Two disks of masses $m_{1}$ and $m_{2}$ are connected by a spring of force constant k. The lower disk pf mass $m_{2}$ lies on a table and the upper disk is vertically above it ( $F i g . 6.203$ ). What vertical force F should be applied ti the upper disk so that when the force is withdraw, the lower disk is lifted off the table?

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Solution

The situation is shown in $F i g . 6.204$ Let the applied force F compresses the spring by x, so $F_{s} = k x$ . After removal of force, the spring recovers its compressed length and further extends by x. At this instant the force exerted by spring is along upward direction. Thus, to lift off the lower disk, normal reaction N becomes zero.
When force is applied and spring is compressed,
$F = F_{s} + m_{1} g$ ..........(i)
When $m_{2}$ leaves contact (we have, N=0) with ground and spring is stretched,
$F_{s} = m_{2} g$ .........(ii)
From Eqs (i) and (ii), we have $F_{m i n} = (m_{1} + m_{2}) g$
So to lift off the lower disk, $F \geq (m_{1} + m_{2}) g$

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