Solve Frictionless Wedge Homework: Find Acceleration

In summary, a 45 degree wedge is pushed along a table with constant acceleration. A block of mass slides without friction on the wedge. The block's acceleration is found by deriving the constraint equation and using a test value.
  • #1
Patta1667
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0

Homework Statement



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.


Homework Equations



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:

[tex] (h-y) = (x-X) \cot(\theta) \implies -y'' = x'' - A \implies A = x'' + y'' [/tex]

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:

[tex] mx'' = \frac{N}{\sqrt{2}} - mA \implies x'' = \frac{N}{m\sqrt{2}} - A [/tex]

[tex] my'' = \frac{N}{\sqrt{2}} - mg \implies y'' = \frac{N}{m\sqrt{2}} - g [/tex]

[tex] \frac{N}{m\sqrt{2}} = x'' + A = y'' + g \implies A = y'' - x'' + g [/tex]

But, [tex] A = x'' + y'' [/tex], so :

[tex] y'' - x'' + g = x'' + y'' \implies x'' = \frac{1}{2} g [/tex]

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

Substituting,

[tex] A = y'' + \frac{1}{2} g [/tex]

Using the test value A = 3g,

[tex] 3g = y'' + \frac{1}{2} g \implies y'' = \frac{5}{2} g [/tex]

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?
 

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  • #2
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
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"
 
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  • #4
ozymandias said:
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"

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:

[tex] A = x'' + y'' [/tex]

[tex] mx'' = \frac{N}{\sqrt{2}} [/tex]

[tex] my'' = \frac{N}{\sqrt{2}} - mg [/tex]

This means:

[tex] x'' = y'' + g \implies A = 2y'' + g [/tex]

Plugging in A = 3g,

[tex] 3g = 2y'' + g \implies y'' = g [/tex]
 
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  • #5
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"
 
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  • #6
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!
 

Related to Solve Frictionless Wedge Homework: Find Acceleration

1. What is the formula for calculating acceleration in a frictionless wedge?

The formula for calculating acceleration in a frictionless wedge is a = (m1 - m2) * g * sin(theta) / (m1 + m2), where m1 and m2 are the masses of the objects on the wedge, g is the acceleration due to gravity, and theta is the angle of the wedge.

2. How do I determine the direction of acceleration in a frictionless wedge?

The direction of acceleration in a frictionless wedge is always in the direction of the applied force. In other words, if the wedge is being pushed from the right, the acceleration will be to the right. If the wedge is being pulled from the left, the acceleration will be to the left.

3. Can I use the same formula for calculating acceleration in a frictionless wedge regardless of the angle?

Yes, the formula for calculating acceleration in a frictionless wedge can be used for any angle. However, it is important to make sure that all units are consistent and that the angle is expressed in radians instead of degrees.

4. Is it necessary to consider the mass and angle of the objects on the wedge when calculating acceleration?

Yes, the mass and angle of the objects on the wedge are important factors in calculating acceleration. The mass affects the magnitude of the acceleration, while the angle affects the direction of the acceleration.

5. Can I use this formula to calculate acceleration in a frictionless wedge on Earth and other planets?

Yes, this formula can be used to calculate acceleration in a frictionless wedge on Earth and other planets. However, the value of g will be different on different planets, so it is important to use the correct value for g in the calculation.

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