(adsbygoogle = window.adsbygoogle || []).push({}); 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

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# Homework Help: Balance of a car on an inclined plane

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