Answer:
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
