Determine normal force and angular acceleration of disk

[JJBladester]

Homework Statement

A 10kg disk contacts an inclined surface. A 11Nm couple M is applied. The coefficient of friciton is 0.4. Determine the normal force at point D, the angular acceleration $\alpha$, and the force of the rod C.

Answers:
N=115.9 Newtons
$\alpha$=2.25rad/s²
C=77.19 Newtons

Homework Equations

$$\sum F_x=m\bar{a}_x$$

$$\sum F_y=m\bar{a}_y$$

$$\sum M_G=\bar{I}\alpha$$

$\bar{x}$ and $\bar{y}$ are the coordinates of the mass center (G) of the disk.

The Attempt at a Solution

First, I drew a free-body diagram which you can see if you click the image above.

From the FBD, my first goal is to find the normal force (N) at point D.

I'm having a hard time doing that because I end up with two equations and 3 unknowns.

I don't konw N, C, or a.

$$\sum F_x=\mu _kNcos(30)+C-Ncos(30)=macos(30)$$

$$N(\mu _k-1)+C=ma$$

$$\sum F_y=\mu_kNsin(30)-mg+Nsin(30)=masin(30)$$

$$-2N=-mg-C$$

$$N=\frac{mg+C}{2}$$

So... I have an expression for N, but it includes another unknown, C (the force of the rod on the disk).

________________________
EDIT:

Figured it out... I needed to set both (ma)'s equal to zero. Also, I did the sin and cos for the frictional force in reverse.

 

[jbsteele]

\sum F_x=C-Ncos(30)-\mu _kNcos(30)=0\sum F_y=Nsin(30)-\mu_kNsin(30)-mg=0N(\mu _k-1)+C=0N=\frac{-C}{\mu_k-1}N=\frac{-(11Nm)(0.4)}{1-0.4}=115.9NNow that I know N, I can solve for C:C=2N-mg=77.19NFinally, I use the moment equation to find \alpha\sum M_G=\bar{I}\alpha(11Nm)=(10kg)(0.5m)^2\alpha\alpha=\frac{11Nm}{(10kg)(0.5m)^2}=2.25rad/s2