Three identical uniform, meter sticks where stacked and clubbed together so as to form a single unit, with 30 cm, 40 cm and x cm, respectively hanging over the edge as shown. The maximum value of x for which the metersticks remains in equilibrium on the table is:
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COM of the system must be equal to 70 cm from left end of the bottom most meter stick for maximum value of x for maintaining equilibrium. If COM is on right side of the edge then torque of gravitational force of the system would produce rotational motion about edge.
Taking O as origin X CM (max) = 70 cm
Using 
210 m = 50 m + 60 m + 20 m + mx
x = 80 m
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