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Asked by safuuu78666 | 06 Feb, 2024, 06:28: PM

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 FL be the reaction force at left side support and FR be the reaction force at right side support .

At equilibrium sum of forces acting on planck is zero.

hence , we get ,    FL + FR - W = 0

FL + FR  =  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 )

FR × ( λ / 2 ) - FL × ( λ / 4 )  = 0

From above expression, we get ,

FL = 2 FR .........................(2)

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

3 FR = 240  or   FR = 80 N  ;

FL = 2 × 80 N = 160 N

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