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.

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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?

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    $\begingroup$ Where’s the question in the problem statement? $\endgroup$
    – Bob D
    May 2, 2020 at 10:09
  • $\begingroup$ 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 $\endgroup$ May 2, 2020 at 12:15

1 Answer 1


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

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


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