Answer:
The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.
Determine Fx."
[tex]F_{x}=-1N.m[/tex]
Explanation:
We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.
torque=cross product of force and position . mathematically this can be express as
[tex]T=r*F[/tex]
Where
[tex]F=F_{x}i+(7N)j-(5N)k[/tex] and the position vector
[tex]r=(2m)i-(3m)j+(2m)k[/tex]
using the determinant method to expand the cross product in order to determine the torque we have
[tex]\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\[/tex]
by expanding we arrive at
[tex]T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\[/tex]
since we have determine the vector value of the toque, we now compare with the torque value given in the question
[tex](4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\[/tex]
if we directly compare the j coordinate we have
[tex]10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m[/tex]
