# Balance of a car on an inclined plane

1. Feb 22, 2012

### okh

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. The attempt at a solution
http://img259.imageshack.us/img259/5361/imagetcq.jpg [Broken]
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

Last edited by a moderator: May 5, 2017
2. Feb 22, 2012

### kushan

You are missing something :D

3. Feb 22, 2012

### okh

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

4. Feb 22, 2012

### kushan

are you getting 8454 ?

5. Feb 22, 2012

### okh

Yes, exactly.

6. Feb 22, 2012

### kushan

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

7. Feb 22, 2012

### okh

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!!!!!

8. Feb 22, 2012

### kushan

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

9. Feb 22, 2012

### ehild

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