Question

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

Answer 1

Answer:

$F_{AB} = 172.1356\\F_{AC} = 258.2033\\F_{AD} = 368.8004$

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:

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

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:

$F_{AB} = 172.1356\\F_{AC} = 258.2033\\F_{AD} = 368.8004$