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In the diagram shown, the normal reaction force between 2 kg and 1 kg is (Consider the surface to be smooth): Given $g=10$ ms$^{-2}$
Detailed Solution
Acceleration of the system up the incline: $a=\frac{60-18-(3+2+1)g\sin30^\circ}{3+2+1}=\frac{60-18-30}{6}=2$ m/s$^2$. For the 1 kg block: $ma=N-18-1\times10\times\frac{1}{2} \Rightarrow 2=N-23 \Rightarrow N=25$ N.