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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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