Coefficient of friction for connected masses on inclined plane with pulley

[bates.384]

Homework Statement

Two objects are connected by a light string that passes over a frictionless pulley as shown in the figure below. m1 = 8.32 kg, m2 = 6.40 kg, and = 56°. When released from rest, m1 accelerates downward at 2.210 m/s2. For this to happen, the coefficient of kinetic friction must be [blank], and to even begin sliding in the first place the coefficient of static friction must be [greater than or less than] than [blank] .

link to figure http://www.webassign.net/userimages/SerPSE8-5-p-030-phi.png?db=v4net&id=208569

Homework Equations

F=ma
Fk=μN

The Attempt at a Solution

Ok here goes. I 'solved' this with great confidence, only getting wrong answers. Hopefully you guys can pinpoint where I went wrong.

So I started with drawing my FBD's and what not, and I want to find the Tension in the string first. I set this up by using m1, since it's hanging and should be fairly straightforward.

mg-T = ma
(8.32)(9.8)-T = (8.32)(2.21)
I got T = 63.1488N

After having that I figured I would use m2's forces in the x direction to solve for μ. I set up the equation by the following:

Fxnet = Tcos34-mgsin34-μmgcos34 = ma, and here I used acos34 since that would be the acceleration in the x direction. I solved and got μ = .1068 but that is the wrong answer. I am very confused.

Thank you in advance for all help!

 

[PhanthomJay]

Homework Statement

Two objects are connected by a light string that passes over a frictionless pulley as shown in the figure below. m1 = 8.32 kg, m2 = 6.40 kg, and = 56°. When released from rest, m1 accelerates downward at 2.210 m/s2. For this to happen, the coefficient of kinetic friction must be [blank], and to even begin sliding in the first place the coefficient of static friction must be [greater than or less than] than [blank] .

link to figure http://www.webassign.net/userimages/SerPSE8-5-p-030-phi.png?db=v4net&id=208569

Homework Equations

F=ma
Fk=μN

The Attempt at a Solution

Ok here goes. I 'solved' this with great confidence, only getting wrong answers. Hopefully you guys can pinpoint where I went wrong.

So I started with drawing my FBD's and what not, and I want to find the Tension in the string first. I set this up by using m1, since it's hanging and should be fairly straightforward.

mg-T = ma
(8.32)(9.8)-T = (8.32)(2.21)
I got T = 63.1488N

this looks good

After having that I figured I would use m2's forces in the x direction to solve for μ. I set up the equation by the following:

Fxnet = Tcos34-mgsin34-μmgcos34 = ma, and here I used acos34 since that would be the acceleration in the x direction. I solved and got μ = .1068 but that is the wrong answer. I am very confused.

Thank you in advance for all help!

The sketch is way out of scale if the interior angle at the top of the incline is 56 degrees, but so be it.
It is most advantageous to choose the x-axis as the axis parallel to the incline, and the y-axis as the axis perpendicular to the incline. This avoids all sorts of headaches when using conventional horiz and vert axes. When you use the x-axis as the axis along the incline, then the x comp of the weight is mgsin34, the friction x comp is umgcos34, but your known tension and acceleration act along the incline.