Normal and friction forces for 1350kg car with cable tension on inclined surface

[fibersnet]

1350kg car resting on a plane surface with its brakes locked.
Unit vector $$e_n = <.231,.923,.308>$$ is perpendicular to the surface. The y-axis points upwards. The direction cosines of a cable supporting the car are $$<-.816,.408,-.408>$$ and the tension in the cable is 1.2KN. Determine the magnitude of the normal and friction forces the car's wheels exert on the surface.

This is what I have so far:

$$F_f = <.231F_x, .923F_y, .308F_z>$$
$$T_{AB} = 1200<-.816, .408, -.408>$$
$$W = -(1350)*(9.8)\hat{j}$$
$$N = |1200|<.231,.923,.308>$$

$$0 = F_f + T_{AB} + N - W$$


But when I solve the equations, I come up with N being 9719N, it is supposed to be about 2500 more.

Thanks!

 

[fibersnet]

Picture

Here is a picture of the diagram and the math.

 

[Leong]

.231N + X -.816*1200=0
.923N+Y-1350*9.8+1200*.408=0
.308N + Z -.408*1200=0
The fourth is correct.

 

[fibersnet]

Got it

Thanks, I figured this out late last night right before i went to bed.

I forgot that <fx,fy,fz> were already in the form of the frictional vector and that the equation .231Fx + .923Fy + .308Fz was just meant to relate the components, they are not the actual components.

Thanks!