Motion analysis of an accelerating wedge and a block

[vanhees71]

I thought the wedge is forced to move with constant acceleration in $x$ direction. For me to analyze it using forces is pretty cumbersome, but with the Lagrange method it's pretty simple.

You can also argue easily with the equivalence principle and calculate the effective "gravitational field" in the rest frame of the wedge and then project the force on the block. You'll get something like $F_{\parallel}=-m (g\sin \alpha + a_0 \cos \alpha)$ where $\alpha$ is the direction of the wedge wrt. the $x$ axis.

 

[jbriggs444]

I will try, but before that, can you tell me if the block and the wedge momentarily lose contact, as the wedge moves towards, and then the mgcosθ makes the block accelerate in -y direction? Is that what happens? I can't visualise what happens.

I am going to try to explain it in a way that appeals to my intuition. Perhaps it will appeal to yours.

Suppose the wedge is accelerated to the right with a huge force. Yes, in that case, the wedge will be pulled out from under the block and the block will simply free-fall to the ground. This scenario is too simple to be interesting. So let's not pull the wedge that hard...

Suppose that the system is already in equilibrium at the start of the exercise with just enough normal force so that the block slides down the wedge at a steady acceleration. In this case there is no sudden jerk or loss of contact when we decide to start looking. This is the assumption that I (and I think @vanhees71) would make. We would consider continuing contact between wedge and block to be a "constraint" on the system.

But you seem to have in mind a starting point where the wedge is held motionless, the block is held motionless and we suddenly stop holding things and apply a fixed acceleration to the wedge. In this situation, the system may not start at equilibrium (*). The block may or may not be pressing as tightly into the wedge as it needs to in order to maintain the correct normal force. You could come up with a microscopic description of how this works. If the normal force is too low, the block moves closer to the wedge and is compressed. With compression, the normal force increases. But inertia carries the block past the equilibrium point and and you have too much normal force. You end up with a microscopic oscillation as the block alternately squeezes closer to the wedge and then rebounds. This oscillation would inevitably damp out. But that level of analysis is way too much work and requires detailed information we do not possess. Take it as a given that (unless the wedge gets pulled out too fast) a stable equilibrium will be attained. Observation shows that it will indeed be attained -- and very quickly.

(*) If we are holding all pieces of a system in place, the forces within the system may be "statically indeterminate". That is, there is no way to tell from the idealized description of the setup what all the forces are. For instance, we might be holding the wedge and block tightly against one another or lightly.

 

[vanhees71]

Yes sure, I made the assumption that the block is always gliding along the block. If it gets detached from the block, it's way more complicated of course!

 

[NoahCygnus]

I thought the wedge is forced to move with constant acceleration in $x$ direction. For me to analyze it using forces is pretty cumbersome, but with the Lagrange method it's pretty simple.

You can also argue easily with the equivalence principle and calculate the effective "gravitational field" in the rest frame of the wedge and then project the force on the block. You'll get something like $F_{\parallel}=-m (g\sin \alpha + a_0 \cos \alpha)$ where $\alpha$ is the direction of the wedge wrt. the $x$ axis.

Isn't ma0cosα a fictitious force? As you're introducing a fictitious force , doesn't that mean the result you'll get from the accelerating frame of wedge will be different than the actual result calculated from an inertial frame?

 

[jbriggs444]

Isn't $ma_0 \cos \alpha$ a fictitious force? As you're introducing a fictitious force , doesn't that mean the result you'll get from the accelerating frame of wedge will be different than the actual result calculated from an inertial frame?

Sure, it's a fictitious force. But Newton's second law still applies. The real physical forces that you can calculate using the accelerating frame are the same as the real physical forces that you would calculate using an inertial frame.

The accelerations that you calculate in the accelerating frame would indeed need to be adjusted (aka "transformed") before they are correct for the ground frame. But adopting the accelerating frame may simplify the analysis enough that it's worth the added effort of a transformation step at the end.

 

[NoahCygnus]

Sure, it's a fictitious force. But Newton's second law still applies. The real physical forces that you can calculate using the accelerating frame are the same as the real physical forces that you would calculate using an inertial frame.

The accelerations that you calculate in the accelerating frame would indeed need to be adjusted (aka "transformed") before they are correct for the ground frame. But adopting the accelerating frame may simplify the analysis enough that it's worth the added effort of a transformation step at the end.

Yes so in the end I'll need to make adjustments to the result obtained from an accelerating frame. I will have to subtract ma0cosα in the end from the result , right ?

 

[zwierz]

Spoiler

perhaps this explanation of what fictitious forces are, would be of some use
https://www.physicsforums.com/threads/normal-force-in-a-curvilnear-motion.913890/#post-5758016

 

[jbriggs444]

Yes so in the end I'll need to make adjustments to the result obtained from an accelerating frame. I will have to subtract ma0cosα in the end from the result , right ?

What result did you start with in the accelerating frame? What end result did you wish to obtain?