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enter image description here

No friction at all anywhere. Let me explain the diagram in steps and the Q.

  1. green colour let us start with. It represent a wedge on which there is mass of 10kg (red colour). That’s it the first picture.

  2. dotted line in white represent when the block touches the bottom of the wedge.

  3. represent the dotted yellow line which represents the direction of acceleration for 10 kg mass. Which is kind of like unknown or not in a symmetry.

  4. Where the Q lies

The Dark yellow line represent displacements. Below one is AC I.e displacement of the wedge. Now , only thing in this Q what I don’t understand is that why is Displacement of mass 10kg BC even though the direction of acc for 10 kg is the dotted line.

If you wish to know why the dotted line , there is an example in my textbook which tells us that. Also , tells us that because of the dotted line kind of acceleration, we use pseudo force for solving such questions.

EDIT: Where a different concept and answer is also used. enter image description here

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  • $\begingroup$ Srijan is the wedge allowed to move? $\endgroup$
    – NMech
    Apr 29 '21 at 13:47
  • $\begingroup$ @NMech Yes. Both wedge and the block is moving. $\endgroup$
    – S.M.T
    Apr 29 '21 at 14:07
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The figure below shall answer your question.

Moving from point B to C, the sliding force $F_S = mgsin\theta = ma$,

since no friction, $a = gsin\theta$,

and $\theta = Tan^{-1}(L_{BC}/L_{AC})$

enter image description here

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  • $\begingroup$ But BC is vertical distance covered. What about the AB distance also covered ? $\endgroup$
    – S.M.T
    Apr 29 '21 at 13:21
  • $\begingroup$ Ratio of distances would not match ratio of displacement right? $\endgroup$
    – S.M.T
    Apr 29 '21 at 13:22
  • $\begingroup$ I feel you took BC because perpendiculardistscnce is always smaller. $\endgroup$
    – S.M.T
    Apr 29 '21 at 13:23
  • $\begingroup$ Also , differentiation of distance is not velocity . So , is that the reason ? $\endgroup$
    – S.M.T
    Apr 29 '21 at 13:23
  • $\begingroup$ @Srijan 1) AB = (AC^2+BC^2)^0.5, or AC/Cos(theta). 2) Why not, displacement is the distance traveled in the x & y axes, that is AC (horiz.) or BC (vert.). 3) I don't understand your question. 4) See this article for ds/dt, abyss.uoregon.edu/~js/glossary/…. $\endgroup$
    – r13
    Apr 29 '21 at 13:42
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In order to find the acceleration the process is similar to this question.

For the inertial observer the angle of the acceleration ($\phi$) with the horizontal axis will be steeper than the angle of the inclined plane ($\theta$) (See red line in the kinetic diagram).

enter image description here

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)
  • $a_W$ is the acceleration of the wedge with respect to the inertial frame. Since the wedge will be accelerating then the acceleration vector is pointing to the right.

The remaining vector $a_{A|W}$ is the acceleration vector with respect to the frame of reference of the Wedge. For a person standing on the Wedge as it is moving, then the acceleration vector $a_{A|W}$ will be parallel to the inclined plane (it will form an angle $\theta$ with the horizontal plane).

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

$$\vec{a}_A = \vec{a}_W + \vec{a}_{A|W}$$

you can add up the accelerations and obtain $a_A$. (apology for the similarities with the other question but the process is exactly the same)

The main difference know is that the motion of the wedge is dependent on the mass A, and on the angle $\theta$. So the diagrams for the wedge are

enter image description here

The resulting equations are the following $$\begin{cases} - N_{1x} = -m a_{Ax}\\ - mg + N_{1y} = -m a_{Ay}\\ N_{1x} = m a_{W}\\ - N_{1y} + N_2 - mg = 0\\ a_{Ax} = a_{Wx} + a_{(A|W)x} a_{Ay} = 0 + a_{(A|W)y} \end{cases}$$

which you can solve for the equations.

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  • $\begingroup$ Hmm K. Thanks a lot. I got it totally. $\endgroup$
    – S.M.T
    Apr 29 '21 at 17:18
  • $\begingroup$ Srijan, my result is not same as the other answer. Just saying. To be more explicit, I don't agree with the other answer. $\endgroup$
    – NMech
    Apr 29 '21 at 17:19
  • $\begingroup$ Yeah. I’m solving the final equations. But your answer looks fine to me. $\endgroup$
    – S.M.T
    Apr 29 '21 at 17:19
  • $\begingroup$ I have got the basic idea of your answer. Also , if you mean the above answer , where do you not agree ? $\endgroup$
    – S.M.T
    Apr 29 '21 at 17:23
  • $\begingroup$ I’ll be going to sleep now. Any Q of anything , ask me. Write it here and I’ll answer it in a few hours. $\endgroup$
    – S.M.T
    Apr 29 '21 at 17:25

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