Physics - NLM Question with Solution | TestHub

The rod presses on the block along the rod's axis, which is perpendicular to the incline. The rod's weight component along that axis is $mg\cos 37^{\circ }$.

For block $M$ at rest, with limiting friction: $Mg\sin 37^{\circ }=\mu \left(Mg\cos 37^{\circ }+mg\cos 37^{\circ }\right)$.

$20(10)(0.6)=0.5(20\cdot 10\cdot 0.8+m\cdot 10\cdot 0.8)\Rightarrow 120=80+4m\Rightarrow m=10\mathrm{kg}$.