I found that if you drag a wooden block across a piece of 150 grit sandpaper, there is more friction than if you do the same with a piece of 50 grit sandpaper. Why is this? I tried to find an answer on the Internet, but they were not specific enough.
[Physics] Why does fine sandpaper cause more friction than coarse sandpaper
frictionhomework-and-exercisesnewtonian-mechanics
Related Solutions
Your understanding that a change in rotational velocity requires a net torque (calculated around any convenient axis) is correct.
The only forces acting are gravity, the normal force from the ramp, and the force of friction between the ramp and the block. Choosing an axis across the ramp through the center of mass of the block makes life simpler; the force of gravity (all of it) acts through this point, and thus exerts no torque.
Next, consider the friction force, that is given by $mg \sin \theta$. This acts up the ramp, at the point of contact of the ramp and block, and thus exerts a torque around the center of mass depending on the dimensions of the block. (Tall blocks are more tippy than squat blocks) In this case, the torque is a clockwise one.
In order for there to be no rotation, there must be a counter-clockwise torque about the CofM. This means that the diagram is wrong; the normal force no longer is distributed uniformly along the base, and cannot be ignored as acting through the CofM. In fact, the normal force shifts towards the front of the block as the ramp slope increases, and when it reaches the front corner, the rear corner lifts and the block tumbles...
I'm guessing your FBD looked something like this, where $F_1$ is the external force you apply. I'm assuming here that the top and bottom block don't slide relative to each other, so the forces at their junction ($F_2$) are equal and opposite.
The net force on the top block is the force you apply, $F_1$, minus the frictional force the bottom block applies to the top block, $F_2$:
$$ F_{top} = F_1 - F_2 $$
Because $F_1 > F_2$ the net force $F_{top} > 0$ and the top block accelerates.
Response to comment:
If the two blocks don't slide relative to each other then their accelerations must be the same so:
$$ \frac{F_{top}}{m_{top}} = \frac{F_{bottom}}{m_{bottom}} $$
We know that $F_{top} = F_1 - F_2$ and $F_{bottom} = F_2$, so:
$$ \frac{F_1 - F_2}{m_{top}} = \frac{F_2}{m_{bottom}} $$
and a quick rearrangement gives:
$$ F_1 = F_2 \frac{m_{top} + m_{bottom}}{m_{bottom}} $$
and since $m_{top} + m_{bottom} > m_{bottom}$ this means $F_1 > F_2$.
Basically $F_2$ is only accelerating the bottom block while $F_1$ is accelerating both blocks, so $F_1$ has to be greater than $F_2$.
Best Answer
Short answer: finer grade sandpaper has more points of physical contact than rougher grade sandpaper, and each point of contact is a source of friction.