A HORIZONTAL FORCE F=mg/3 IS APPLIED ON THE UPPER SURFACE OF A UNIFORM CUBE OF MASS m AND SIDE a WHICH IS RESTING ON A ROUGH HORIZONTAL SURFACE HAVING u=1/2.THE DISTANCE B/W LINES OF ACTION OF mg and normal reaction is?

Dear vardaan

let distance between line of action of mg and N is x

so verticle force balance 

N=mg

clearly mg/3 is less than the maximum static friction force .

so only mg/3 static friction force will act .

now torque balance from the center of the cube

mg/3 *a/2  + mg/3 *a/2  = N*x

mga/3 =N*x

mga/3 = mgx

 x =a/3

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