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On balancing the moments about centre of gravity, we get
mg (a/2) + mg(a/3) = mg (a/2+x)
Therefore, on solving we get, x=a/3.
Ans is (b)
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 co-eff. of friction = 1/2. The distance between the lines of action of mg and normal reaction is:
A. a/2
B. a/3
C. a/4
D. none of these
Please explain the answer to me.
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2 Answers
1111See, the point of application of the normal reaction is displaced through x bec to produce a clockwise torque abt the centre of mass that may balance the anticlockwise torque produced by the frictional force.
In other words, the normal reaction is acted at a distance x bec a horizontal force F is applied. So for =lbm, N is acted at a distance x in anticlockwise direction.
For toppling (rolling ) of the body, slipping of normal reaction is required at a point or the line of contact to the max value.
