5. Determine the tension developed in AB, AC, and AD cables for equilibrium of the crate. The 390-lb crate in is equilibrium.C3 ft2ft 3ft-GoB6 ft3 ftD4 ft

Answered: Determine the tension developed in AB,…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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5. Determine the tension developed in AB, AC, and AD cables for equilibrium of the crate.

The 390-lb crate in is equilibrium.
C
3 ft
2ft 3ft-
Go
B
6 ft
3 ft
D
4 ft

Expert Solution

Step 1

Here, A(0,6,0), B(-3,0,2), C(2,0,3), D(0,9,4)

Unit vectors in directions of cables will be,

${\hat{u}}_{A B} = \frac{- 3 \hat{i} - 6 \hat{j} + 2 \hat{k}}{\sqrt{{(- 3)}^{2} + {(- 6)}^{2} + 2^{2}}} = \frac{- 3}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{2}{7} \hat{k} {\hat{u}}_{A C} = \frac{2 \hat{i} - 6 \hat{j} + 3 \hat{k}}{\sqrt{2^{2} + {(- 6)}^{2} + 3^{2}}} = \frac{2}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{3}{7} \hat{k} {\hat{u}}_{A D} = \frac{0 \hat{i} + 3 \hat{j} + 4 \hat{k}}{\sqrt{0^{2} + 3^{2} + 4^{2}}} = \frac{3}{5} \hat{j} + \frac{4}{5} \hat{k}$

Tensions in cables will be given by,

$T_{A B} = T_{A B} (\frac{- 3}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{2}{7} \hat{k}) T_{A C} = T_{A C} (\frac{2}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{3}{7} \hat{k}) T_{A D} = T_{A D} (\frac{3}{5} \hat{j} + \frac{4}{5} \hat{k})$

Weight of crate will be,

$W = - 390 \hat{k}$

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