A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector vector r = (2.00 m)ihat - (3.00 m)jhat + (2.00 m)khat, the force is vector F = Fxihat + (7.00 N)jhat - (5.30 N)khat and the corresponding torque about the origin is ô = (1.90 N·m)ihat + (-1.60 N·m)jhat + (-4.30 N·m)khat. Determine Fx.

Answer 1

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