A uniform rectangular wood block of mass $M$, with length $b$ and height $a$, rests on an incline as shown. The incline and the wood block have a coefficient of static friction, $\mu_s$. The incline is moved upwards from an angle of zero through an angle $\theta$. At some critical angle the block will either tip over or slip down the plane. Determine the relationship between $a$, $b$, and $\mu_s$ such that the block will tip over (and not slip) at the critical angle. The box is rectangular, and $a \neq b$.
[asy]
draw((-10,0)--(0,0)--20/sqrt(3)*dir(150));
label("$\theta$",(0,0),dir(165)*6);
real x = 3;
fill((0,0)*dir(60)+(-x*sqrt(3),x)--(3,0)*dir(60)+(-x*sqrt(3),x)--(3,3)*dir(60)+(-x*sqrt(3),x)--(0,3)*dir(60)+(-x*sqrt(3),x)--cycle,grey);
draw((0,0)*dir(60)+(-x*sqrt(3),x)--(3,0)*dir(60)+(-x*sqrt(3),x)--(3,3)*dir(60)+(-x*sqrt(3),x)--(0,3)*dir(60)+(-x*sqrt(3),x)--cycle);
label("$a$",(0,0)*dir(60)+(-x*sqrt(3),x)--(3,0)*dir(60)+(-x*sqrt(3),x));
label("$b$",(3,3)*dir(60)+(-x*sqrt(3),x)--(3,0)*dir(60)+(-x*sqrt(3),x),dir(60));
[/asy]

(A) $\mu_s > a/b$
(B) $\mu_s > 1-a/b$
(C) $\mu_s >b/a$
(D) $\mu_s < a/b$
(E) $\mu_s < b/a-1$
1. **Determine the condition for slipping:**
   For the block to not slip, the static friction force must be greater than or equal to the component of the gravitational force parallel to the incline. The frictional force is given by:
   \[
   f_s = \mu_s N
   \]
   where \( N \) is the normal force. The gravitational force component parallel to the incline is:
   \[
   mg \sin \theta
   \]
   The normal force \( N \) is:
   \[
   N = mg \cos \theta
   \]
   Therefore, the condition for no slipping is:
   \[
   \mu_s N \geq mg \sin \theta
   \]
   Substituting \( N = mg \cos \theta \):
   \[
   \mu_s mg \cos \theta \geq mg \sin \theta
   \]
   Dividing both sides by \( mg \cos \theta \):
   \[
   \mu_s \geq \tan \theta
   \]

2. **Determine the condition for tipping:**
   For the block to tip over, the torque about the edge of the block must be considered. The block will tip when the line of action of the gravitational force passes through the edge of the block. The critical angle \(\theta_c\) for tipping can be found by considering the geometry of the block.

   The height \(a\) and the base \(b\) of the block form a right triangle with the incline. The block will tip when the center of mass is directly above the edge of the block. This occurs when:
   \[
   \tan \theta_c = \frac{b}{a}
   \]

3. **Compare the conditions for slipping and tipping:**
   To ensure the block tips rather than slips, the condition for tipping must occur before the condition for slipping. Therefore, we need:
   \[
   \tan \theta_c \leq \mu_s
   \]
   Substituting \(\tan \theta_c = \frac{b}{a}\):
   \[
   \frac{b}{a} \leq \mu_s
   \]
   This simplifies to:
   \[
   \mu_s \geq \frac{b}{a}
   \]

Thus, the relationship between \(a\), \(b\), and \(\mu_s\) such that the block will tip over (and not slip) at the critical angle is:
\[
\mu_s \geq \frac{b}{a}
\]

The final answer is \( \boxed{ \mu_s \geq \frac{b}{a} } \)