# Frictionless Wedge

1. Jun 17, 2009

### Patta1667

1. The problem statement, all variables and given/known data

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 its acceleration.

2. Relevant equations

3. The attempt at a solution

I quick sketched up the problem in the attached picture. It seems very simple but I'm left with the wrong answer in the end when I use a test value - I'll state later. By looking at the block's position and wedge's position at a later time with respect to the start point, one can derive the following constraint equation:

$$(h-y) = (x-X) \cot(\theta) \implies -y'' = x'' - A \implies A = x'' + y''$$

where h is the height of the block.

The test value is "If A = 3g, y'' = g " and so I need to eliminate the x'' term from the constraint equation using the force diagram. Taking y and x to be positive to the north and east, we have:

$$mx'' = \frac{N}{\sqrt{2}} - mA \implies x'' = \frac{N}{m\sqrt{2}} - A$$

$$my'' = \frac{N}{\sqrt{2}} - mg \implies y'' = \frac{N}{m\sqrt{2}} - g$$

$$\frac{N}{m\sqrt{2}} = x'' + A = y'' + g \implies A = y'' - x'' + g$$

But, $$A = x'' + y''$$, so :

$$y'' - x'' + g = x'' + y'' \implies x'' = \frac{1}{2} g$$

(by this point I seriously doubt the validity of my answer, which calls x'' independent of A)

Substituting,

$$A = y'' + \frac{1}{2} g$$

Using the test value A = 3g,

$$3g = y'' + \frac{1}{2} g \implies y'' = \frac{5}{2} g$$

This does not match the book's answer of y'' = g, and my value of x'' seems intuitively meaningless. What have I done wrong with the forces?

#### Attached Files:

• ###### block.jpg
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2. Jun 18, 2009

### bucher

Well, by drawing a free body diagram of the mass "m", I'm seeing a horizontal force by the wedge and the vertical gravitational force. Since F = m*a, the two forces correlate with an acceleration. One acceleration is horizontal "A" and the other is vertical "g". The angle of the wedge doesn't matter much since there is no friction.

I may be wrong, of course, but it seems like its the answer.

3. Jun 18, 2009

### ozymandias

Smells like Kleppner & Kolenkow to me ... :)

It took me a while to zero in on it because it's so subtle.
Your horizontal force equation is the problem. There shouldn't be a -mA there. x is the distance to the mass as measured in an inertial system (and NOT from the wedge's origin), and you're also working in an inertial system, so -mA shouldn't be there.

-----
Assaf
http://www.physicallyincorrect.com" [Broken]

Last edited by a moderator: May 4, 2017
4. Jun 18, 2009

### Patta1667

Yup, I figured K & K looked like a good intro to mechanics choice. Did you encounter this book yourself or from classes?

Thanks! I see the mistake now, it works:

$$A = x'' + y''$$

$$mx'' = \frac{N}{\sqrt{2}}$$

$$my'' = \frac{N}{\sqrt{2}} - mg$$

This means:

$$x'' = y'' + g \implies A = 2y'' + g$$

Plugging in A = 3g,

$$3g = 2y'' + g \implies y'' = g$$

Last edited by a moderator: May 4, 2017
5. Jun 19, 2009

### ozymandias

I found it in the library in my first ever semester at uni.
Ah, the memories :). It's known for its extra-tough problems. If you can solve all of them, you can easily get into an honors class, I think. I personally think it's superb, but it's not easy.
Another good book to have is David Morin's book, as it has many problems with detailed solutions (although, on the other hand, I wouldn't be able to resist the temptation and check out the solutions - sometimes having no solutions is better in a sense!).

-----
Assaf
http://www.physicallyincorrect.com" [Broken]

Last edited by a moderator: May 4, 2017
6. Jun 20, 2009

### Patta1667

I'm heading into my senior year of high, but I'll be taking some sort of general physics class at a local uni in the fall so I need to prepare bigtime! I'll make sure to write down Morin's name, thanks!