Balance of a car on an inclined plane

[okh]

Homework Statement

A 900 kg car goes up along an inclined road by 30 degrees characterized by an acceleration of 0.25 m/s^2. If friction coefficient is 0.5, find the magnitude of the force that moves the car.
Solution: F = 12285.6 N

Homework Equations

The Attempt at a Solution

http://img259.imageshack.us/img259/5361/imagetcq.jpg
Sum of all forces is: m *a = 900 * 0.25 = 225
Then Ffriction is = 0.5 * FwPerpendicular = 0.5 * sin(60) * 900 * 9.8 = 3819 N
Then Fwparallel is = sin(60) * 900 * 9.8 = 4410
F = 3819 + 4410 + 225 = 8454 N != solution

 

[kushan]

You are missing something :D

 

[okh]

You are missing something :D

Do you mean in the problem statement ( ) or in my attempt?

 

[kushan]

are you getting 8454 ?

 

[okh]

are you getting 8454 ?

Yes, exactly.

 

[kushan]

From where did you get this question
because as far as I know ( you are doing correctly (hopefully)

 

[okh]

From where did you get this question
because as far as I know ( you are doing correctly (hopefully)

The teacher dictated us that problem. However I've just asked my schoolmates and the teacher said the solution was wrong and the right one was 8454!

 

[kushan]

Then why you posted this things here ( : ) ) i am still trying make a smiley smile oh god

 

[ehild]

Then Ffriction is = 0.5 * FwPerpendicular = 0.5 * sin(60) * 900 * 9.8 = 3819 N
Then Fwparallel is = sin(60) * 900 * 9.8 = 4410

F = 3819 + 4410 + 225 = 8454 N != solution

FwPerpendicular is mgcos(60).
Figure out which force acts in the direction of acceleration, up along the incline and which one acts against it.

ehild