Question

Three cables AB, AC, and AD are supporting a 70-m tall tower. The magnitude of force in each cable is 2 kN. Find the total force the cables exert on the tower (magnitude and direction angle).

Answer 1

Explanation:

The coordinates of points are:

A (0, 70, 0), B (40, 0, 0), C (40, 0, 40) D (60, 0, 60).

The position vectors for corresponding cables are:

$r_{AD} =(-60-0)i+(0-70)j+(-60-0)k\\r_{AD} =-60i-70j-60k\\r_{AC} =(-40-0)i+(0-70)j+(40-0)k\\r_{AC} =-40i-70j+40k$

$r_{AB} =(40-0)i+(0-70)j+(0-0)k\\r_{AB} =40i-70j+0k$

the unit vectors for these positions are:

$u_{AD} =r_{AD} /[r_{AD} ]=(-60/110)i-(70/110)j-(60/110)k\\=-0.5455i-0.6364j-0.5455k$

$u_{AC} =r_{AC} /[r_{AC} ]=-(40/90)i-(70/90)j+(40/90)k\\=-0.444i-0.778j+0.444k$

$u_{AB} =r_{AB}/[ r_{AB} ]=(40/80.6)j-(70/80.6)j+0k\\=0.4963i-0.8685j+0k$

The factors are:

$F_{AB} =[F_{AB} ]u_{AB} =0.9926i-1.737j+0k\\\\F_{AC}= [F_{AC]} u_{AC} =-0.8888i-1.5556j+0.8888k$

$F_{AD} =[F_{AD}] u_{AD} =-1.0910i-1.2728j-1.0910k$

so The resultant force exerted on tower by cables are:

$F_{R} =F_{AB}+ F_{AC}+ F_{AD} \\=-0.9875i-4.5648j-0.2020k\\magnitude=[F_{R} ]=4.674kN$