Find the acceleration of a block relative to incline plane in non and inertial frames?

Question

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:

Above is the FBD I drew.

I have written some points and want to confirm them:

1. From an inertial frame. The acceleration of the block should be mg+ma?
2. 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?

EDIT:

I solved it this way now. (Different way).It is a photo. So , I hope it is clear to just read. Only thing I got wrong is that when I added a pseudo force.

My equation becomes :

N(normal by the wedge in the lift)-$m(g+a)$sin $\theta$ - ma = $m*(velocity of block wrt inclined plane = $a_v$).

$m(g+a) =N.$

$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.

Answer 1

As shown in this thread, the acceleration of a mass on an incline equals $g*sin\theta$. When the inclination is lowered with a deacceleration of -$a_{dec}$, from relativity, it means the mass is moving backward from the original frame, thus, $a = (g - (-a_{dec}))sin\theta = (g + a_{dec})sin\theta$.

Answer 2

The diagrams are the following

On the mass (lets call it A) act only the weight and the reaction from the inclined floor ($N_1$).

Also, from the kinetic diagram you see the accelerations. As you can see on the inertial Frame you have:

• $a_A$ the acceleration of A with respect to the inertial frame. (the direction is not known)
• $a_E$ is the acceleration of the elevation with respect to the inertial frame. Since the elevator is decelerating during its descent (velocity is downwards), then the acceleration vector is pointin upwards.

The remaining vector $a_{A|E}$ is the acceleration vector with respect to the frame of reference of the elevator. For a person standing in the elevator as it is moving, then the acceleration vector $a_{A|E}$ will be parallel to the inclined plane.

So you have two vectors $a_E$ and $a_{A|E}$ , that you know their directions. Therefore because of the equation:

$$\vec{a}_A = \vec{a}_E + \vec{a}_{A|E}$$

you can add up the accelerations and obtain $a_A$.

Now if you want to actually find the numerical value:

$$\begin{cases} - N_{1x} = -m a_{Ax}\\ - mg + N_{1y} = -m a_{Ay}\\ \vec{a}_A = \vec{a}_E + \vec{a}_{A|E} \end{cases}$$

However, the acceleration can be rewritten as:

$$\vec{a}_A = \vec{a}_E + \vec{a}_{A|E} = \begin{cases} a_{Ax} = a_{Ex} + a_{(A|E)x}\\ a_{Ay} = a_{Ey} + a_{(A|E)y} \end{cases} \Rightarrow \begin{cases} a_{Ax} = 0 - a_{(A|E)}\cos(\theta)\\ a_{Ay} = a_{Ey} - a_{(A|E)}\sin(\theta) \end{cases} $$

Therefore the above is:

$$\begin{cases} - \color{red}{N_{1}}\sin(\theta) = - m (- \color{blue}{a_{(A|E)}}\cos(\theta))\\ - mg + \color{red}{N}_{1}\cos(\theta) = - m (a_{Ey} - \color{blue}{a_{(A|E)}}\sin(\theta)) \end{cases}$$

The only unknowns are $\color{red}{N_{1}}$ and $ \color{blue}{a_{(A|E)}}$, which you can now solve for.