Block sliding down wedge problem

[freshcoast]

Homework Statement

Homework Equations

Conservation of energy

The Attempt at a Solution

All I'm thinking for this problem is to just use conservation of energy, since the cube is frictionless and it is just sliding down the wedge, aren't the only forces that are at play is just gravity?

I don't see any other way how the mass or the length of the wedge that can be implemented.

So far I've got

Vf = sqrt(2gH)

and since he is asking for m --> 0/ ∞ , I'm just thinking if m --> ∞ > M(mass of the wedge) well the wedge would just break?

 

[DEvens]

Well, ask this question. Can the block ever come out of contact with the wedge? That is, can the wedge ever move fast enough to the left such that the block loses contact with it?

If the block can fly then when it contacts the horizontal surface it will be doing something very different to if it can't fly.

Also, frictionless means the forces are all normal to the surfaces. So the block and the wedge have a normal force. And the wedge has a force normal to the table.

 

[consciousness]

Imagine the situation. The wedge will also move. Therefore we must consider the velocity of the wedge of conserving energy also. This means a new variable and now we need another equation.

Taking the wedge+mass the system what is the net horizontal force acting on it? What can we apply then for the system in the horizontal direction?

As the block doesn't lose contact with the wedge what does this imply for the relative velocity of the mass normal to the inclination of the wedge?

 

[freshcoast]

Taking the wedge+mass the system what is the net horizontal force acting on it? What can we apply then for the system in the horizontal direction?

For the mass+wedge system, if we set the the coordinate system to be the center of the cube with +x in the direction of the cube, the horizontal forces acting on the system would be the horizontal component of gravity from the cube, the horizontal component of gravity from the wedge and the horizontal component of the normal force in the (-x) direction. We can then apply summation of forces in the horizontal direction? but I don't see where this leads me to. Solve for acceleration?

As the block doesn't lose contact with the wedge what does this imply for the relative velocity of the mass normal to the inclination of the wedge?

the velocity would be constant? Can you re-phrase this question I don't think I'm understanding it

 

[haruspex]

There are two main approaches possible.
1. The usual free body approach using inertial frames. Introduce unknowns for all the forces and accelerations on each body and see what equations you can write down. If you have more unknowns than equations then you're missing an equation.
2. Consider the centre of mass of the whole system. This can be useful because there are no external horizontal forces. What does that tell you about the horizontal displacement of the mass centre? You will still need other equations.

 

[consciousness]

For the mass+wedge system, if we set the the coordinate system to be the center of the cube with +x in the direction of the cube, the horizontal forces acting on the system would be the horizontal component of gravity from the cube, the horizontal component of gravity from the wedge and the horizontal component of the normal force in the (-x) direction. We can then apply summation of forces in the horizontal direction? but I don't see where this leads me to. Solve for acceleration?


the velocity would be constant? Can you re-phrase this question I don't think I'm understanding it

You are considering the horizontal forces on the block. I was talking about the net horizontal force on the whole system ie the wedge AND the block. This will be ZERO as the gravitational force and the normal force are the only EXTERNAL forces acting and they are both vertical. And this implies?

You don't need to think about the velocity yet first make the two equations.

 

[freshcoast]

You are considering the horizontal forces on the block. I was talking about the net horizontal force on the whole system ie the wedge AND the block. This will be ZERO as the gravitational force and the normal force are the only EXTERNAL forces acting and they are both vertical. And this implies?

You don't need to think about the velocity yet first make the two equations.

Okay, so from what I am understanding so far, you are saying if I treat the wedge and the block as a whole system, or just one mass, that there are components of forces in the horizontal and direction, and that the external forces are just the normal and gravitational force in the vertical direction. This implies that the normal force is equal to just the force of gravity of the system.

Now, when I treat each object as just an individual system, would the normal force in that system equal to the force of gravity of mass + wedge system? Because from there I'm thinking I will be able to create an equation for acceleration of the cube.

 

[consciousness]

Okay, so from what I am understanding so far, you are saying if I treat the wedge and the block as a whole system, or just one mass, that there are components of forces in the horizontal and direction, and that the external forces are just the normal and gravitational force in the vertical direction. This implies that the normal force is equal to just the force of gravity of the system.

Now, when I treat each object as just an individual system, would the normal force in that system equal to the force of gravity of mass + wedge system? Because from there I'm thinking I will be able to create an equation for acceleration of the cube.

Yes the external forces are the normal and the gravitational force. An easy way to see this to neglect the internal forces ie the forces acting between the components of the system like the normal reaction between the wedge and the block in this case.

But the Normal force is NOT equal to the force of gravity! Some parts of the system (the block) show acceleration in the vertical direction so net vertical force isn't zero. Gravitational and normal forces aren't action-reaction pairs! That's a common misconception!

This is the reason we took the horizontal direction. The net force along it is zero. Now list the principles you can apply. This problem can be solved in many ways!

 

[sankalpmittal]

Approach 1: Follow haruspex's ideas. Use concept of centre of mass or usual free body diagrams from inertial/non inertial frames.

Approach 2: Synchronizing the concepts of conservation of linear momentum( in horizontal) and the usual law of conservation of energy (mechanical) are advantageous over here.