In the diagram shown,the normal reaction forc | vedclass

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(A) First,calculate the acceleration of the system. The total mass is $M = 3 + 2 + 1 = 6\,kg$.The forces acting along the incline are $F_1 = 60\,N$ (u

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$25$

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(A) First,calculate the acceleration of the system. The total mass is $M = 3 + 2 + 1 = 6\,kg$.The forces acting along the incline are $F_1 = 60\,N$ (upwards) and $F_2 = 18\,N$ (downwards).The component of gravity acting downwards for all blocks is $Mg \sin 30^{\circ} = 6 	imes 10 	imes 0.5 = 30\,N$.The net force is $F_{	ext{net}} = 60 - 18 - 30 = 12\,N$.The acceleration is $a = \frac{F_{	ext{net}}}{M} = \frac{12}{6} = 2\,ms^{-2}$ (upwards).Now,consider the $1\,kg$ block. Let $N$ be the normal reaction force between the $2\,kg$ and $1\,kg$ blocks.The forces on the $1\,kg$ block are $N$ (upwards),$F_2 = 18\,N$ (downwards),and $mg \sin 30^{\circ} = 1 	imes 10 	imes 0.5 = 5\,N$ (downwards).Applying Newton's second law: $N - 18 - 5 = m 	imes a \implies N - 23 = 1 	imes 2 \implies N = 25\,N$.

In the diagram shown,the normal reaction force between the $2\,kg$ and $1\,kg$ blocks is (Consider the surface to be smooth) $.........N$ (Given $g = 10\,ms^{-2}$)

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