A block of mass m lying on a rough horizontal | vedclass

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(B) For the block to be in equilibrium,the forces must balance in both horizontal and vertical directions.Resolving force $Q$ into components:Horizont

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$\frac{P+Q \sin 	heta}{m g+Q \cos 	heta}$

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(B) For the block to be in equilibrium,the forces must balance in both horizontal and vertical directions.Resolving force $Q$ into components:Horizontal component = $Q \sin 	heta$Vertical component = $Q \cos 	heta$Vertical equilibrium:$N = mg + Q \cos 	heta$Horizontal equilibrium:$f_r = P + Q \sin 	heta$Since $f_r \le \mu N$,the minimum coefficient of friction $\mu$ is given by:$\mu = \frac{f_r}{N} = \frac{P + Q \sin 	heta}{mg + Q \cos 	heta}$

$A$ block of mass $m$ lying on a rough horizontal plane is acted upon by a horizontal force $P$ and another force $Q$ inclined at an angle $\theta$ to the vertical. The minimum value of the coefficient of friction between the block and the surface for which the block will remain in equilibrium is:

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