Two blocks A of 6 kg and B of 4 kg are placed in contact with each other as shown in figure.

There is no friction between A and ground, and between both the blocks. The coefficient of friction between B and ground is 0.5. A horizontal force F is applied on A. Find the minimum and maximum values of F, which can be applied so that both blocks can move combinely without any relative motion between them. 

$\mathrm{F}-\mathrm{Nsin}{37}^{\circ }=6\mathrm{a}\Rightarrow \mathrm{F}-\frac{3\mathrm{N}}{5}=6\mathrm{a} ...\left(\mathrm{i}\right)$

               $\begin{array}{l}{\mathrm{N}}_2=4\mathrm{g}-\mathrm{Ncos}{37}^{\circ }=40-\frac{4\mathrm{N}}{5} ...\left(\mathrm{ii}\right)\\ \mathrm{Nsin}{37}^{\circ }-\mathrm{f}=4\mathrm{a} ...\left(\mathrm{iii}\right)\end{array}$

From Eqs. (i) and (iii): $\mathrm{F}-\mathrm{f}=10\mathrm{a}$

$\begin{array}{l}\Rightarrow \mathrm{F}-{\mathrm{mN}}_2=10\mathrm{a} ...\left(\mathrm{iv}\right)\\ \Rightarrow \mathrm{F}-\mathrm{m}\left[40-\frac{4\mathrm{N}}{5}\right]=10\mathrm{a} ...\left(\mathrm{v}\right)\end{array}$

Put the value of N from Eq. (i) in (v) and also put the value of m to get $\mathrm{a}=\frac{5\mathrm{F}-60}{42}$

Now to start motion : $\mathrm{a}>0\Rightarrow \mathrm{F}>12\mathrm{N}$.

So the minimum force F to just start the motion is 12 N.

Now maximum F will be when N₂ just becomes zero. Then from Eqs. (ii): N = 50 N.

From Eqs. (i) and (iv), we get F = 75 N (by putting N₂ = 0)

If we apply F > 75 N, then B will start sliding up on A, but we do not want this.