In the figure,a block of weight 60 , N is pla | vedclass

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(C) The block of weight $60 \, N$ is on a rough surface with coefficient of friction $\mu = 0.5$. The maximum frictional force $F_{max}$ that can act

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$30 \, N$

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(C) The block of weight $60 \, N$ is on a rough surface with coefficient of friction $\mu = 0.5$. The maximum frictional force $F_{max}$ that can act on the block is given by $F_{max} = \mu N = 0.5 	imes 60 \, N = 30 \, N$.For the block not to slip,the tension $T_1$ in the horizontal string must be less than or equal to $F_{max}$. Thus,$T_1 = 30 \, N$.At the junction point,the forces are in equilibrium. Let $T_2$ be the tension in the inclined string at an angle of $45^{\circ}$ to the horizontal.The horizontal component of $T_2$ balances $T_1$: $T_2 \cos 45^{\circ} = T_1 = 30 \, N$.$T_2 	imes \frac{1}{\sqrt{2}} = 30 \implies T_2 = 30\sqrt{2} \, N$.The vertical component of $T_2$ balances the weight $W$: $W = T_2 \sin 45^{\circ}$.$W = 30\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 30 \, N$.Therefore,the maximum weight $W$ is $30 \, N$.

In the figure,a block of weight $60 \, N$ is placed on a rough surface. The coefficient of friction between the block and the surface is $0.5$. Find the maximum weight $W$ such that the block does not slip on the surface.

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