A factory worker pushes a 29.7kg crate a distance of 5.0m along a level floor at constant velocity by pushing downward at an angle of 32$^\circ$ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.24.
What magnitude of force must the worker apply to move the crate at constant velocity?
So because constant velocity \begin{align}\sum F&=0\\ F_x\times \cos{(360^\circ-32^\circ)} - f_k &= 0\\ F_x\times \cos{(360^\circ-32^\circ)} - (mg)\times 0.24 &= 0\\ F_x\times \cos{(328^\circ)} - 69.8544\,\mathrm N &= 0\\ F_x &= 69.8544\,\mathrm N/\cos{(328^\circ)}\\ F_x&=82.3708\,\mathrm N \end{align}
That turns out to be wrong, the expected answer is $f_k$ the force of kinetic friction. Why? Shouldn't I be able to use $\sum F=0$ in this problem to find the answer?
I also tried to find $\sqrt{F_x^2+F_y^2} -f_k$ but that didn't help either.