AS Mechanics

Hi, could someone please explain what the equations of motion for each of the blocks here are, given the lift is accelerating upwards by 2ms-2?

Is this correct?:
P: R1-10g=10*2
Q: R2-20g=20*2
Lift/whole system: T-250g-10g-20g=280*2

So the R2 force acts from the floor, onto Q. And R1 acts from Q onto P? And weight acts on each of the blocks, but not onto the lift? Why don't we include the reaction forces when looking at the whole system? I am just trying to understand how each of these forces work and what they act on, to help me with these kinds of questions.

Thanks

If its your sketch, Id have put the R1 arrow starting at the PQ boundary and the R2 arrow starting at the QFloor boundary. Also maybe the weights act at the COM (centre of the blocks) rather than the floor, but this is less important. Agree for P, but for Q youd need to include P (30g and 30*2) as R2 provides the force to accelerate the combined system P+Q upwards. Agree for the whole system.

You dont include the internal reactions when you consider a single "blob". Here a blob is P+Q for the R2 force and P+Q+Lift for the tension. You only consider external forces acting on the blob and its resultant motion (or not). As an example, if you considered just the Q block, then the external forces would be
R2 - R1 - 20g = 20*2
as P would be pushing down on Q with a force of R1, just as P is being pushed up with a force R1 (a force pair). From the force equation for P, this obviously becomes
R2 - (10g + 10*2) - 20g = 20*2
Or
R2 - 30g = 30*2
so the same as modelling PQ single blob.

Thanks for your reply, that clears up a lot of the misconceptions I had. However, could you please explain why you ignore the internal forces when modelling as a 'blob', e.g. let's say the whole system: T-250g-10g-20g=280*2. It makes sense not to include R1 and R2, but I still don't understand why? Thanks.

For simplicity, just consider the lift/masses in equilibrium so the right hand side is zero (its not important, but simplifies the argument a bit). As in the previous post, you could model just the lift, so as you only consider external forces acting on the object, you'd get
T - R2 - 250g = 0
as Q pushes down on the lift with a force of magnitude R2 (though the lift also pushes up on Q with an equal but opposite force - newton 3). Subbing for R2 using the previous results gives
T - 280g = 0
as youd have got if youd modelled the Lift+P+Q as a single object. You model external forces acting on the object. Internally, you have equal but opposite force pair(s) acting on different parts of the object so overall the force pair(s) sum to zero.

As a thought experiment for the Lift+P+Q blob, would you include R1 and R2 if they were glued together? What about if they the metal was welded together? What about if they were simply put one on top of another on? The reality is there is no difference between these scenarios for the combined object as it acts as a single object and the two external forces (tension and gravity) act on the combined object.

The force balance equation
T - 280g = 0
occurs where the string is attached to the lift. At that point, there is a force pair between the end of the string and the top of the lift (which includes P & Q). The top of the lift doesnt care about how the masses are "attached" inside the lift.

That makes more sense now - thanks

Going back to your original sketch, Id be tempted to do seperate sketches for the blob(s) and the external forces that act on it/them. So if the blob is P, then forget about the lift and Q and just mark on R1 and 10g weight. If the blob is P+Q, then again ignore the lift and its R2 and 30g weight. Similarly for the lift its T and 280g. It would be slightly more work, but would make clear about what blob you were modelling.

Sorry, another small question - when you said that modelling internal forces would cancel because they have an equal and opposite force, could you please explain this a bit more? Wouldn't that also cancel out the weight of the blocks?

Talking about modelling.. the internal forces is a bit of a rabbit hole that you don't particularly want/need to go down, a bit like thinking about the atomic repesentation of what tension is
https://en.wikipedia.org/wiki/Tension_(physics)
At a level, its just not necessary. However, a bit like the P/Q and P+Q distinction above. You could imagine a 30kg mass PQ which was theoretically divided horizontally into a 20kg cylinder Q and a 10kg cylinder P. Obviously you could/should treat them as a single body of mass 30kg, but you could imagine the 20kg base pushing up the 10kg top and vice versa. The two forces are equal and opposite (force pair)
https://www.physicsclassroom.com/class/newtlaws/Lesson-4/Identifying-Action-and-Reaction-Force-Pairs

So considering the P & Q blobs seperately, the equations are
P: R1 - 10g = 10*2
Q: R2 - R1 - 20g = 20*2
Summing them the R1s cancel (force pair-equal and opposite) and you get the expected PQ blob
PQ: R2 - 30g = 30*2

A similar scenario is you analyse the motion of a car towing a caravan (or similar). Unless you need to model the compression/tension in the tow bar, its often easier to simply consider the car+caravan blob directly. If you did model them seperately then combine, the tension in the tow bar would be equal and opposite on the two blobs and when you combine them, it would cancel and just leave you with the car+caravan equation, but derived in a more complex way.