Question

An arrangement of masses and pulleys is shown in the figure. Strings connecting masses $A$ and $B$ with the pulleys are horizontal and all pulleys and strings are light. Friction coefficient between the surface and the block $B$ is $0.2$ and between blocks $A$ and $B$ is $0.7$. The system is released from rest. Use $(g=10{\text{m/s}}^2)$. Find the magnitude of frictional force (in $\text{N}$) between block $A$ and $B$.

Answer 1

First, we assume that the system is moving with common acceleration $a$ and that there is no slipping between block $A$ and $B$.
FBD of the system:


Here, if block $C$ moves with acceleration $a$, then block $A,B\&D$ will also move with the same acceleration $a$. (from observation)
$T=$ tension in the left string
$T^{\prime }=$ tension in right string
$f=$ friction force acting between $A$ & $B$ block
$f^{\prime }=$ friction force between surface and block $B$.
From FBD of $6\text{kg}$ (block $C$):
$6g-T=6a$
$T=6(g-a)$ ... (1)
From FBD of $1\text{kg}$ (block $D$):
$T^{\prime }-g=1\times a$,
$T^{\prime }=g+a$ ... (2)

Assume block $A$ & $B$ as a system, making FBD. (we can neglect the internal friction force between $A$ and $B$.


$T-T^{\prime }-f^{\prime }=(6+3)a$
$\Rightarrow T-T^{\prime }-0.2(6+3)g=9a$
$\Rightarrow T-T^{\prime }=1.8g+9a$ ...... (3)
From eqn. (1) and (2), we get
$6g-6a-(g+a)=T-T^{\prime }$
i.e $6g-6a-g-a=1.8g+9a$ (from (3))
$\Rightarrow 5g-7a=1.8g+9a$
$\Rightarrow a=2{\text{m/s}}^2$

Now, for block $A$,
$T-f=6\times a$


From eq. (1),
$6(g-a)-f=6a$
$6\times 10-12\times 2=f$
$\Rightarrow f=36\text{N}$
Maximum friction force that can act between block $A$ and block $B$ $=\mu \left(mg\right)=0.7\times (6\times 10)=42\text{N}=f_{max}$
Since, $f<f_{max}$, so both the blocks will move together without slipping. Hence our assumption was correct.