A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force $\mathrm{F}$ is…

A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force $\mathrm{F}$ is applied normal to one of the faces at a point that is directly above the centre of the face, at a height $3 \mathrm{a} / 4$ above the base. The minimum value of $\mathrm{F}$ for which the cube begins to topple an edge is (assume that cube does not slide)
  1. $\frac{\mathrm{mg}}{3}$
  2. $\frac{\mathrm{mg}}{2}$
  3. $\frac{2 \mathrm{mg}}{3}$
  4. $\frac{3 \mathrm{mg}}{4}$

Solution

For toppling about edge $x$ x At the moment of toppling the normal force pass through axis $\mathrm{xx}$ '. $ \mathrm{F}_{\min } \frac{3 \mathrm{a}}{4}=\mathrm{mg} \frac{\mathrm{a}}{2} \text { or } \mathrm{F}_{\min }=\frac{2 \mathrm{mg}}{3} $

Asked in: BITSAT 2009

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