# Block sliding on accelerating wedge

• jbunniii
In summary, the problem involves a 45 degree wedge being pushed along a table with constant acceleration A. A block of mass m slides on the wedge without friction, and the task is to find the block's acceleration. The forces acting on the block are its weight (mg) pointing downward and a normal force (N) pointing 45 degrees upward and to the right. This normal force can be decomposed into an upward component of N/sqrt(2) and a rightward component of N/sqrt(2). By analyzing the forces in the vertical and horizontal directions, the equations N/sqrt(2) - mg = m(ddot)y and N/sqrt(2) = m(ddot)x are obtained. To find another equation
jbunniii
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## Homework Statement

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.

##F=ma##

## 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.

#### Attachments

• Kleppner2.11.png
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Last edited:
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.

Last edited:
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.

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).

1 person
haruspex said:
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!

## 1. How does the acceleration of the wedge affect the block sliding on it?

The acceleration of the wedge affects the block sliding on it by changing the magnitude and direction of the net force acting on the block. As the wedge accelerates, the block experiences a changing force due to the acceleration of the wedge, which causes the block to accelerate as well.

## 2. What factors influence the friction between the block and the wedge?

The friction between the block and the wedge is influenced by the coefficient of friction between the two surfaces and the normal force acting on the block. The coefficient of friction is a measure of how rough or smooth the surfaces are, while the normal force is the perpendicular force exerted by the wedge on the block.

## 3. How does the angle of the wedge affect the motion of the block?

The angle of the wedge affects the motion of the block by changing the direction of the net force acting on the block. As the angle of the wedge increases, the force acting on the block becomes more horizontal, causing the block to accelerate more in the horizontal direction.

## 4. What is the relationship between the mass of the block and its acceleration?

The relationship between the mass of the block and its acceleration is described by Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that as the mass of the block increases, its acceleration decreases, and vice versa.

## 5. How does the presence of friction affect the motion of the block on the wedge?

The presence of friction affects the motion of the block on the wedge by opposing the motion of the block and causing it to slow down. This is because friction is a force that acts in the opposite direction of motion, and the block must overcome this force to continue moving on the wedge.

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