(somewhat complex inclined plane problem) why is this wrong?

In summary: Has it moved horizontally?Suppose you mark on the wedge the initial position of the block. After the system has been moving a little... where is the block with respect to the mark on the wedge? Has it moved horizontally?In summary, the conversation discusses the problem of finding the acceleration of a wedge with a block on it in a frictionless system. Three equations have been derived, but the fourth equation is incorrect and the correct equation is a2 = asin(theta). The discussion also explores the relationship between the horizontal component of the normal force and the perpendicular acceleration, and clarifies that the block and wedge are not moving together in the horizontal direction,
  • #1
EddiePhys
131
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Note:- All surfaces given here are frictionless.
To find:-
Acceleration a

Relevant eqs:-
F = ma

Attempt at solution:-
The equations I've gotten so far:-
1) N1sin(theta) = m2a
2) m1gsin(theta) = m1a1
3) m1gcos(theta) - N1 = m1a2

So far, 3 eqs, 4 unknowns.

For the 4th eq, I did a2sin(theta) = a i.e the horizontal component of a2 = a.
However, this is wrong. The correct last eq is a2 = asin(theta) i.e. the component of a perpendicular to the plane = a2. Why is my answer wrong? I don't see the difference in relating them in these different ways.

Note:- I've verified for a fact that my answer is incorrect(a2sin(theta) = a)
 
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  • #2
Can you add a problem statement that specifies what it is that's to be calculated?
 
  • #3
gneill said:
Can you add a problem statement that specifies what it is that's to be calculated?

done
 
  • #4
EddiePhys said:
done

Try using conservation of energy and momentum. As both the wedge and the block are accelerating, it's difficult to use the forces between them.
 
  • #5
PeroK said:
Try using conservation of energy and momentum. As both the wedge and the block are accelerating, it's difficult to use the forces between them.

Tbh, this post wasn't intended to ask for ways to solve the above problem. I've already come up with 3 eqs but the 4th one I've come up with is wrong. I'm trying to understand where I went wrong and why the correct answer is...well, correct.
 
  • #6
EddiePhys said:
Tbh, this post wasn't intended to ask for ways to solve the above problem. I've already come up with 3 eqs but the 4th one I've come up with is wrong. I'm trying to understand where I went wrong and why the correct answer is...well, correct.

Analyse the motion from the reference frame of the wedge.
 
  • #7
EddiePhys said:
I did a2sin(theta) = a
The actual acceleration of the wedge is a. It has a component a2 in the direction of a2 (or the surfaces would not stay touching) and a component normal to a2.
If you try to work the other way around, yes a2 can be thought of as contributing a2 sinθ to a, but it is not the only contributor.
 
  • #8
haruspex said:
The actual acceleration of the wedge is a. It has a component a2 in the direction of a2 (or the surfaces would not stay touching) and a component normal to a2.
If you try to work the other way around, yes a2 can be thought of as contributing a2 sinθ to a, but it is not the only contributor.

What are the other contributors?
 
  • #9
PeroK said:
Analyse the motion from the reference frame of the wedge.

I've already solved the problem using that method. I'm just trying to solve it from an inertial reference frame too.
 
  • #10
EddiePhys said:
What are the other contributors?
a has a component normal to a2.
The point is that the relationship between a and a2 is determined by the plane of conact between the mass and the wedge. For the mass and the wedge to remain in the same relationship in that plane the two accelerations must have the same component normal to it. a2 is entirely normal to that plane, whereas only a component of a is normal to it.
 
  • #11
EddiePhys said:
I've already solved the problem using that method. I'm just trying to solve it from an inertial reference frame too.

The point is that ##a_2' = 0##, where ##a_2'## is the normal acceleration of the block in the wedge's frame. Hence, in the inertial frame ##a_2 = A_2## where ##A_2## is the acceleration of the wedge in the normal direction. Hence ##a_2 = asin \theta## (by resolving ##a## into normal and tangential components).
 
  • #12
haruspex said:
a has a component normal to a2.
The point is that the relationship between a and a2 is determined by the plane of conact between the mass and the wedge. For the mass and the wedge to remain in the same relationship in that plane the two accelerations must have the same component normal to it. a2 is entirely normal to that plane, whereas only a component of a is normal to it.

