NEET Class neet Answered

Initially supports are provided at extreme left and extreme right edge of the plank as shown in left side figure.

At each support point , we are given that the reaction force F = 120 N .

Let W be the weight acting on center of mass .

At equilibrium sum of forces acting on planck is zero.

Hence  we get ,   2F - W = 0

(upward direction force is positive and downward direction force is negative )

W = 2 F = 2 × 120 N = 240 N .

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When left side support is moved towards center of mass so that left support is at a distance λ/4

from centre of mass as shown in figure, we have following equations.

Let F_L be the reaction force at left side support and F_R be the reaction force at right side support .

At equilibrium sum of forces acting on planck is zero.

hence , we get ,    F_L + F_R - W = 0

F_L + F_R  =  W = 240 N  ............................(1)

At equilibrium sum of moment of forces acting on planck is zero.

( Counterclockwis moment is positive and clockwise moment is negative )

F_R × ( λ / 2 ) - F_L × ( λ / 4 )  = 0

From above expression, we get ,

 F_L = 2 F_R .........................(2)

By substituting F_L from eqn.(2) , we rewrite eqn.(1) as

3 F_R = 240  or   F_R = 80 N  ;   

F_L = 2 × 80 N = 160 N