A 1,800 kg car is parked on a road that has an elevation angle of 7°. Suppose the coefficient of static friction of the kinds of rubber and asphalt involved is 0.65. Which is approximately the force of static friction between the tires and the road?
Question options:

2,200 N

9,300 N

11,350 N

18,000 N

I got A, but I was told I was wrong.

Question

A 1,800 kg car is parked on a road that has an elevation angle of 7°. Suppose the coefficient of static friction of the kinds of rubber and asphalt involved is 0.65. Which is approximately the force of static friction between the tires and the road?
Question options:

2,200 N

9,300 N

11,350 N

18,000 N

I got A, but I was told I was wrong.

fandude727 · Answer

B?

anonymous · Answer

How did you get that @Fandude727 ?

fandude727 · Answer

wait

fandude727 · Answer

F = 1800kg x 9.80N/kg x sin7° F = 2150.0N 
I think A is right

anonymous · Answer

Haha @Fandude727 That's exactly what I thought, but I was told I was wrong!

fandude727 · Answer

but it doesn't match any of the options

anonymous · Answer

Yep, and 2200 isn't correct either. :(

fandude727 · Answer

Hmm,

fandude727 · Answer

Well, I'm completely stumped

anonymous · Answer

@Fandude727 well thank you for trying :) I'm very stumped too haha!

fandude727 · Answer

You're welcome.

fandude727 · Answer

can you retake that question?

anonymous · Answer

Hmm? I believe so, and I really just want to know where I went wrong.

fandude727 · Answer

So do I.

fandude727 · Answer

@Artkid101 My brother says it's D.

anonymous · Answer

@Fandude727 How did he get that? Do you know?

fandude727 · Answer

No, he sat down at the computer for like 10 minutes and said it was D.

anonymous · Answer

@Fandude727 Thank you so much, and thank you to your brother! :)

fandude727 · Answer

You're welcome, I'm glad you got it to work!

anonymous · Answer

It was c :)

fandude727 · Answer

Awesome, glad to help you! :D