Calculate Magnitude of Moment in 3D

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SUMMARY

The discussion focuses on calculating the magnitude of the moment about point O due to the force acting on point A in a 3D space. The tension in cable AB is given as 1.5 kN, and the coordinates of points A and B are specified as A = (1.6, 0, 2.4) and B = (2.3, 1.4, 0). The correct calculation involves using the cross product of the position vector OA and the force vector TAB, leading to a final moment magnitude of 3.58, correcting an earlier miscalculation of 3.523.

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  • Understanding of 3D vector operations, including cross products.
  • Familiarity with the concept of moments in physics.
  • Knowledge of tension forces and their representation in vector form.
  • Ability to perform dot products and vector normalization.
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  • Explore vector normalization techniques for accurate force representation.
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Homework Statement



The turnbuckle is tightened until the tension in cable AB is 1.5 kN. Calculate the magnitude of the moment about point O of the force acting on point A.


Homework Equations


ƩMO = sqrt(Mx2 + My2 + M2)


The Attempt at a Solution



First I identified the coordinates of points A and B:

A = (1.6, 0, 2.4)
B = (2.3, 1.4, 0)

OA = r = <1.6, 1.4, 0>
AB = <0.7, 1.4, -2.4>

|AB| = 2.865

λAB = AB/|AB|
λAB = 0,2443i + 0.4877j - 0.8377k

Then I took the dot product with the tension in the cable AB and the unit vector of AB

TABAB = 0.36645i + 0.73305J - 1.25655K

ƩMO = rXTAB

(0-1.75932)i -(-2.81796)J + (1.17288)k

ƩMO = 3.523

But it says my answer is wrong and I don't know where I'm going wrong.

Suggestions would be appreciated
 

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I've just checked my working and found that I miscalculated the dot product.

<1.6, 0, 2.4> X <0.36645, 0.73305, -1.25655>

= -1.75932i + 2.88996J + 1.17288K

When I take the magnitude of this I get 3.58 which is the correct answer.
 

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