Two blocks A and B of masses m_A = 1,kg and m_B = 3,kg are kept on table as shown in figure. coeffic

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4-2.Friction

hard

Two blocks $A$ and $B$ of masses $m_A = 1\,kg$ and $m_B = 3\,kg$ are kept on the table as shown in figure. The coefficient of friction between $A$ and $B$ is $0.2$ and between $B$ and the surface of the table is also $0.2.$ The maximum force $F$ that can be applied on $B$ horizontal, so that the block $A$ does not slide over the block $B$ is ........ $N$. [Take $g = 10\,m/s^2$ ]

$8$

$16$

$12$

$40$

(JEE MAIN-2019)

Solution

$\begin{array}{l}
{M_A} = 1\,kg,\,{M_B} = 3kg\\
{\mu _{AB}} = 0.2\\
{\mu _B} = 0.2\\
{F_{\max }} = \left( {{M_A} + {M_B}} \right) \times 0.2 \times 10\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {{M_A} + {M_B}} \right) \times 0.2 \times \\
\,\,\,\,\,\,\,\,\,\,\, = 4 \times 2 + 4 \times 2 = 16
\end{array}$