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.

 

[nasu]

If you consider A and B as a system, the only net external force acting on it is the tension on the rope. In this case the mass of the system is $m_A+m_B$.
$$T=(m_A+m_B)a$$
If you consider the blocks A and B separetely, the net force on B is the resultant of the tension in the rope and the normal force produced by A acting on B. In this case you only take $m_B$ when writing Newton's second law.But in this case you have another equation for B as well:
$$ T-F_N=m_A a $$
$$F_N=m_B a$$
If you add these two equations you get the equation for the system.
Both points of view give the same result but if you want to calculate the normal force you need to consider the blocks separately.