Question about friction and normal force

[Supernejihh]

Homework Statement

A 3.5 kg block is pushed along a horizontal floor by a force F of magnitude 15 N at an angle of 40 degrees DOWNWARD. The coefficient of kinetic friction between the block and the floor is 0.25.

My question is about the normal force. I know the normal force is given by:

N - mg - Fsin(theta) = 0.

N = mg + Fsin(theta)

The correct normal force they obtain in the problem is from (3.5)(9.8) + 15sin(40)

I am wondering why they chose the angle as 40 when it was directed down into the 4th quadrant. Why isn't the angle then -40 degrees (negative 40 degrees) because it is in the 4th quadrant, below the horizontal.

Picture: http://www.google.com/imgres?q=a+3....=rc&dur=655&sig=111188412953169759186&page=1&

Homework Equations

N = mg + Fsin(theta)

The Attempt at a Solution

(3.5)(9.8) + 15sin(40)
and
(3.5)(9.8) + 15sin(-40)

 

[nasu]

When you choose "-" sign for the third term (first equation of your post) you already took into account the fact that the force is downwards, the angle is in the fourth quadrant etc.

 

[Supernejihh]

Thanks nasu. Much appreciated