Nothing Special   »   [go: up one dir, main page]
Scribd Logo
Upload

Welcome to Scribd!

  • 111 views

    ch2 97 100

    Uploaded by

    Gonçalo Cruzeiro
    The document provides the solution steps to determine the tensions in cables AB and AD for a problem involving a plate stabilized by three cables (AC, AB, and AD) attached at point A. It is given that the tension in cable AC is 27 lb and that the resultant force at A must be vertical. The solution finds expressions for the individual forces of each cable in terms of their tensions and unit vector directions. Setting the horizontal and vertical components of the summed forces equal to zero results in two equations, which are solved to determine the tensions in cables AB (245 lb) and AD (257 lb).

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as PDF, TXT or read online from Scribd

    ch2 97 100

    Uploaded by

    Gonçalo Cruzeiro
    111 views6 pages
    The document provides the solution steps to determine the tensions in cables AB and AD for a problem involving a plate stabilized by three cables (AC, AB, and AD) attached at point A. It is given that the tension in cable AC is 27 lb and that the resultant force at A must be vertical. The solution finds expressions for the individual forces of each cable in terms of their tensions and unit vector directions. Setting the horizontal and vertical components of the summed forces equal to zero results in two equations, which are solved to determine the tensions in cables AB (245 lb) and AD (257 lb).

    Original Title

    ch2_97_100

    Copyright

    © Attribution Non-Commercial (BY-NC)

    Available Formats

    PDF, TXT or read online from Scribd

    Did you find this document useful?

    Is this content inappropriate?

    The document provides the solution steps to determine the tensions in cables AB and AD for a problem involving a plate stabilized by three cables (AC, AB, and AD) attached at point A. It is given that the tension in cable AC is 27 lb and that the resultant force at A must be vertical. The solution finds expressions for the individual forces of each cable in terms of their tensions and unit vector directions. Setting the horizontal and vertical components of the summed forces equal to zero results in two equations, which are solved to determine the tensions in cables AB (245 lb) and AD (257 lb).

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    111 views6 pages

    ch2 97 100

    Uploaded by

    Gonçalo Cruzeiro
    The document provides the solution steps to determine the tensions in cables AB and AD for a problem involving a plate stabilized by three cables (AC, AB, and AD) attached at point A. It is given that the tension in cable AC is 27 lb and that the resultant force at A must be vertical. The solution finds expressions for the individual forces of each cable in terms of their tensions and unit vector directions. Setting the horizontal and vertical components of the summed forces equal to zero results in two equations, which are solved to determine the tensions in cables AB (245 lb) and AD (257 lb).

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    You are on page 1of 6

    PROBLEM 2.

    97
    For the semicircular ring of Problem 2.91, determine the magnitude and direction of the resultant of the forces exerted by the cables at B knowing that the tensions in cables BD and BE are 220 N and 250 N, respectively.

    SOLUTION
    For the solutions to Problems 2.91 and 2.92, we have
    TBD = (120 N ) i + (140 N ) j + (120 N ) k TBE = (120 N ) i + (150 N ) j (160 N ) k

    Then:
    R B = TBD + TBE

    = ( 240 N ) i + ( 290 N ) j ( 40 N ) k and


    R = 378.55 N

    RB = 379 N

    cos x =

    240 = 0.6340 378.55

    x = 129.3
    cos y = 290 = 0.7661 378.55

    y = 40.0
    cos z = 40 = 0.1057 378.55

    z = 96.1

    101

    PROBLEM 2.98
    To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in AB is 920 lb and that the resultant of the forces exerted at A by cables AB and AC lies in the yz plane, determine (a) the tension in AC, (b) the magnitude and direction of the resultant of the two forces.

    SOLUTION
    Have
    TAB = ( 920 lb )( sin 50 cos 40i cos 50 j + sin 50 sin 40 j) TAC = TAC ( cos 45 sin 25i sin 45 j + cos 45 cos 25 j)

    (a) R A = TAB + TAC

    ( RA ) x

    or

    =0

    ( RA ) x

    = Fx = 0:

    ( 920 lb ) sin 50 cos 40 TAC cos 45 sin 25 = 0


    TAC = 1806.60 lb TAC = 1807 lb

    (b)

    ( RA ) y

    = Fy : ( 920 lb ) cos 50 (1806.60 lb ) sin 45

    ( RA ) y

    = 1868.82 lb

    ( RA ) z

    = Fz :

