Mechanics question: Three masses and a pulley on a tabletop

[Zayan]

Homework Statement

Find mass of block C such that there is no relative motion between B and A

Relevant Equations

F=ma.

The diagram is below

.


My approach is first finding acceleration of block C and hence B because it would be equal. So writing equations Mcg-T=Mc*a.

I'm confused about the second equation.
I first wrote T = Mb*a. My intuition is that because Only tension is acting on block B towards right, it would be the one responsible for horizontal motion of block B. So solving we get a= Mcg/2m+mc. Using this as acceleration of frame I considered pseudo force on block A causing normal reaction with block B hence writing friction equals gravitational force on A. This gives the value of Mc as 2m/(u-1).

But going back to the second equation, I could also write equation for the whole system, i.e, a= Mcg/mc+ma+mb. Using this approach the answer comes out to be 10m/u-1. Which of the approach is correct and why so.

 

[PeroK]

It doesn't matter whether block A is nailed to block B, held to block B by friction (or, sliding down block B). As far as an external horizontal force is concerned, A and B form a single object of their combined mass.

It is the same if you are pulling a cart. You have to accelerate the cart and everything inside it (unless things can slide backwards).

 

[Zayan]

It doesn't matter whether block A is nailed to block B, held to block B by friction (or, sliding down block B). As far as an external horizontal force is concerned, A and B form a single object of their combined mass.

It is the same if you are pulling a cart. You have to accelerate the cart and everything inside it (unless things can slide backwards).

Makes sense. Also, upon further inspection I think I forgot that when B is accelerating, there is also normal force between A and B which is opposing the tension. So to avoid calculating internal forces we consider it a system.