The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is .......N.

(Take, g = 10 ms^-2 )

Given, coefficient of static friction, µ = 0.5

Value of acceleration due to gravity, g = 10 ms^-2

For complete system to move together,

                F = ma

Here, m is total mass.

               F = (1 + 2) a = 3a                       ...(i)

From free body diagram of 1 kg block,

Balance forces in horizontal direction,

                  ${\mathrm{F}}_{\mathrm{s}}=\mathrm{\mu R}=\mathrm{m}\text{'}\mathrm{a} ...\left(\mathrm{ii}\right)$

Balance forces in vertical direction,

                $\mathrm{R}=\mathrm{m}\text{'}\mathrm{g}$

Put value of R in Eq. (ii),

$\begin{array}{l} \mathrm{\mu m}\text{'}\mathrm{g}=\mathrm{m}\text{'}\mathrm{a}\\ \Rightarrow 0.5\times 1\times 10=1\times \mathrm{a}\\ \Rightarrow \mathrm{a}=5{\mathrm{ms}}^{-2}\end{array}$

Put the value of a in Eq. (i), we get

          F = 3 × 5 = 15 N

Thus, the maximum horizontal force required to move block together is 15 N.