A block B,lying on a table,has weight w. The | vedclass

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(B) Given that,weight of block $B = w$.Coefficient of static friction between block $B$ and the table $= \mu$.Let the maximum weight of block $A$ be $

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$\mu w 	an 	heta$

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(B) Given that,weight of block $B = w$.Coefficient of static friction between block $B$ and the table $= \mu$.Let the maximum weight of block $A$ be $w_A$.For the system to be in equilibrium,the forces acting on the knot and block $B$ must be balanced.From the free-body diagram $(FBD)$ of the knot,the vertical component of the tension $T$ must balance the weight of block $A$:$w_A = T \sin 	heta$  --- $(i)$From the $FBD$ of block $B$,the horizontal component of the tension $T$ must be balanced by the limiting friction $f$:$f = \mu N = \mu w = T \cos 	heta$  --- (ii)Dividing equation $(i)$ by equation (ii),we get:$\frac{w_A}{\mu w} = \frac{T \sin 	heta}{T \cos 	heta}$$w_A = \mu w 	an 	heta$Thus,the maximum weight of block $A$ is $\mu w 	an 	heta$.

$A$ block $B$,lying on a table,has weight $w$. 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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