    ( 920 lb ) sin 50 sin 40 + (1806.60 lb ) cos 45 cos 25 ( RA ) z


    = 1610.78 lb

    RA = (1868.82 lb ) j + (1610.78 lb ) k
    Then:
    RA = 2467.2 lb RA = 2.47 kips

    102

    PROBLEM 2.98 CONTINUED


    and cos x = cos y = cos z = 0 =0 2467.2

    x = 90.0 y = 139.2 z = 49.2

    1868.82 = 0.7560 2467.2


    1610.78 = 0.65288 2467.2

    103

    PROBLEM 2.99
    To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in AC is 850 lb and that the resultant of the forces exerted at A by cables AB and AC lies in the yz plane, determine (a) the tension in AB, (b) the magnitude and direction of the resultant of the two forces.

    SOLUTION
    Have TAB = TAB ( sin 50 cos 40i cos 50 j + sin 50 sin 40 j) TAC = ( 850 lb )( cos 45 sin 25i sin 45 j + cos 45 cos 25 j) (a)

    ( RA ) x

    =0

    ( RA ) x

    = Fx = 0: TAB sin 50 cos 40 ( 850 lb ) cos 45 sin 25 = 0


    TAB = 432.86 lb TAB = 433 lb

    (b)

    ( RA ) y

    = Fy : ( 432.86 lb ) cos 50 ( 850 lb ) sin 45

    ( RA ) y

    = 879.28 lb

    ( RA ) z

    = Fz :

    ( 432.86 lb ) sin 50 sin 40 + (850 lb ) cos 45 cos 25 ( RA ) z


    = 757.87 lb

    R A = ( 879.28 lb ) j + ( 757.87 lb ) k
    RA = 1160.82 lb RA = 1.161 kips

    cos x = cos y =

    0 =0 1160.82

    x = 90.0 y = 139.2 z = 49.2

    879.28 = 0.75746 1160.82


    757.87 = 0.65287 1160.82

    cos z =

    104

    PROBLEM 2.100
    For the plate of Problem 2.89, determine the tension in cables AB and AD knowing that the tension if cable AC is 27 lb and that the resultant of the forces exerted by the three cables at A must be vertical.

    SOLUTION
    With:
    JJJG AC = ( 45 in.) i ( 48 in.) j + ( 36 in.) k

    AC =

    ( 45 in.)2 + ( 48 in.)2 + ( 36 in.)2

    = 75 in.

    TAC = TAC AC = TAC

    JJJG AC 27 lb ( 45 in.) i ( 48 in.) j + ( 36 in.) k = AC 75 in.

    TAC = (16.2 lb ) i (17.28 lb ) j + (12.96 ) k and


    JJJ G AB = ( 32 in.) i ( 48 in.) j + ( 36 in.) k

    AB =

    ( 32 in.)2 + ( 48 in.)2 + ( 36 in.)2

    = 68 in.

    TAB = TAB AB = TAB

    JJJ G AB T = AB ( 32 in.) i ( 48 in.) j + ( 36 in.) k AB 68 in.

    TAB = TAB ( 0.4706i 0.7059 j + 0.5294k ) and


    JJJG AD = ( 25 in.) i ( 48 in.) j ( 36 in.) k

    AD =

    ( 25 in.)2 + ( 48 in.)2 + ( 36 in.)2

    = 65 in.

    TAD = TAD AD = TAD

    JJJG AD T = AD ( 25 in.) i ( 48 in.) j ( 36 in.) k AD 65 in.

    TAD = TAD ( 0.3846i 0.7385 j 0.5538k )

    105

    PROBLEM 2.100 CONTINUED


    Now R = TAB + TAD + TAD
    = TAB ( 0.4706i 0.7059 j + 0.5294k ) + (16.2 lb ) i (17.28 lb ) j + (12.96 ) k

    + TAD ( 0.3846i 0.7385 j 0.5538k )


    Since R must be vertical, the i and k components of this sum must be zero. Hence:

    0.4706TAB + 0.3846TAD + 16.2 lb = 0


    0.5294TAB 0.5538TAD + 12.96 lb = 0 Solving (1) and (2), we obtain:
    TAB = 244.79 lb, TAD = 257.41 lb

    (1) (2)

    TAB = 245 lb TAD = 257 lb

    106

    You might also like