Question

In the arrangement shown in figure, $m_A=m_B=2\text{kg}$. String is massless and pulley is frictionless. Block $B$ is resting on a smooth horizontal surface, while friction coefficient between block $A$ and $B$ is $0.5$. The maximum horizontal force $F$ that can be applied so that block $A$ does not slip over the block $B$ ($g=10\text{m}/s^2$) is

• A. $25\text{N}$
• B. $20\text{N}$
• C. $30\text{N}$
• D. $40\text{N}$

Answer 1

The correct option is B $20\text{N}$

Net horizontal force due to tension in the string on block $B$ is zero. Hence, the given figure (a) can be replaced by figure (b).


Maximum value of friction $f_{max}=\mu m_{A}g=0.5\times 2\times 10=10\text{N}$

Block $B$ moves due to friction alone. Therefore, maximum common acceleration of the two blocks can be
$a_{max}=\frac{f_{max}}{m_B}=\frac{10}{2}=5\text{m/s}{}^2$

But $a_{max}=\frac{F_{max}}{m_A+m_B}$
$\Rightarrow F_{max}=(m_A+m_B)a_{max}=4\times 5=20\text{N}$