Question

In the arrangement shown in figure, $m_{A} = m_{B} = 2 \,kg$ 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

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

Answer 1

Answer: (D)

Maximum value of friction on $A, f_{\text{max}}=\mu mg$
$=0.5 \times 2 \times 10=10\,N$
Hence, the block B is subjected to this frictional force in the forward direction. Since the block A does not slip on B, both of them move with a common acceleration. The acceleration of B is the same as the acceleration of A
To determine acceleration of ‘a ’, we find that $m_{B}a_{\text{max}} =f_{\text{max}}$
$a_{\text{max}}=\frac{10}{2}=5\,m/s^{2}$
$F_{\text{max}}=\left(m_{A}+m_{B}\right)a_{\text{max}}$
$=4\times5=20\,N$