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 $g = 10\,m/s^2 ]$
JEE MAIN 2019, Difficult
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${M_A} = 1 kg, {M_B} = 3\ kg$
${\mu _{AB}} = 0.2$
${\mu _B} = 0.2$
${F_{\max }} = \left( {{M_A} + {M_B}} \right) \times 0.2 \times 10$
$+ \left( {{M_A} + {M_B}} \right) \times 0.2 \times$
$ = 4 \times 2 + 4 \times 2 = 16$
${\mu _{AB}} = 0.2$
${\mu _B} = 0.2$
${F_{\max }} = \left( {{M_A} + {M_B}} \right) \times 0.2 \times 10$
$+ \left( {{M_A} + {M_B}} \right) \times 0.2 \times$
$ = 4 \times 2 + 4 \times 2 = 16$
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