Question

Problems (e) and (f) refer to the tower held in place by three cables. OOON 400 N 800 N 16 m OUTB 18 m (e) Write the 600 N force in cable DC, acting at point D on the tower, as a Cartesian vector. TDC = (f) Given the position vectors DA = 8i + 10j + 24k meters and DB =-6i+4j - 24k, determine the angle between the cables DA and DB.

Answer 1

Solution:

Co-ordinates

O (0, 0, 0) m

D (0, 0, 24) m

C (16, - 18, 0) m

Unit vector along DC :

Upc = (16-0)i + (-18 - 0)j + (0-24)k (16 - 0)2 + (-18 –0)2 + (0 – 24)2

Upc = 0.470588235 i - 0.529411764 j - 0.705882352 k

(e) Force in cable DC in Cartesian vector :

TDC = Tp.upa

Tpc = 600 x 0.470588235 i – 0.529411764 ; -0.705882352 k

Tpc = 282.3529412 i - 317.6470588 ; - 423.5294118 k N

(f) Angle between DA and DB

DA = 8i + 10j + 24k

DB = -6i + 4j - 24k

Angle (cos ) = DADB DADBI

cos = - (81 + 10j + 24k).(-6i + 4k – 24k) (82 +102 +242) × (-62) +42 + (-242)

cos @ = -0.856677114

A = 148.945512