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.

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