Physics Inclined Question

[stitchattack]

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An older gentleman drags a wooden crate (m=10 kg) containing his childhood toys out of storage. The 120 N force F that he applies on the crate makes an angle of 30° with the horizontal. The coefficient of kinetic friction between the well-polished floor and wooden crate is experimentally determined to be 0.1. What is the frictional force f on the crate? (cos 30° = .866, sin 30° = .5)
A) 2.4 N
B) 3.8 N
C) 7.2 N
D) 9.8 N

Answer is B.

I got the answer by making an educated guess but I need help with the concept. So, to find out the frictional F, it's coefficient of kinetic friction x normal F.

To find the normal F: forces up = forces down....hence 120 x sin 30 + N = mg.

Question is, why is it sin 30 instead of cos 30? I thought with inclines, N points up while mgcos points down. I'm just confused on what the mgsin direction is suppose to be versus the mgcos. Sorry, if my question doesn't make any sense...I'm not sure how else to state it.

 

[Cawolf]

I think you are making a trig error - look at this sketch of the problem and see how the sine of the angle is the upwards component of force.

Edit: I just saw you were talking about an angle to the gravitational force. This block is on flat ground per the problem - the force is being applied at an incline.

 

[stitchattack]

@Cawolf Thank you! Omg..I made such a silly mistake thinking that it was actually on an incline instead of the ground. This makes so much more sense now.