# Movement of blocks stacked together

In the given figure three blocks A, B & C are arranged as shown and a horizontal force $$F$$ is applied to C. Block A is connected to wall with a horizontal string. The tension in the string is $$T$$ and the friction between A and B and that between B & C are $$f_1$$ & $$f_2$$ respectively. The coefficient of friction between A & B and B & C is $$μ$$. The floor is friction less. The acceleration of block B is $$a$$. Mass of A, B & C are $$2m$$, $$m$$ & $$3m$$ respectively. The force F is $$3μmg$$. Tension in string can take any value. Now I know $$f_{1_{max}}$$ = $$2μmg$$ and $$f_{2_{max}}$$= $$3μmg$$. So when F acts due to friction between B and C the net force on C is $$0$$ i.e. it is just begins to move. On block B due to max friction from C and max friction from A the net force comes out to be $$μmg$$ in forward direction. Therefore there should be slipping and B should move with acceleration $$μg$$. A remains at rest. But the acceleration on B is actually $$\frac {μg}4$$. Why does this happen? From what I'm able to guess, as maximum friction is acting between B and C, they behave as one unit of mass $$4m$$ thus having net force of $$μg$$ and so they move with $$\frac {μg}4$$. Can someone please confirm this or correct me?

• Where’s the question in the problem statement? – Bob D May 2 '20 at 10:09
• The question was to find tension for A, both frictions f1 and f2 and acceleration of block B. I was able to find everything other than the acceleration and wanted help for that – Shaurya Goyal May 2 '20 at 12:15

We can assume that $$A$$ is stationary and so $$T=f_{1max}=2\mu mg$$ (otherwise if $$T < f_1$$ then $$A$$ moves to the right and the string extends until $$T=f_1=f_{1max}$$).

If $$B$$ is moving to the right with acceleration $$a$$ then

$$f_2-f_1=ma \\ \Rightarrow f_2 = ma + f_1 = ma + f_{1max} = ma + 2 \mu mg$$

Suppose $$C$$ is moving to the right with acceleration $$a'$$. Then

$$F - f_2 = 3ma' \\ \Rightarrow f_2 = F - 3ma' = 3\mu mg - 3ma'$$

We know that $$a' \ge a$$. But if $$a' > a$$ then $$f_2 = f_{2max}=3 \mu mg$$ and so $$a'=a=0$$. So we can assume that $$a'=a$$ (i.e. there is no relative motion between $$B$$ and $$C$$), in which case

$$f_2 = 3 \mu mg - 3ma \\ \Rightarrow ma + 2 \mu mg = 3\mu mg - 3ma \\ \Rightarrow 4ma = \mu mg \\ \Rightarrow a = \frac {\mu g} {4}$$