A cube of side length a is resting on an incl | vedclass

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(C) For a cube of side $a$ to topple,the line of action of the gravitational force $mg$ must pass outside the base of the cube. This occurs when the a

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$\mu > 1$

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(C) For a cube of side $a$ to topple,the line of action of the gravitational force $mg$ must pass outside the base of the cube. This occurs when the angle of inclination $	heta$ exceeds $45^{\circ}$.For the cube to topple before sliding,the condition for toppling must be met before the condition for sliding is met.The condition for sliding is $mg \sin 	heta > \mu mg \cos 	heta$,which simplifies to $\mu To ensure it topples first,the cube must not slide at the angle where it is on the verge of toppling (i.e.,$	heta = 45^{\circ}$).Therefore,at $	heta = 45^{\circ}$,we must have $\mu > 	an 45^{\circ}$.Since $	an 45^{\circ} = 1$,the condition is $\mu > 1$.

$A$ cube of side length $a$ is resting on an inclined plane. What must be the value of the coefficient of friction $\mu$ between the cube and the plane so that the cube topples before sliding?

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