A block can slide on a smooth inclined plane of inclination $\theta$ kept on the floor of a lift. When the lift is descending with retardation i.e. $ a \; m/s^2 $. Find the acceleration of the block relative to the in inclined plane.
How I solved till now:
I have written some points and want to confirm them:
- From an inertial frame. The acceleration of the block should be mg+ma?
- Since it asks for acceleration of $m$ relative to the incline, I can take a non inertial frame which is sitting on the inclined plane at the top. So, I have marked a person there. For that person, the lift will be at rest. So, then the acceleration of the block would be $mg sin \ \theta$. Since, I view it from NIF. Then, I add a pseudo force which will be $-ma$ (a is acceleration of lift). I have assumed that the person also has an acceleration $a m/s^2$ as given in Q (then only the person sees the lift at rest). Then, the pseudo force is always negative. So, it acts in the direction towards the NIF on the mass $m$. So, we get:
$mg sin\ \theta$ - $ma sin \ theta$ = m*(relative acc of block wrt inclined plane).
Therefore, answer is $(g-a)sin \ \theta$.
I have solved this much. But my answer is wrong. The correct answer is $g+a= sin \ \theta$
Also, if there is a way to solve it in inertial frame, it would be great to know, since most answers solve it with non inertial. Is there a way to solve with inertial frame also?
My equation becomes :
N(normal by the wedge in the lift)-$m(g+a)$sin $\theta$ - ma = $m*(velocity of block wrt inclined plane = $a_v$).
$m(g+a) $- $m(g+a)$sin $\theta$ - $ma$= $m* a_0. $
This is what I got.
So , if I don’t include the N and -ma. Then , my answer comes to be right.