The horizontal component of the normal force due to the block is the net force causing wedge to accelerate towards the right so why is saying that the horizontal component of the perpendicular acceleration = the acceleration of the wedge? Don't they both HAVE to be equal if they're moving together?
 
  • #13
EddiePhys said:
The horizontal component of the normal force due to the block is the net force causing wedge to accelerate towards the right so why is saying that the horizontal component of the perpendicular acceleration = the acceleration of the wedge? Don't they both HAVE to be equal if they're moving together?
We were discussing accelerations, not forces.
Yes, the horizontal component of the normal force corresponds to the acceleration a, but what is its relationship to acceleration a2?
 
  • #14
haruspex said:
We were discussing accelerations, not forces.
Yes, the horizontal component of the normal force corresponds to the acceleration a, but what is its relationship to acceleration a2?

Well, that's what I said.
The horizontal component of the normal force due to the block is the net force causing wedge to accelerate towards the right so why is saying that the horizontal component of the perpendicular acceleration = the acceleration of the wedge wrong? Don't they both HAVE to be equal if they're moving together?
 
  • #15
EddiePhys said:
why is saying that the horizontal component of the perpendicular acceleration = the acceleration of the wedge? Don't they both HAVE to be equal if they're moving together
They are not moving together in the horizontal direction. They are moving together in the direction perpendicular to the plane of the wedge.
 
  • #16
haruspex said:
They are not moving together in the horizontal direction. They are moving together in the direction perpendicular to the plane of the wedge.

Wait, what? How are they not moving together in the horizontal? If they were not, wouldn't the block leave the surface of the wedge?
 
  • #17
EddiePhys said:
Wait, what? How are they not moving together in the horizontal? If they were not, wouldn't the block leave the surface of the wedge?
Suppose you mark on the wedge the initial position of the block. After the system has been moving a little while, where is the block in relation to the mark? Have both moved the same horizontal distance?
 
  • #18
haruspex said:
Suppose you mark on the wedge the initial position of the block. After the system has been moving a little while, where is the block in relation to the mark? Have both moved the same horizontal distance?

Oh yeah. The block's net acceleration will actually be towards the left. That clears things up, thanks!
 

FAQ: (somewhat complex inclined plane problem) why is this wrong?

Why is there a discrepancy between the theoretical and experimental values for the inclined plane problem?

The discrepancy between theoretical and experimental values for the inclined plane problem can be due to various factors such as air resistance, friction, and measurement errors. Air resistance can affect the motion of the object on the inclined plane, leading to a difference in the final velocity. Friction between the object and the inclined plane can also cause a difference in the acceleration of the object, resulting in a discrepancy between the theoretical and experimental values. Additionally, measurement errors can also contribute to the difference as they can affect the accuracy of the data collected during the experiment.

How can we improve the accuracy of our results for the inclined plane problem?

To improve the accuracy of results for the inclined plane problem, it is essential to reduce the effects of external factors such as air resistance and friction. This can be achieved by conducting the experiment in a controlled environment, such as a vacuum chamber, where air resistance is eliminated. Additionally, using lubricants on the inclined plane can reduce friction and ensure smoother motion of the object. It is also crucial to take multiple measurements and calculate the average to minimize the impact of measurement errors.

Is it possible to have a negative acceleration in the inclined plane problem?

No, it is not possible to have a negative acceleration in the inclined plane problem. According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In the inclined plane problem, the net force acting on the object is always in the direction of motion, leading to a positive acceleration. Therefore, a negative acceleration is not physically possible in this scenario.

Can the angle of inclination affect the results of the inclined plane problem?

Yes, the angle of inclination can affect the results of the inclined plane problem. As the angle of inclination increases, the gravitational force acting on the object also increases, leading to a higher acceleration. This can result in a difference between the theoretical and experimental values. Additionally, a steeper angle of inclination can also increase the effects of air resistance and friction, further impacting the results.

How does the mass of the object affect its motion on the inclined plane?

The mass of the object plays a crucial role in its motion on the inclined plane. According to Newton's second law of motion, the acceleration of an object is inversely proportional to its mass. This means that a heavier object will have a lower acceleration compared to a lighter object on the same inclined plane. Therefore, the mass of the object can affect the results of the inclined plane problem, and it is essential to keep it constant during the experiment to minimize discrepancies.

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