Question

The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire

Answer 1

`F_(AB) = 172.1356\\F_(AC) = 258.2033\\F_(AD) = 368.8004`

Step-by-step explanation:

Using the diagram (see attachment) we extract the following position vectors:

`Vector (OA) = 6i + 0j + 0k \\Vector (OB) = 0i + 3j + 2k \\Vector (OC) = 0i - 2j + 3k`

Next step is to find unit vectors
`u_(AB) ,u_(AC), u_(AD), u_(AE)` as follows:

`u_(AB) = (vector(AB))/(magnitude(AB)) \\= \frac{OB - OA}{magnitude({vector(OB - OA))} }\\=(-6i +3j+2k)/(√(6^2 + 3^2+2^2) ) \\\\=-0.857 i +0.429j+0.286k\\\\u_(AC) = (vector(AC))/(magnitude(AC)) \\= \frac{OC - OA}{magnitude({vector(OC - OA))} }\\=(-6i -2j-3k)/(√(6^2 + 2^2+3^2) ) \\\\=-0.857 i -0.286j+0.429k\\\\u_(AD) = +1i\\u_(AC) = -1k`

Using the diagram we find the corresponding vectors Forces:

`F_(AB) = F_(AB) i + F_(AB)j +F_(AB)k\\F_(AC) = F_(AC) i + F_(AC)j +F_(AC)k\\F_(AD) = F_(AD) i + F_(AD)j +F_(AD)k\\W = -160 k`

Equation of Equilibrium:

`Sum of forces = 0\\F_(AB). u_(AB) + F_(AC).u_(AC) + F_(AD).u_(AD) + W = 0\\(-0.857F_(AB)i + 0.429F_(AB)j +0.286F_(AB)k) + (-0.857F_(AC)i - 0.286F_(AC)j +0.429F_(AC)k) + (+1F_(AD) i) + (-160k) = 0`

Comparing i , j and k components as follows:

`-0.857F_(AB) -0.857F_(AC) +1F_(AD) = 0\\+ 0.429F_(AB) - 0.286F_(AC) = 0\\+0.286F_(AB) +0.429F_(AC) = 160`

Solving Above equation simultaneously we get:

`F_(AB) = 172.1356\\F_(AC) = 258.2033\\F_(AD) = 368.8004`