Block tied to a fixed string on an accelerating wedge

[PeroK]

Now when I calculate the net acceleration in the ground frame, Along the incline I get this:

Acc = a - acosø

What gives rise to the extra term acosø

It's your equation, so you'd better tell us how you got that equation and what it's supposed to mean.

The full equations have already been posted in post #13.

I'm struggling to see why there are still any questions.

 

[Null_Void]

It's your equation, so you'd better tell us how you got that equation and what it's supposed to mean.

The full equations have already been posted in post #13.

I'm struggling to see why there are still any questions.

I understand your approach using the displacement of the block. It makes total sense.

But from the perspective of forces ( which i obviously understand are not a great way to solve certain problems from your post),

It seems like since the wedge moves to the left with an acceleration a, the block too moves along to the left with the same acceleration.
One component results into acosø opposite to the direction of motion of the block along the plane.
While the other component keeps it in contact with the wedge.

Thus we have,
Acc parallel to incline = a - acosø
Acc normal to incline = asinø

From my previous understanding of such problems which did not involve strings but rather only systems of blocks and wedges,
I remembered that just because the wedge accelerates horizontally with some acceleration, it doesn't mean that the block is directly affected by the motion. ( Such as having an acceleration of the same magnitude in the same direction as the wedge in the ground frame).
But since that obviously Isn't the case here and while I once again acknowledge your advice with a whole heart, I'm simply curious as to why this happens. The string obviously changes things, I'm just trying to get a deeper understanding of how.

 

[PeroK]

The string obviously changes things, I'm just trying to get a deeper understanding of how.

I don't think anything deep is going on. If the wedge moves, then the block can't stay where it is. It must slide down the wedge. If you don't see that, then I guess you don't see it.

The force of gravity on the block is important, of course - that's why there is a limit on how quickly the wedge can accelerate.

 

[kuruman]

Because the wedge is accelerating, it's tricky to calculate the contact force between the block and the wedge.

Not if you first use kinematics to find the acceleration of the block in the inertial frame as done above. Assuming a coordinate system such that $\mathbf {\hat x}$ is down the incline and $\mathbf {\hat y}$ is perpendicular and away from the incline (see figure on the right.)

From the drawing,
$$\beta =(90^{\circ}-\alpha/2) \implies
\begin{cases}\cos\!\beta=\sin(\alpha/2) \\ \sin\!\beta=\cos(\alpha/2)
\end{cases}$$The acceleration of the block relative to the ground is $$\begin{align} & \mathbf a_{\text{bg}}= a_{\text{bg}}(\cos\!\beta~ \mathbf {\hat x}-\sin\!\beta~ \mathbf{\hat y})= 2a \cos(\alpha/2)\left[\sin(\alpha/2)~\mathbf {\hat x}-\cos(\alpha/2)~ \mathbf{\hat y}\right] \nonumber \\
& = a\left[\sin\!\alpha~\mathbf {\hat x}-(1+\cos\!\alpha)~\mathbf{\hat y}\right] \nonumber
\end{align}$$
The standard FBD of a block on an incline yields
$$\begin{align} & T=mg\sin\!\alpha+ma \sin\!\alpha \nonumber \\
& N=mg\cos\!\alpha-ma(1+\cos\!\alpha). \nonumber
\end{align}$$ This post has been edited. See posts #35 and #36.

I'm struggling to see why there are still any questions.

Assuming a fixed value of the incline angle $\alpha$, I can see the following question to be explored using the above equations.
As the acceleration $a$ is increased, which of the two occurrences below happens first and at what value of $a$?

1. The block is detached from the surface.
2. The string goes slack.

 

[Null_Void]

Not if you first use kinematics to find the acceleration of the block in the inertial frame as done above. Assuming a coordinate system such that $\mathbf {\hat x}$ is down the incline and $\mathbf {\hat y}$ is perpendicular and away from the incline,

The acceleration of the block relative to the wedge is $~\mathbf a_{\text{bw}}=2a\cos(\alpha/2)~\mathbf {\hat x}$

The acceleration of the wedge relative to the ground is $~\mathbf a_{\text{wg}}=-a\cos(\alpha)~\mathbf {\hat x}-a\sin(\alpha)~\mathbf {\hat y}$

The acceleration of the block relative to the ground is $$\mathbf a_{\text{bg}}=\mathbf a_{\text{bw}}+\mathbf a_{\text{wg}}=[2a\cos(\alpha/2)-a\cos(\alpha)]\mathbf {\hat x}-a\sin(\alpha)~\mathbf {\hat y}.$$The standard FBD of a block on an incline yields
$$\begin{align} & T=mg\sin(\alpha)+m[2a\cos(\alpha/2)-a\cos(\alpha)] \nonumber \\
& N=mg\cos(\alpha)-ma\sin(\alpha). \nonumber
\end{align}$$

Assuming a fixed value of the incline angle $\alpha$, I can see the following question to be explored using the above equations.
As the acceleration $a$ is increased, which of the two occurrences below happens first and at what value of $a$?

1. The block is detached from the surface.
2. The string goes slack.

I think you made a mistake. The acceleration of block w.r.t to ground is 2acosα/2.

 

[kuruman]

I think you made a mistake. The acceleration of block w.r.t to ground is 2acosα/2.

You are correct. I looked at the wrong equation. I will fix the error.

 

[Null_Void]

You are correct. I will fix the error.

Nevertheless, a huge Thank you for your time!

 

[kuruman]

Nevertheless, a huge Thank you for your time!

Error fixed. Thank you for spotting it.