\(=0.1 \times \frac{50 \times \sqrt{3}}{2}\)
\(=2.5 \sqrt{3} \mathrm{~N}\)
\({F}_1=\mathrm{mg} \sin \theta+\mathrm{f}_{\mathrm{K}} \)
\(=25+2.5 \sqrt{3}\)
\(\mathrm{F}_2=\mathrm{mg} \sin \theta-\mathrm{f}_{\mathrm{K}}\)
\(\quad=25-2.5 \sqrt{3}\)
\(\therefore \mathrm{F}_1-\mathrm{F}_2=5 \sqrt{3} \mathrm{~N}\)