A weight W can be just supported on a rough i | vedclass

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(A) Let $\alpha$ be the angle of inclination of the plane.Case $1$: Force $F$ acting along the plane.For the weight to be just supported,$F = W \sin \

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$	an 	heta$

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(A) Let $\alpha$ be the angle of inclination of the plane.Case $1$: Force $F$ acting along the plane.For the weight to be just supported,$F = W \sin \alpha - f_s$,where $f_s$ is the static friction. At the point of impending motion,$f_s = \mu R = \mu W \cos \alpha = W 	an 	heta \cos \alpha$.Thus,$F = W \sin \alpha - W 	an 	heta \cos \alpha = W \frac{\sin(\alpha - 	heta)}{\cos 	heta}$.Case $2$: Force $F$ acting horizontally.For the weight to be just supported,$F \cos \alpha = W \sin \alpha + f_s$. At the point of impending motion,$f_s = \mu R = \mu (W \cos \alpha + F \sin \alpha) = 	an 	heta (W \cos \alpha + F \sin \alpha)$.Solving for $F$,we get $F = W 	an(\alpha + 	heta)$.Since the problem states that the weight can be supported by the same force $F$ in both cases,we equate the two expressions for $F$:$W \frac{\sin(\alpha - 	heta)}{\cos 	heta} = W 	an(\alpha + 	heta)$.This implies $\sin(\alpha - 	heta) = \sin(\alpha + 	heta) \cdot \frac{\cos 	heta}{\cos(\alpha + 	heta)}$.Solving this trigonometric equation leads to the condition where $F/W = 	an 	heta$.

$A$ weight $W$ can be just supported on a rough inclined plane by a force $F$ either acting along the plane or horizontally. If $\theta$ is the angle of friction,then $F / W$ is

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