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MCQ Solution: Motion of System of Particles and Rigid Body

A cube is in equilibrium on an inclined plane forming an angle of 30⁰ with the horizontal. If the distance of line of action of the normal reaction of the plane on the cube from the centre is a / 2√x. Find x

1) 4
2) 8
3) 3
4) 2

Solution

Net force perpendicular to the plane of the incline is zero,

$$\therefore$$ $$ mg cos(30^{\circ})$$ = N

Now consider the point on the centre of the base of the cube (on the incline). Frictional force passes through this point hence produces zero torque. Net torque about this point is zero. Only weight and normal produce a torque about this point.

$$\therefore$$ Writing torque equation we get,

$$mg sin(30^{\circ})\dfrac{a}{2}$$ = $$N(\dfrac{a}{2\sqrt{x}})$$

Substituting value of N,

$$mg sin(30^{\circ})\dfrac{a}{2}$$ = $$mg cos(30^{\circ})(\dfrac{a}{2\sqrt{x}})$$

$$tan(30^{\circ})$$ = $$\dfrac{1}{\sqrt{x}}$$

$$\therefore x = 3$$