In the arrangement shown in figure $\left[\sin {37}^{\circ }=\frac{3}{5}\right]$

$\begin{array}{l}10\mathrm{gsin}{37}^{\circ }=10\times 10\times 0.6=60\mathrm{N}\\ 4\mathrm{g}=4\times 10=40\mathrm{N}\\ \text{ Since }60\mathrm{N}>40\mathrm{N}\end{array}$

Block of mass 10 kg has a tendency to move down the plane. Therefore force of friction on 10 kg is up the plane.

$\mathrm{\mu mgcos}\mathrm{\theta }=0.7\times 10\times 10\times 0.8=56\mathrm{N}={\mathrm{F}}_1\text{ (say) }$

Net pulling force:

${\mathrm{F}}_2=(60-40)\mathrm{N}=20\mathrm{N}$

Since ${\mathrm{F}}_2<{\mathrm{F}}_1,$ system will remain at rest.

$\begin{array}{l} \mathrm{T}=40\mathrm{N}\\ \mathrm{T}+\mathrm{f}=60\mathrm{N}\\ \therefore \mathrm{f}=20\mathrm{N}\end{array}$