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21 February, 00:16

The turnbuckle is tightened until the tension in the cable AB equals 2.3 kN. Determine the vector expression for the tension T as a force acting on member AD. Also find the magnitude of the projection of T along the line AC.

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  1. 21 February, 01:29
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    a) 0.83984 i + 0.41992 j - 2.0996 k KN

    b) T_ac = 1.972888 KN

    Explanation:

    Given:

    - The tension in cable AB = 2.3 KN

    Find:

    a) Determine the vector expression for the tension T as a force acting on member AD.

    b) Also find the magnitude of the projection of T along the line AC.

    Solution:

    part a)

    - Find unit vector AB:

    vector (AB) = 2 i + j - 5 k

    mag (AB) = sqrt (2^2 + 1^2 + 5^2)

    mag (AB) = sqrt (30)

    unit (AB) = (1 / sqrt (30)) * (2 i + j - 5 k)

    - Find Tension vector:

    vector (T) = unit (AB) * 2.3 KN

    = 0.83984 i + 0.41992 j - 2.0996 k

    - The projection of T onto AC can be found from the dot product of vector T to unit vector (AC)

    - For unit vector (AC)

    vector (AC) = 2 i - 2 j - 5 k

    mag (AC) = sqrt (2^2 + 2^2 + 5^2)

    mag (AC) = sqrt (33)

    unit (AC) = (1 / sqrt (33)) * (2 i - 2 j - 5 k)

    - Compute the projection:

    T_ac = vector T. unit (AC)

    T_ac = (0.83984 i + 0.41992 j - 2.0996 k). (1 / sqrt (33)) * (2 i - 2 j - 5 k)

    T_ac = 0.2923947572 - 0.146973786 - 1.827467232

    T_ac = 1.972888 KN
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