B Confusion with orientation of coordinate axis in inclined plane

[rudransh verma]

In that problem the constraint is different. As the incline moves too, the relationship between the components of the acceleration of the block is different in the frame attached to the ground.

First of all to be clear we do measure everything relative to ground in Newtonian mechanics. Right?
Is it $\tan \theta=\frac{a_1\sin \theta}{a_1\cos \theta-a_2}$ where $a_1$ is acceleration along the incline and $a_2$ the horizontal one?

 

[nasu]

First of all to be clear we do measure everything relative to ground in Newtonian mechanics. Right?
Is it $\tan \theta=\frac{a_1\sin \theta}{a_1\cos \theta-a_2}$ where $a_1$ is acceleration along the incline and $a_2$ the horizontal one?

We measure it in respect to any convenient reference frame. The ground is quite common but it is nothing special about it. But in order to apply Newton's laws without "fictitious" forces the reference frame must be inertial. The ground is pretty good aproximation of inertial frame. It does not men that non-inertial frame are not useful in solving problems. Sometimes is quicker to solve in a non-inertial frame.

 

[rudransh verma]

We measure it in respect to any convenient reference frame. The ground is quite common but it is nothing special about it. But in order to apply Newton's laws without "fictitious" forces the reference frame must be inertial. The ground is pretty good aproximation of inertial frame. It does not men that non-inertial frame are not useful in solving problems. Sometimes is quicker to solve in a non-inertial frame.

To make it easier for you the problem is like this. The wedge is initially and rest and the moving block hits it setting it into motion. The block moves up with decreasing velocity and at the top both the wedge and the block have same velocity. The constraint net acceleration's angle I calculated is shown above.Tell me if I am right?

 

[nasu]

To make what easier for me?

 

[rudransh verma]

To make what easier for me?

You did not tell me whether the relationship between the components of the acceleration in moving incline case is right. I thought you did not get what the problem is.
In post #31 is the expression right?

 

[nasu]

If a₂ is non-zero the relationship is not true.
If a₂=0 you just have the definition of the tangent function.
So it is either wrong or useless, depending on the values of a₂. You don't need a diagram to realize this.

 

[rudransh verma]

If a2 is non-zero the relationship is not true.

Then what is the constraint in this case?
“A block of mass m is pushed towards a movable wedge of mass ηm and height h, with a velocity u. All surfaces are smooth. The minimum u for which the block will reach the top of the wedge is”
How is the constraint extracted from the question?

 

[rudransh verma]

Do you see why it is independent of the coordinate axes we use?

Yeah! Because it’s part of the constraint.

 

[gmax137]

Yeah! Because it’s part of the constraint.

What I was thinking is, the normal force is by definition the force perpendicular to the surface. That's what "normal" means in this context. So the normal force is equal and opposite to the component of gravity perpendicular to the surface. This is unrelated to what axes you subsequently choose for the analysis.

The notion that the inclined plane is "hard" such that the mass doesn't sink into the plane is in my opinion a "given" in problems such as these. Calling it a "constraint" is probably correct but to dwell on that idea overmuch misses the point of these problems - namely to correctly identify the forces, create the free body diagram, and practice trigonometry skills.

 

[nasu]

So the normal force is equal and opposite to the component of gravity perpendicular to the surface. This is unrelated to what axes you subsequently choose for the analysis.

But this is not true if the incline is mobile. This is why is better to see where the "rule" comes from rather than applying it blindly as an universal condition. It's not important what you call it but it's important to realize that it something you need besides N's laws to solve the motion.

 

[gmax137]

But this is not true if the incline is mobile.

Ahh, OK, I was not thinking about the mobile incline problem. That one is certainly a level of sophistication up from the current thread, at least as it started.