Why is using Newton's First Law failing me?

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Murtuza Tipu
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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∘ 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
∑ f=0
Fx *cos(360-32) -fk = 0
Fx *cos(360-32) -(mg)(24) = 0
Fx *cos(328) - 69.8544N = 0
Fx = 69.8544N/cos(328∘)
Fx = 82.3708N

That turns out to be wrong, the expected answer is fk the force of kinetic friction. Why? Shouldn't I be able to use ∑F=0 in this problem to find the answer?

I also tried to find the square root of Fx2 + Fy2 but that didn't help either.
 
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You didn't use the right friction force. Start with the definition of friction and maybe draw yourself a free body diagram.
 
Murtuza Tipu said:
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∘ 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
∑ f=0
Fx *cos(360-32) -fk = 0
I'm going to just use ##32^\circ## below; it doesn't make a difference. Your equation isn't correct. You should have
$$F_x - f_k = 0$$ or
$$F\cos 32^\circ - f_k = 0.$$ The x-component of F is ##F_x = F\cos 32^\circ##. Writing ##F_x \cos 32^\circ## doesn't make logical sense.

Why? Shouldn't I be able to use ∑F=0 in this problem to find the answer?
Yes, you can. Your thinking here is correct. It's just your execution is flawed for the reason paisello2 has hinted at.
 
You will save a lot of time and grief if you draw a free body diagram. have you done that?