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Question: Three identical uniform, meter sticks where stacked and clubbed together so as to form a single unit...

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 meter sticks remains in equilibrium on the table is :

A

10 cm

B

40 cm

C

50 cm

D

80 cm

Answer

80 cm

Explanation

Solution

COM of the system must be equal to 70 cm from left end of the bottom most metre 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 XCM(max) = 70 cm

Using XCM=m1x1+m2x2+m3x3 m1+m2+m3\mathrm { X } _ { \mathrm { CM } } = \frac { \mathrm { m } _ { 1 } \mathrm { x } _ { 1 } + \mathrm { m } _ { 2 } \mathrm { x } _ { 2 } + \mathrm { m } _ { 3 } \mathrm { x } _ { 3 } } { \mathrm {~m} _ { 1 } + \mathrm { m } _ { 2 } + \mathrm { m } _ { 3 } }

70 cm=m⋅(50 cm)+m(50+10)cm+m(x−30+50)cm3 m70 \mathrm {~cm} = \frac { \mathrm { m } \cdot ( 50 \mathrm {~cm} ) + \mathrm { m } ( 50 + 10 ) \mathrm { cm } + \mathrm { m } ( \mathrm { x } - 30 + 50 ) \mathrm { cm } } { 3 \mathrm {~m} }

210 m = 50 m + 60 m + 20 m + mx

x = 80 m