The tension in cable da has a magnitude of tda=6.27 lb. find the cartesian components of tension tda, which is directed from d t

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Oliga [24] · Answer 1 · 2022-05-13T00:56:19+03:00

Complete Question

The Complete Question is attached below

We have that the Cartesian components of tension $T_{da}$ is

$T_{DA}=-4.433i-3.49j+2.735k$

From the Question we are told that

$M_{da}=6.27 lb\\\\w=9.50ft\\\\d=6.60ft\\\\h=4.50ft$

$\vec {DA}=-4.7i-3.7j+2.9k)ft$

$\vec {DB}=-1.9i-3.7j+1.9k)ft\\\\\vec {DC}=-1.9i+5.8j-1.6k)ft$

Generally the equation for $T_{DA}$ is mathematically given as

$T_{DA}=\phi_{DA}* M_{da}$

Where

$\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{(-4.7)^2+(-3.7)^2+(2.9)^2}\\\\\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}$

Therefore

$T_{DA}=\phi_{DA}* M_{da}$

$T_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}* 6.27$

$T_{DA}=-4.433i-3.49j+2.735k$

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