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in Laws of motion by (48.4k points)

In the given fig, coordinates of the three cable attachment points B, C, and D are (-3.3, -4.5, 0) m, (1.1, -5.3, 1) m, and (1.6, -5.4, -1) m, respectively. What are the tensions in cables OB, OC, and OD?

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The position vectors from O to pts B, C, and D are

Dividing by the magnitudes, we obtain the unit vectors

Using these unit vectors, we obtain the equilibrium equations

From the solution of  TOA = 6774 N. Solving these equations, we obtain

TOB = 3.60 kN, TOC = 1.94 kN, TOD = 2.02 kN.

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