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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:

enter image description here 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:

enter image description hereI 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.

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  • $\begingroup$ I am on the mobile now, so I can't really draw properly and reply (I will do later), but I couldn't help asking who is "Sir Tom". Also, I would suggest that you clean up a bit the grammar of the post, to make it easier to read. $\endgroup$
    – NMech
    Apr 29 at 8:25
  • $\begingroup$ @NMech Sir Tom? $\endgroup$
    – S.M.T
    Apr 29 at 8:26
  • $\begingroup$ @NMech I edited it . I hate this auto type sometimes. $\endgroup$
    – S.M.T
    Apr 29 at 8:30
  • $\begingroup$ @NMech Np. Whenever you’re free, have a look at it. Thank you, $\endgroup$
    – S.M.T
    Apr 29 at 8:31
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    $\begingroup$ @NMech 👍🏻 Done. I think it’s better now. $\endgroup$
    – S.M.T
    Apr 29 at 8:36
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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, $a = (g - (-a_{dec}))sin\theta = (g + a_{dec})sin\theta$.

enter image description here

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  • $\begingroup$ Answer is (g+a)sin $\theta$ . $\endgroup$
    – S.M.T
    Apr 29 at 17:09
  • $\begingroup$ Actually , the left is descending . So , it means acc is decreasing. Then , second statement reads that retardation is happening. That means retardation of the descending is happening. Therefore , it mean initial acceleration is negative , but rate of change of acceleration is positive. Therefore , g+a $\endgroup$
    – S.M.T
    Apr 29 at 17:11
  • $\begingroup$ @Srijan Yeah, I missed the word "relative" in the subject title. Answer corrected. $\endgroup$
    – r13
    Apr 29 at 17:24
  • $\begingroup$ Did you get my solution I wrote using NIF or pseudo forces? $\endgroup$
    – S.M.T
    Apr 29 at 17:26
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    $\begingroup$ @Srijan I've added a sketch. In relative terms, the two mass are leaving apart with an acceleration/deacceleration a, thus it can be (a+g) or (a-g) depending on where you stand (on A or B) and looking out for the other (B or A). This fits the meaning of "pseudo" - which is used to mark something that superficially appears to be (or behaves like) one thing, but is something else. $\endgroup$
    – r13
    Apr 29 at 18:27
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The diagrams are the following

enter image description here

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.

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