A box on a moving wedge, on an inclined plane

Lagrangian using generalized coordinates.In summary, the conversation is about finding the Lagrangian and Euler-Lagrange equations for a system involving two blocks, one at rest and one with an initial upward velocity. The initial attempt at finding the Lagrangian is deemed oversimplified, and suggestions are given to approach it using generalized coordinates and considering the constraints of the system. The potential energies are also noted to have different zero points, and it is advised to define a coordinate system and calculate velocities before writing the Lagrangian using generalized coordinates.
  • #1
Demon117
165
1

Homework Statement


The situation is indicated in the diagram below. Block m is at rest, and the block M has an initial velocity upward. Need to find the Lagrangian of this system, and then the Euler-Lagrange equations.

The Attempt at a Solution



This is the Lagrange that I came up with from the beginning, but it seems like I am oversimplifying it:

[itex]T=(1/2)m(\dot{x_{1}}^{2}+\dot{y_{1}}^{2})+(1/2)M(\dot{x_{2}}^{2}+\dot{y_{2}}^{2})[/itex]

[itex]V= mgy_{1}cos(\beta) + Mgy_{2}cos(\alpha)[/itex]

[itex]L=(1/2)m(\dot{x_{1}}^{2}+\dot{y_{1}}^{2})+(1/2)M(\dot{x_{2}}^{2}+\dot{y_{2}}^{2}) - mgy_{1}cos(\beta) - Mgy_{2}cos(\alpha)[/itex]

Like I said, this seems over simplified and I feel like there should be some cross terms in the kinetic energies. The second attempt that I have made is try to come up with velocity vectors. In this case I have the velocity of the wedge with respect to the incline [itex]v_{w}=\left(\partial _tx_2\right)\hat{i}-\left(\partial _ty_2\right)\hat{j}[/itex] and the velocity of the box with respect to the wedge, with respect to the incline [itex]v_{B}=\left(\partial _tx_2\right)\hat{i}-\left(\partial _ty_2\right)\hat{j}+\left(\partial _tx_1\right)\hat{i}'-\left(\partial _ty_1\right)\hat{j}'[/itex].


Geometry has always been an extreme weakness of mine. It seems like I am approaching this in all of the wrong ways. Any suggestions would be helpful.
 

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  • #2
You have to write your generalized coordinates at first, then you would have some cos sine dependence in the velocity. That is because the box is constrained to move in a certain way and by not introducing gen. coordinates you would simply get Newtons 2nd law.
Also your potential energies have different zero points?
Define a coordinate system
 

Related to A box on a moving wedge, on an inclined plane

1. How does the angle of the inclined plane affect the motion of the box?

The angle of the inclined plane affects the motion of the box by increasing or decreasing the force of gravity acting on the box. The steeper the angle, the greater the force of gravity and the faster the box will accelerate down the plane.

2. What is the relationship between the mass of the box and its acceleration on the inclined plane?

The relationship between the mass of the box and its acceleration on the inclined plane is inverse. As the mass of the box increases, its acceleration decreases. This is due to the increased force of gravity acting on the box.

3. How does the velocity of the wedge affect the motion of the box on the inclined plane?

The velocity of the wedge does not directly affect the motion of the box on the inclined plane. However, the velocity of the wedge does affect the angle of the inclined plane, which in turn affects the motion of the box.

4. What is the role of friction in the motion of the box on the inclined plane?

Friction plays a significant role in the motion of the box on the inclined plane. It acts in the direction opposite to the motion of the box, slowing it down and eventually bringing it to a stop. The amount of friction is dependent on the materials of the box and the inclined plane, as well as the angle of the plane.

5. How does the weight of the box on the wedge affect the normal force between the box and the inclined plane?

The weight of the box on the wedge affects the normal force between the box and the inclined plane. As the weight of the box increases, the normal force also increases, as the weight is acting perpendicular to the inclined plane. This normal force counteracts the force of gravity and allows the box to remain on the inclined plane without sliding down.

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