Determine the moment of the force F about the Aa axis. Express the result as a cartesian vector !

Question

Coordinates for the Aa axis ends are:  A = { 6i + 8j + 4k } meters   ;   a = { - 6i - 5j + 8k } meters

Force vector F = { 0.5i -2j +7k } kn   ;   the position of the force vector begins at { 5i - 4j }

Using the equation moment M = u * { r x F } ; r is the position vector from Aa to F.

The unit vector used is found by the position vector rAa divided by the magnitude of rAa

Please show all details to the solution of the problem; M = u dot { r cross F } ; show M in cartesian coordinates.

Comments

Although I can't answer your question, I can see that, because you submitted it twice, you are seeking a solution quickly.

I'm puzzled about U, the unit vector. Does aA represent a cross-product of a and A? It can't be a dot product because U is described as a vector, not a scalar. The cross product would then appear to produce units of square metres, and why then would U be a unit vector? Both a and A are defined as vectors, but are we defining U as the product of the magnitudes of a and A and in which direction would be the unit vector? I appreciate i, j and k are unit vectors, but I don't follow how U is a unit vector. Can you explain these points for other users?

Answer 1

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