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(B) Let the total mass of the board and the added weight be $M_{total} = (M + m)$.When the board is tilted at an angle $	heta$ with the horizontal,th

Vedclass · Accepted Answer

$	an 	heta$

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(B) Let the total mass of the board and the added weight be $M_{total} = (M + m)$.When the board is tilted at an angle $	heta$ with the horizontal,the force of gravity $(M + m)g$ acting on the center of mass can be resolved into two components:$1$. $A$ component perpendicular to the board: $(M + m)g \cos 	heta$,which is balanced by the normal reaction $N$.$2$. $A$ component parallel to the board: $(M + m)g \sin 	heta$,which tends to make the board slide down.For the board to be in equilibrium,the frictional force $f$ must balance the component of gravity acting parallel to the board.Thus,$f = (M + m)g \sin 	heta$.Since the board is on the verge of sliding,the frictional force is the limiting friction,$f = \mu N$.Substituting $N = (M + m)g \cos 	heta$,we get:$f = \mu (M + m)g \cos 	heta$.Equating the two expressions for $f$:$(M + m)g \sin 	heta = \mu (M + m)g \cos 	heta$$\mu = \frac{\sin 	heta}{\cos 	heta} = 	an 	heta$.Therefore,the coefficient of friction is $	an 	heta$.

$A$ board is balanced on a rough horizontal semicircular log. Equilibrium is obtained with the help of the addition of a weight to one of the ends of the board when the board makes an angle $\theta$ with the horizontal. The coefficient of friction between the log and the board is:

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