# Block sliding on accelerating wedge (1 Viewer)

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#### jbunniii

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1. The problem statement, all variables and given/known data
This is problem 2.11 from Kleppner and Kolenkow, 2nd edition. I have completed the problem but I'm not 100% sure I did it right and would like to check my understanding.

A $45$ degree wedge is pushed along a table with constant acceleration $A$. A block of mass $m$ slides without friction on the wedge. Find the block's acceleration.

See attached figure.

2. Relevant equations

$F=ma$

3. The attempt at a solution
The figure specifies the directions of the coordinates: $y$ points upward and $x$ points to the right. The acceleration $A$ is also to the right. The origin $(x,y) = (0,0)$ can be taken at some arbitrary fixed location (not moving with the wedge).

I began by looking at the forces acting on the block. There are two forces. First, there is the weight of the block, $W = mg$, pointing downward. Second, there is a normal force $N$ pointed $45$ degrees upward and to the right, because of the $45$ degree wedge on which it sits. We can decompose the normal force into an upward component of $N / \sqrt{2}$ and a rightward component of $N / \sqrt{2}$.

Then in the vertical direction, we have $N/\sqrt{2} - mg = m\ddot y$.

In the horizontal direction, we have $N/\sqrt{2} = m\ddot x$. I believe this acceleration must include A. In other words, $\ddot x$ is the total horizontal acceleration of the block due to both gravity and $A$. If $A$ were excluded, then this would effectively mean that the $(x,y)$ coordinates would not be stationary but would instead be accelerating to the right by $A$. Therefore we would not have an inertial coordinate system, so Newton's laws would not be applicable. Please correct me if my reasoning is wrong - this is clearly the crux of the problem.

Assuming the above is right, we have two equations and three unknowns, and we have no explicit dependence on $A$. To get another equation including $A$, we look at how $x$ and $y$ are connected. Since the slope of the wedge is $-1$, we would have $y = -x + b$ if the wedge was not moving. $b$ is some constant depending on the location of the origin. It will go away when we differentiate, which is why we don't care where the origin is, as long as it is fixed.

The wedge is moving, so we have to take that into account. If we create a new coordinate system $x', y'$ which moves with the wedge, then $y' = -x' + b'$ is true in that coordinate system, where $b'$ is again some constant. Of course $y' = y$. The offset between $x'$ and $x$ grows quadratically with time, but we really only need to know the difference in the second derivatives: $\ddot x' = \ddot x - A$ since the $(x',y')$ coordinate system moves with the wedge and hence does not "see" the acceleration. We can now differentiate $y' = -x' + b'$ to obtain $\ddot y' = -\ddot x'$, and substituting, we end up with $\ddot y = -(\ddot x - A) = A - \ddot x$

Assuming that is right, the hard part is done and now we just need to solve the system of 3 equations and 3 unknowns:
\begin{align} \frac{N}{\sqrt{2}} - mg &= m\ddot y\\ \frac{N}{\sqrt{2}} &= m\ddot x\\ \ddot y &= A - \ddot x\\ \end{align}
This is straightforward: we end up with
$$\ddot y = \frac{A - g}{2}$$
and
$$\ddot x = \frac{A + g}{2}$$
The text gives the hint that if $A = 3g$ then $\ddot y$ should be $g$, and indeed that is the case: $\ddot y = (3g - g)/2 = g$. By the way, that's an UPWARD acceleration of $g$, which is kind of cool.

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#### jbunniii

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It is interesting to think about what it would feel like if you were sitting on the wedge in place of the block.

Consider the case $A = g$. In that case, as the answer shows, we would have $\ddot y = 0$ and hence the net vertical force on you would be zero. Does it mean you would feel weightless, at least for as long as the guy pushing the wedge can keep increasing the speed linearly?

You would also have $\ddot x = g$, so the net horizontal force is $mg$ (your nominal weight). Thus you would feel the full effect of the horizontal acceleration of the wedge.

If $A = 2g$ it would be even weirder: $\ddot y = g/2$ so you're rising, and $\ddot x = 3g/2$ so you're no longer experiencing the full horizontal acceleration of the wedge.

Interesting also to note that the normal force is
$$N = \sqrt{2}m\ddot x = m \frac{A + g}{\sqrt{2}}$$
which grows linearly with $A$ as expected, but the real fun would happen if $A< -g$ (i.e. the acceleration is in the leftward direction with a magnitude of $g$ or higher) in which case you lose contact with the wedge altogether.

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#### jbunniii

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Oops, scratch that comment about $\ddot y = 0$ meaning you would feel weightless. You also have $\ddot y = 0$ when standing on a horizontal floor. I guess you would also need zero vertical normal force ($N = 0$ in addition to $\ddot y = 0$) in order to feel weightless. That won't happen in this problem regardless of what $A$ is.

#### haruspex

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The OP all looks correct. There is an easier way to get the third equation. Just consider the component of acceleration normal to the wedge. This must equal A/sqrt(2).

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#### jbunniii

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The OP all looks correct. There is an easier way to get the third equation. Just consider the component of acceleration normal to the wedge. This must equal A/sqrt(2).
I had to think about this for a few minutes, but now I see what you mean. The block can only move along the surface of the wedge, so it can't add any acceleration of its own in the direction normal to the wedge. Thus the acceleration (of both the block and the wedge) in that direction is $A/\sqrt{2}$ as you said.

On the other hand, the acceleration of the block is $(\ddot x, \ddot y)$, and the projection/component of this acceleration in the direction normal to the wedge is the inner product with the unit vector in that direction: $(\ddot x, \ddot y) \cdot (1/\sqrt{2}, 1/\sqrt{2}) = \ddot x / \sqrt{2} + \ddot y / \sqrt{2}.$ Setting this equal to $A/\sqrt{2}$ gives us the third equation, $\ddot x + \ddot y = A$. Very nice, thanks!

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