# Block sliding on accelerating wedge

**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 block, then ##y’ = -x’ + b## is true in that coordinate system. 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 ##\ddot x’## 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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