Incline plane, two masses, a pulley

[jimz]

Homework Statement

A string through an ideal pulley connects two blocks, one (mass=2m) is on an inclined plane with angle theta to horizontal and coefficient of kinetic friction mu. The other block (mass=m) hangs off the pulley on the high side of the inclined plane.

(looks like this, except no acceleration and the block on the plane is twice the mass of the hanging block:
http://session.masteringphysics.com/problemAsset/1010976/24/MLD_2l_2_v2_2_a.jpg)

Find the angle theta that allows the mass to move at a constant speed.

Homework Equations

First, I'm not even sure which direction the blocks will move, and I'm not going to figure it out but I suspect they can move at a constant velocity either way if we vary theta and mu. For now I'll assume that the 2m mass will move down the plane towards the earth. If that's wrong a the friction sign will be changed but the same idea here.

for the 2m mass, with +x along the plane pointing down, and y perpendicular pointing up :

\sum F_{2mx}=F_{gravityx}-F_{frictionx}-T
\sum F_{2mx}=2mg sin \theta-(\mu)(2mg cos \theta)-T

\sum F_{2my}=N-F_{gravityy}
\sum F_{2my}=N-2mg cos \theta

for the m mass, with +y in the up direction:
\sum F_{my}= T-F_{gravity}
\sum F_{my}= T-mg

The Attempt at a Solution

At a constant velocity, neither block is accelerating, so a=0

I believe all I need to do is focus on one mass, I'll try block on the plane (mass=2m) with the x direction along the inclined plane.
\sum F_{2mx}=2mg sin \theta-(\mu)(2mg cos \theta)-T=2m\ddot{x}=0

Can I then assume T=mg from the hanging mass with no acceleration?

2mg sin \theta-(\mu)(2mg cos \theta)-mg=2m\ddot{x}=0
mg(2sin \theta-(\mu)(2cos \theta)-1)=0
sin \theta-(\mu)(cos \theta)-\frac{1}{2}=0
-(\mu)(cos \theta)=\frac{1}{2}-sin \thetaAnd then I'm kind of stuck and unsure I even set it up right.

Thanks!

 

[IITian]

Is this your homework?

 

[jimz]

yes, it's a problem from our book. Does that matter?

I now think I set it up correctly and it's now just a math problem but I still can't get anywhere closer to the solution for theta.