Help with Forces in 2 Dimensions

[MatthewBann]

Homework Statement

Jim moves a 2kg brick along the ceiling by applying a force at an angle of 24.6 degrees. If the brick moves with no acceleration and the coefficient of kinetic friction between the brick and the ceiling is 0.467, what is the magnitude of the applied force?

Homework Equations

F(parallel) = F(app)cos24.6
F(perpendicular) = F(app)sin24.6
F(normal) = F(perpendicular) - F(g)
F(g) = (2.0)(9.8) = 19.6N
F(net) = 0 = F(parallel) - F(kineticfriction)
F(kineticfriction) = (coeff-kin-fric)F(normal) = 0.467(F(app)sin24.6 - 19.6)

The Attempt at a Solution

Since there is no net force, I set the Force due to Friction equal to the Applied force and then attempted to solve for applied force. See here:

0.467(F(app)sin24.6 - 19.6) = F(app)cos24.6

Doing so, I get F(app) = -12.233N. This answer makes no sense, and I don't see my error. My students are confused and lose confidence in me when I can't solve my own problems...and it concerns me. Help, please?

 

[Dew.J]

No acceleration implies that Fk = Fx

Breakinging F down into it's components we get:
Fcos\theta = Fx
Fsin\theta = Fy

Now setting Fx = Fk = (2kg * 9.81 m/s - Fy)

Now you have three equations and three unknowns and can solve for F.

I made +y in the direction of gravity, make sure you check your convention so you don't get the directions mixed