Question

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

• A. $25N$
• B. $40N$
• C. $30N$
• D. $20N$

Answer 1

The correct option is D $20N$

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

Maximum value of friction between A and B $f_{max}=\mu m_{A}g=\left(0.5\right)\left(2\right)\left(10\right)=10N$

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

$\therefore F=(m_A+m_B)a_{max}F=(2+2)\left(5\right)=20N$