Mg cos theta in context of incline plane

[negation]

For a mass on an incline plane, is the downward force mg cos Θ or -mg cos Θ?
I'm inclined to state there should be a negative but from my lectures, the negative does not appear to be stipulated. I was wondering if it's was a mistake by the lecturer.

mg cos Θ + F_N = 0
mg cos Θ = -F_N

the downward force must be of equal magnitude to the normal force-but opposite direction.
This makes sense mathematically.

However, geometrically, it does make sense for the down ward force to be -mg cos Θ.

Could someone clarify?

 

[ModusPwnd]

It depends! Are you defining up or down to be negative? You are free to do either. But the great thing about vectors is that it doesn't matter what coordinate frame you choose. And you have just found this out. Regardless of your choice of positive or negative you have found that the normal force must be in an opposite direction of the gravitational force. I am not sure if that clarify's completely but it should help.

 

[Doc Al]

Using a sign convention where into the plane is negative, the component of the weight normal to the surface would be -mg cosθ:
-mg cosθ + F_N = 0
F_N = mg cosθ (positive, thus out of the plane)

 

[negation]

It depends! Are you defining up or down to be negative? You are free to do either. But the great thing about vectors is that it doesn't matter what coordinate frame you choose. And you have just found this out. Regardless of your choice of positive or negative you have found that the normal force must be in an opposite direction of the gravitational force. I am not sure if that clarify's completely but it should help.

Using a sign convention where into the plane is negative, the component of the weight normal to the surface would be -mg cosθ:
-mg cosθ + F_N = 0
F_N = mg cosθ (positive, thus out of the plane)

I am defining down to be negative.

I am suspecting the lecturer committed a blunder by forgetting the negative sign.