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Question

(A) Let $T_1$ be the tension in the horizontal string and $T_2$ be the tension in the inclined string at angle $	heta$.For the hanging block of mass

Vedclass · Accepted Answer

$\mu=\cot 	heta$

Vedclass · Answer

(A) Let $T_1$ be the tension in the horizontal string and $T_2$ be the tension in the inclined string at angle $	heta$.For the hanging block of mass $M$,the vertical equilibrium gives $T_2 \sin 	heta = Mg$ ... $(i)$.For the point $O$,the horizontal equilibrium gives $T_1 = T_2 \cos 	heta$ ... (ii).For the block on the floor,the horizontal force is $T_1$ and the limiting friction is $f_s = \mu N = \mu Mg$.For equilibrium,$T_1 = f_s = \mu Mg$ ... (iii).From (ii) and (iii),$T_2 \cos 	heta = \mu Mg$ ... (iv).Dividing $(i)$ by (iv),we get $\frac{T_2 \sin 	heta}{T_2 \cos 	heta} = \frac{Mg}{\mu Mg}$.Therefore,$	an 	heta = \frac{1}{\mu}$,which implies $\mu = \cot 	heta$.

$A$ wooden block of mass $M$ lies on a rough floor. Another wooden block of the same mass is hanging from the point $O$ through strings as shown in the figure. To achieve equilibrium,the coefficient of static friction between the block on the floor and the floor itself is:

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