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

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(A) To find the condition for equilibrium,we analyze the forces acting on the block.$1$. The vertical forces are the weight $mg$ acting downwards,the

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$\frac{P + Q\sin	heta}{mg + Q\cos	heta}$

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(A) To find the condition for equilibrium,we analyze the forces acting on the block.$1$. The vertical forces are the weight $mg$ acting downwards,the vertical component of force $Q$ which is $Q\cos	heta$ acting downwards,and the normal reaction $R$ acting upwards.So,$R = mg + Q\cos	heta$.$2$. The horizontal forces are the force $P$ acting in one direction and the horizontal component of force $Q$ which is $Q\sin	heta$ acting in the same direction.The total horizontal force tending to move the block is $F_{net} = P + Q\sin	heta$.$3$. For the block to remain in equilibrium,the frictional force $f$ must balance the net horizontal force,i.e.,$f = P + Q\sin	heta$.$4$. The maximum static friction is given by $f_{max} = \mu R = \mu(mg + Q\cos	heta)$.$5$. For equilibrium,the applied horizontal force must be less than or equal to the limiting friction: $P + Q\sin	heta \leq \mu(mg + Q\cos	heta)$.Therefore,the coefficient of friction $\mu$ must satisfy: $\mu \geq \frac{P + Q\sin	heta}{mg + Q\cos	heta}$.

$A$ 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 block will remain in equilibrium,if the coefficient of friction between it and the surface is

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