# Analysing a lift moving up with acceleration from the inertial frame of the ground

A frictionless inclined plane,making angle $$θ$$ with the horizontal, of length $$L$$ (inclined plane is a right-angled triangle, $$L$$ is length of hypotenuse) is kept in a lift which is moving up with an acceleration $$a$$. A block of mass $$m$$ lies on the inclined plane. Find the time the block takes to move to the bottom of the inclined plane.

Coming to the frame of the lift,I apply a force $$ma$$ downwards on the block. This gives the force component along the plane of the incline $$m(g+a)sin (θ)$$ and time taken to move down a length $$L$$ will be:-

\sqrt{\frac{2L}{(g+a) sin θ }}

My question is that if I do this working from ground frame, the acceleration the block posses is $$a$$ in vertically upward direction and $$g$$ in vertically downward direction. As the block can only move vertically up and along the plane of the incline, resolving Component of $$g$$ along the plane of incline:- $$g sin (θ)$$ down the plane of incline Resolving components of $$a$$ along the incline I have:- $$a sin θ$$ up the plane of incline So acceleration of Block along plane of incline is:- $$(g-a) sin θ$$ The only forces acting on the block along plane of the incline is that of $$mg sin θ$$

So first doubt is how does the block posses an acceleration less than $$g sin θ$$ if no other force acts on that plane Because perpendicular to the plane of incline, the equation of motion will be:-

$$N$$ - $$mg cos θ$$ = $$m a cos θ$$

Where $$N$$ is normal reaction by the surface of incline acting in the block.

Now my second question concerns with the concept of pseudo forces.

While solving the question from the frame of the life, I added a force of $$ma$$ vertically downwards.

Now I know pseudo forces are but accelerations that a non-inertial frame perceives. Basically,if I were on the lift,moving up with an acceleration $$a$$, I would add an acceleration vector of magnitude $$a$$ vertically downwards. Even on the block. So to tackle the problem by the line of reasoning, i first need the acceleration of the block with respect to the frame of ground.

I want to draw an FBD by this explaination. As I can’t realize the acceleration of the block with respect to the frame of ground, I can’t really know what it’s acceleration be from the frame of the lift and just using $$ma$$ downwards is a method, not explaination because the whole concept of pseudo forces depends upon relative motion. If I’m moving with an acceleration, then I will observe I’m at rest while everything else possesses this acceleration in the opposite direction than the one I was seen moving from the ground. I want to work this problem from this approach but these conceptual doubts block me from doing that(pun intended)