Solve 3D Force Systems Homework | F_A, F_B & F_C

[sami23]

Homework Statement

A system of cables supports a bucket. The dimensions in the figure are as follows: x_1 = 4.80 ft, x_2 = 1.70 ft, y_1 = 1.10 ft, y_2 = 3.30 ft, z_1 = 2.55 ft, and z_2 = 3.10 ft. If the bucket and its contents have a combined weight of W_1 = 17.5 lb, determine F_A, F_B, and F_C, the tensions in cable segments DA, DB, and DC, respectively.

Homework Equations

vec_F = F(unit vector)
unit vector = vec_r / |r|
[tex]\Sigma[/tex]F = 0
F_DA + F_DB + F_DC + W = 0

The Attempt at a Solution

coordinates: A(4.8,0,2.55) B(1.7,0,0) C(0,3.3,3.10) D(1.7,1.1,0)

r_DA = [(4.8-1.7)i +(0-1.1)j + (2.55-0)k] / [tex]\sqrt{(3.1^2+1.1^2+2.55^2)}[/tex]
vec_F_DA = F_DA [0.74483i -0.2643j + 0.61269k]

r_DB = [(1.7-1.7)i +(0-1.1)j + (0-0)k] / [tex]\sqrt{(1.1^2)}[/tex]
vec_F_DB = F_DB [-1j]

r_DC = [(0-1.7)i +(3.33-1.1)j + (3.1-0)k] / [tex]\sqrt{(1.7^2+2.2^2+3.1^2)}[/tex]
vec_F_DC = F_DC [-0.40826i + 0.52834j +0.74448k]

In the end... my 3 equations with the Foces unknown:
1.) 0.74483F_DA - 0.4826F_DC = 0
2.) -0.2643F_DA - F_DB +0.52834F_DC = 0
3.) 0.61269F_DA + 0.74448F_DC - 17.5 = 0

I solved for F_DA in 1.) and substituted F_DC in 3.) and using back substitution in 2.) I got:
F_DA = 44.1 lb
F_DB = 28.6 lb
F_DC = 15.7 lb

but it was wrong. What am I doing wrong?

 

[collinsmark]

Hello Sami23,

In the end... my 3 equations with the Foces unknown:
1.) 0.74483F_DA - 0.4826F_DC = 0
2.) -0.2643F_DA - F_DB +0.52834F_DC = 0
3.) 0.61269F_DA + 0.74448F_DC - 17.5 = 0

Check the number above in red. It's a 'typo' of some sort.

I solved for F_DA in 1.) and substituted F_DC in 3.) and using back substitution in 2.) I got:
F_DA = 44.1 lb
F_DB = 28.6 lb
F_DC = 15.7 lb

But in both cases, with or without the typo, I'm not coming up with same solution as you. Show us your work on how you found the tensions.