accelerationforceshomework-and-exercisesnewtonian-mechanics
The reason the block accelerates down the inclined plane is that the resultant of $N$ and $mg$ acts down the incline.
However, why is the force causing the block to accelerate down the incline often cited as the component of gravity $mg\sin\theta$ parallel to the plane.
I'm not quite sure how this works.
I'm not sure I entirely understand the question. This is what I think you're asking; please ignore the rest of this answer if I've misunderstood you.
If the plane is stationary (and I assume there is no friction) then a block on the plane will feel a force down the plane of $mg \space \sin\theta$, so it will accelerate down the plane. If we push the plane so it accelerates at the same rate as the block, then the block won't move relative to the plane. What force on the plane is required to do this?
Assuming I've understood you correctly, the mistake you've made is to assume that the force you apply to the plane, $F$, is the same as the force the block exerts on the plane.
The force $f$ that the block exerts on the plane is not the same as the force $F$ that you exert on the plane.
When you apply a force F to the inclined plane it starts accelerating at some acceleration $a$ given by $a = F/(M + m)$. If you want the block to stay still relative to the plane the acceleration $a$ along the plane must be the same as the acceleration down the plane due to gravity:
$$ a \space \cos\theta = g \space \sin\theta $$
or:
$$ \frac{F}{M + m} \space \cos\theta = g \space \sin\theta $$
so:
$$ F = (M + m)g \space \tan\theta $$
If you equate the torque about the center of the cylinder you will find that only two force can produce torque. therefore:-
$$F_t*R=F_f*R$$, and the tension is equal to the friction. Then you write the equation in x direction along the slope of the incline :-
$$F_t*cos(\theta)+F_f-m*g*sin(\theta)=o $$
here you replace friction with tension and you will have the answer in required form.
Best Answer
Since acceleration is a vector you can decompose it in the coordinate system you find convenient. If you define a cartesian coordinate system whose axis are along the normal to the plane and the plane itself you see there is a component of the acceleration $g\sin\theta$ along the plane. This is why the block accelerate in this direction. Notice that along the normal axis, $N$ cancels $mg\cos\theta$ and the block does not leave the plane.