Simple friction problem? Maybe not (multiple inclines)

[I'msorry]

Hello everyone!

Homework Statement

A weightless, friction-less pulley and rope system connect two blocks, called A and B. A has a mass of 2.2 kg, and B has a mass 4.0 kg. Both blocks are on a scalene triangle, with the pulley located at the triangle's apex. Block A is on an incline of 51 degrees, and block b is on an incline 21 degrees. The coefficient of kinetic friction between each block and the surface is 0.30.

Here is the visual of the problem: http://i.imgur.com/k8YSshe.jpg

Now, so far, so good. Here is where things get tricky:

If mA moves up, and mB moves down, determine their acceleration.

Homework Equations

Force (friction) = coefficient of friction * Normal force
ƩF = ma

The Attempt at a Solution

I assigned the conventional "tilted" x-y axis, with the x-axis being parallel to the incline under block B.

X-direction for block B:

mass*g*sin21' - co of fr*cos21'*mass*g =
14.048 - 10.979 = 3.069

X-direction for block A: (all negative because going against movement, as defined in the problem)(also equal to tension)
-(mass*g*sin51') - co of fr*cos51'*mass*g =
-16.755 -4.07 = -20.825


Which can't possibly be correct. To be honest, I've never done a problem with two different inclines like this before, but the numbers just seem off to me. I have absolutely no idea what to do from here, which means I'm probably grievously wrong somehow.

This problem has made me so confused. Thank you for even considering helping me!

 

[Simon Bridge]

I assigned the conventional "tilted" x-y axis, with the x-axis being parallel to the incline under block B.

pointing the +x-axis for A "up" the incline and the +x-axis for B "down" the incline?
You did free-body diagrams for these?

mass*g*sin21' - co of fr*cos21'*mass*g =

... this is not all the forces - what about the tension in the rope? You also left off the =ma part. Same with block A.

Notice that both blocks must have the same acceleration.
Do your algebra as far as possible without subbing the numbers in.

 

[I'msorry]

Sorry, I should have been more clear with my axis: + is down the incline.

To make my thought process more clear (well, hopefully):

Block a: massa*g*sin21' - co of fr*cos21'*massa*g - Ft= massa*a
Block b: -(mass*g*sin51') - co of fr*cos51'*mass*g + Ft = ma
Rearrange block b:
Ft = (mass*g*sin51') + co of fr*cos51'*mass*g + massb*a

Plug in:
massa*g*sin21' - co of fr*cos21'*massa*g - (massb*g*sin51') - co of fr*cos51'*massb*g + massb*a= massa*a
3.069 - 20.825 -2.2a = 4a

Which is a negative acceleration.

 

[Simon Bridge]

Sorry, I should have been more clear with my axis: + is down the incline.

Still unclear: you have two inclines - pointing in opposite directions.
Did you put the x-axis pointing down the inclines in both cases?

 

[I'msorry]

Yes, I can see how that is confusing! For block B, positive is down the incline. For block A, positive is up the incline, since (well, this is my reasoning) it is attached by a pulley and I am only using it to find the tension.

 

[Simon Bridge]

All right - then, provided you have not made an arithmetic error, what is the negative acceleration telling you?
Notice that the problem statement does not tell you that the acceleration will be in the +x direction.

 

[I'msorry]

Aaaaaand it's negative. Thank you for your help!

 

[Simon Bridge]

If this is a long-answer, then there is usualy some credit given for commenting on what the numerical answer means.

If the acceleration is negative but the motion is positive - what is happening?