Block B lying on a table weighs W. The coeffi | vedclass

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(B) Let the weight of block $A$ be $W^{\prime}$.For the system to be in equilibrium, the tension $T_1$ in the horizontal cord connected to block $B$ m

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$\mu W 	an 	heta$

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(B) Let the weight of block $A$ be $W^{\prime}$.For the system to be in equilibrium, the tension $T_1$ in the horizontal cord connected to block $B$ must be equal to the limiting friction force, so $T_1 = \mu W$.Now, consider the equilibrium of the knot. Let $T_2$ be the tension in the cord inclined at an angle $	heta$ to the horizontal, and $T_3$ be the tension in the vertical cord connected to block $A$. Thus, $T_3 = W^{\prime}$.Resolving the forces at the knot:Horizontal component: $T_2 \cos 	heta = T_1 = \mu W$Vertical component: $T_2 \sin 	heta = T_3 = W^{\prime}$Dividing the two equations:$\frac{T_2 \sin 	heta}{T_2 \cos 	heta} = \frac{W^{\prime}}{\mu W}$$	an 	heta = \frac{W^{\prime}}{\mu W}$$W^{\prime} = \mu W 	an 	heta$

Block $B$ lying on a table weighs $W$. The coefficient of static friction between the block and the table is $\mu$. Assume that the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is

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