT is minimum when 90.0 = (b) Have ( ) ( )70 lb sin 180 70 30BCT =

    68.93 lb=

    68.9 lbBCT=

    66

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    PROBLEM 2.65

    Collar A shown in Figure P2.65 and P2.66 can slide on a frictionless

    vertical rod and is attached as shown to a spring. The constant of the

    spring is 660 N/m, and the spring is unstretched when h = 300 mm.Knowing that the system is in equilibrium when h = 400 mm, determine

    the weight of the collar.

    SOLUTION

    Free-Body Diagram: Collar A

    Have: ( )s AB ABF k L L=

    where:

    ( ) ( )2 2

    0.3 m 0.4 m 0.3 2 mAB ABL L = + =

    0.5 m=

    Then: ( )660 N/m 0.5 0.3 2 msF =

    49.986 N=

    For the collar:

    ( )4

    0: 49.986 N 05

    yF W = + =

    or 40.0 NW =

    67

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    PROBLEM 2.66

    The 40-N collar A can slide on a frictionless vertical rod and is attached

    as shown to a spring. The spring is unstretched when h = 300 mm.Knowing that the constant of the spring is 560 N/m, determine the value

    ofh for which the system is in equilibrium.

    SOLUTION

    Free-Body Diagram: Collar A( )

    2 20: 0

    0.3y s

    hF W F

    h

    = + =+

    or 240 0.09shF h= +

    Now.. ( )s AB ABF k L L=

    where

    ( )

    2 20.3 m 0.3 2 mAB AB

    L h L = + =

    Then: ( )2 2560 0.09 0.3 2 40 0.09h h h + = +

    or ( ) 214 1 0.09 4.2 2 mh h h h + =

    Solving numerically,

    415 mmh =

    68

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    PROBLEM 2.67

    A 280-kg crate is supported by several rope-and-pulley arrangements as

    shown. Determine for each arrangement the tension in the rope. (Hint:

    The tension in the rope is the same on each side of a simple pulley. Thiscan be proved by the methods of Chapter 4.)

    SOLUTION

    Fr ee-Body Diagram of pulley

    (a)

    (b)

    (c)

    (d)

    (e)

    ( )( )20: 2 280 kg 9.81 m/s 0yF T = =

    ( )1

    2746.8 N2

    T =

    1373 NT =

    ( )( )20: 2 280 kg 9.81 m/s 0yF T = =

    ( )1

    2746.8 N2

    T =

    1373 NT =

    ( )( )20: 3 280 kg 9.81 m/s 0yF T = =

    ( )1

    2746.8 N

    3

    T =

    916 NT =

    ( )( )20: 3 280 kg 9.81 m/s 0yF T = =

    ( )1

    2746.8 N3

    T =

    916 NT =

    ( )( )20: 4 280 kg 9.81 m/s 0

    y

    F T = =

    ( )1

    2746.8 N4

    T =

    687 NT =

    69

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    PROBLEM 2.68

    Solve parts b and d of Problem 2.67 assuming that the free end of the

    rope is attached to the crate.

    Problem 2.67: A 280-kg crate is supported by several rope-and-pulley

    arrangements as shown. Determine for each arrangement the tension inthe rope. (Hint: The tension in the rope is the same on each side of a

    simple pulley. This can be proved by the methods of Chapter 4.)

    SOLUTION

    Fr ee-Body Diagram of pulley

    and crate

    (b)

    (d)

    ( )( )20: 3 280 kg 9.81 m/s 0yF T = =

    ( )1

    2746.8 N3

    T =

    916 NT =

    ( )

    ( )

    20: 4 280 kg 9.81 m/s 0yF T = =

    ( )1

    2746.8 N4

    T =

    687 NT =