Maximizing Friction Coefficient for Block on a Wedge: A Calculus Approach

[Silicon-Based]

Homework Statement

A block of mass m is placed on a rough wedge inclined at an angle α to the horizontal, a distance d up the slope from the bottom of the wedge. The coefficient of kinetic friction between the block and wedge is given by µ_0x/d, where x is the distance down the slope from the starting point. Calculate the maximum value of µ_0 which will allow the block to reach the bottom of the wedge.

2. The attempt at a solution
After drawing a diagram and resolving the forces I found the acceleration of the block in terms of x:
$$a(x)=g(\sin(\theta)-x\frac{µ_0}{d}\cos(\theta))$$
I don't know how to proceed further, given that the acceleration is defined over x rather than t, which prevents me from simply integrating this expression. I suspect the chain rule could be useful (a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}), but here again, I don't know how x depends on t.

 

[haruspex]

Homework Statement

A block of mass m is placed on a rough wedge inclined at an angle α to the horizontal, a distance d up the slope from the bottom of the wedge. The coefficient of kinetic friction between the block and wedge is given by µ_0x/d, where x is the distance down the slope from the starting point. Calculate the maximum value of µ_0 which will allow the block to reach the bottom of the wedge.

2. The attempt at a solution
After drawing a diagram and resolving the forces I found the acceleration of the block in terms of x:
$$a(x)=g(\sin(\theta)-x\frac{µ_0}{d}\cos(\theta))$$
I don't know how to proceed further, given that the acceleration is defined over x rather than t, which prevents me from simply integrating this expression. I suspect the chain rule could be useful (a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}), but here again, I don't know how x depends on t.

If you rewrite the equation as $\ddot x+Bx+C=0$, does it remind you of anything?
Alternatively, use that chain rule, but remember that dx/dt is v.

 

[Silicon-Based]

Update:

If I were to integrate the acceleration with respect to time I would get:
$$v(x,t)=g(\sin(\theta)-x\frac{\mu_0}{d}\cos(\theta))t$$ Knowing that the maximum occurs when v(x=d)=0, this equation gives \mu_0=\tan(\theta).

Alternatively, the chain rule gives a=\frac{dv}{dx}v, so $$\int a dx=\int v dv$$ Again, applying the above condition, I end up with \mu_0=2g\tan(\theta). Which one is correct?

 

[haruspex]

Update:

If I were to integrate the acceleration with respect to time I would get:
$$v(x,t)=g(\sin(\theta)-x\frac{\mu_0}{d}\cos(\theta))t$$ Knowing that the maximum occurs when v(x=d)=0, this equation gives \mu_0=\tan(\theta).

Alternatively, the chain rule gives a=\frac{dv}{dx}v, so $$\int a dx=\int v dv$$ Again, applying the above condition, I end up with \mu_0=2g\tan(\theta). Which one is correct?

Neither.
The first is wrong because you cannot integrate x wrt t merely by multiplying it by t. x depends on t.
The second cannot be right because it is dimensionally wrong. You have a dimensionless number on the left and an acceleration on the right. But the method looks ok, so please post your working.