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

Given the position vector r and torque vector τ, we can determine the force vector F by using the equation for torque: τ = r x F

First, we need to find the cross product r x F. Since we are given the torque vector and all components of r and F except Fx, we can write the cross product's components as:

τ_i = r_j * F_k - r_k * F_j
τ_j = r_k * F_i - r_i * F_k
τ_k = r_i * F_j - r_j * F_i

Now, we can plug in the given values and solve for Fx.

τ_i = (1.90 N·m) = (-3.00 m) * (-5.30 N) - (2.00 m) * (7.00 N)
τ_j = (-1.60 N·m) = (2.00 m) * Fx - (2.00 m) * (-5.30 N)
τ_k = (-4.30 N·m) = (2.00 m) * (7.00 N) - (-3.00 m) * Fx

From the second equation:
-1.60 N·m = 2.00 m * Fx + 10.60 N·m
Now we can solve for Fx:
Fx = (-12 N·m) / (2.00 m) = -6 N

Therefore, Fx = -6 N.