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 $μ_{s} = 1 / 2$ . The distance between lines of action of 'mg' and normal reaction 'N' is given as $a / x$ . Find $x$ .

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Solution

Horizontally applied force(F) on the upper surface of cube = mg/3;
weight of cube = mg
maximum value of friction = mg( $μ$ ) = $\frac{m g}{2}$
So , cube will have no translational motion and F will try to topple the cube about the forward lower edge. So for equilibrium
mg a/2 - F a - Nx = 0 ;
a = edge length, N = normal force on cube by surface = mg
mg a/2 - mg/3 a - mg x = 0
x = a/6
x = distance of N from lower forward edge
mg is acting at 'a/2' distance from same edge so , distance(d) between lines of action of mg and N
d = a/2-a/6
d =a/3

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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 μs=1/2. The distance between lines of action of 'mg' and normal reaction 'N' is given as a/x. Find x.

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