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# When the particle has the position vector r = (2.00 m)i − (3.00 m)j + (2.00 m)k, the force is F = Fxi + (7.00 N)j − (7.10 N)k?

A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r = (2.00 m)i − (3.00 m)j + (2.00 m)k, the force is F = Fxi + (7.00 N)j − (7.10 N)k and the corresponding torque about the origin is vector t = (7.30 N · m)i + (6.60 N · m)j + (2.60 N · m)k.

Determine Fx.

Maybe it is the wording of the problem but I have no idea what to do, please help!

### 1 Answer

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- NCSLv 72 months agoFavorite Answer
τ = r x F

where τ, r and F are vectors

and x represents the cross product

so here

(7.30, 6.60, 2.60) = (2.00, -3.00, 2.00) x (F, 7.00, -7.10)

When I plug that into wolfram alpha, I get

F = -3.80 (N)

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