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 = (1.00 N · m)i hat + (0.400 N · m)j + (−0.400 N · m)k. Determine Fx. -I know that you have to find the vector components of F x r, and I did for j and k since i cancels out. But I keep getting the wrong answer. Please provide the solution and the correct answer. I only have three attempts left.

Answer 1

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."

$F_{x}=-1N.m$

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

$T=r*F$

Where

$F=F_{x}i+(7N)j-(5N)k$  and the position vector

$r=(2m)i-(3m)j+(2m)k$

using the determinant method to expand the cross product in order to determine the torque we have

$\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]$

by expanding we arrive at

$T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k$

since we have determine the vector value of the toque, we now compare with the torque value given in the question

$(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k$

if we directly compare the j coordinate we have

$10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m$