Question

The coefficient of static friction is $0.2$ between block $A$ of mass $2\text{kg}$ and the table as shown in the figure. What would be the maximum mass value of block $B$ so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless $(g=10{\text{m/s}}^2)$

• A. $2.0\text{kg}$
• B. $4.0\text{kg}$
• C. $0.2\text{kg}$
• D. $0.4\text{kg}$

Answer 1

The correct option is D $0.4\text{kg}$

For equilibrium
FBD


$\begin{array}{}T=mg& \text{(1)}\end{array}$
From FBD of $2kg$ block
$\Rightarrow f=T,$ where $f$ is the friction force
$f=\mu N$, where $N=2g$
$\begin{array}{}f=\mu 2g& \text{(2)}\end{array}$
from above equations,
$\mu 2g=mg$
$m=0.2\times 2=0.4kg$
This is the maximum mass of the block $B$ for the system to be in rest.