<p>Can any1 explain about frictional, perpendicular forces and parallel forces.....and solving for each one in an inclined plane problem.</p>
<p>it all comes down to one thing: free-body diagrams.</p>
<p>Basically.
On an inclined plane, draw a mg (gravitational force). This will be your hypotenuse. Using that as your hypotenuse draw a right angled triangle. You obviously now have mgsin0 and mgcos0. mgsin0 is now your force applied or force downwards. mgcos0 is now you Fn.</p>
<p>^ truth.</p>
<p>draw and label your force vectors, hopefully you’ll get an idea.
mgsinX is acceleration due to gravity down the ramp, mgcosX is the normal force</p>
<p>the frictional force is found by (u)(Fn)</p>
<p>GAY posted before be >_></p>
<p>so suppose they didn’t give u the mu.
they just gave u the angle as 30 degrees, m = 2kg, a=2m/s^2. The block is sliding down the plane, wat is the Ff?</p>
<p>Ok.
formula:
a=net F/ m
net F = ma = 4
net F = mgsin0 - Ff
4 = 20sin30 - Ff
Ff=6</p>
<p>If you don’t know how I got net F = mgsin0 - Ff, it’s because now parallel to your plane, you have TWO forces. One is Ff, the opposite is x component of the gravitational force.</p>
<p>ahhh, thx so much jzhang722!</p>