Thread: Friction coefficient/ Incline/ Forces - Physics help?

A 1200kg mass rests on a track inclined at 30 degrees to the horizontal.
A force P inclined at 18 degrees TO THE TRACK (48 degrees to the horizontal) causes the carriage to move up the track at constant velocity (i.e. no acceleration). Coefficient of friction is 0.2; determine P. Use gravity as 9.8 ms^-1.
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Constant velocity means no net forces act on the mass.
Parallel forces: Gravity = (m)(g)(sin30) = 5880N
Frictional force = (0.2)(Normal force)
That's all I've got so far :/
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I wasn't taught this in my horrible school so I have only a rough idea of how to work these type of questions. Could you include all formulae used and just solve it for me? I can learn from the working.
~Thanks.

1.) Draw a picture (aka- free body diagram) to help visualize the problem.

2.) Break up the forces into necessary components (in green). Note: I have chosen the coordinate system, as indicated in the picture, to simplify the problem.

3.) Newton's Second Law (for x):

4.) Newton's Second Law (for y):

5. Substitute back into the equation derived for :

6. Solve for :

Hopefully that will clear things up! Good luck!

There was typo in part 4; last line should have sin18, not sin30.
This was great, though! Thanks SO MUCH! I was terrible at this topic, as my high-school teacher did nothing all year, so I had to try and self-teach. I think your one comment has taught me the entire thing perfectly. Rep!

Lol sorry about the typo! I'm glad it helped! If you have anymore questions feel free to send me a message, and I'll do my best to help.

Last thing, just to make sure I've got this. I'll post here in case someone else ever makes use of this. Same type of question, but the mass = 2kg, angle of incline of plane = 20 degrees, and there's a force Y on the mass that is 25 degrees TO THE PLANE. Coefficient of friction is still 0.2. Just want to make sure my math is right; is the answer approximately 10.5N? This isn't really vital, so no need to bother if you're busy.
Here's what I really want to know: if the mass is just on the verge of slipping DOWN the plane (but still stationary, therefore net force = 0), is everything is still the same, except in the equation for parallel forces you'd put:
Ycos25 - mg(sin20) + (0.2)(n) = 0?
The change is just to add friction, instead of subtracting it, right?

That's right. Friction always opposes motion, so in this case it will be pointing up the plane because the mass "wants" to move down. Another change, however, is that we no longer use the kinetic friction force, but instead, the static friction force because the block is not moving yet.

Thanks, man. You really saved the day on this one!