A block of mass 15 , kg is resting on a rough | vedclass

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(A) The block is in equilibrium,so the net force acting on it is zero. We consider the forces along the inclined plane ($x$-axis) and perpendicular to

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$\frac{1}{2}$

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(A) The block is in equilibrium,so the net force acting on it is zero. We consider the forces along the inclined plane ($x$-axis) and perpendicular to it ($y$-axis).The forces acting on the block are:$1$. Weight $mg = 15 	imes 10 = 150 \, N$ acting vertically downwards.$2$. Normal reaction $N$ perpendicular to the plane.$3$. Tension $T = 50 \, N$ acting horizontally.$4$. Limiting friction $f = \mu N$ acting up the plane.Resolving forces along the $x$-axis (parallel to the plane):$\Sigma F_x = 0$$f + T \cos 45^{\circ} = mg \sin 45^{\circ}$$\mu N + 50 \cos 45^{\circ} = 150 \sin 45^{\circ} \quad ...(i)$Resolving forces along the $y$-axis (perpendicular to the plane):$\Sigma F_y = 0$$N = mg \cos 45^{\circ} + T \sin 45^{\circ}$$N = 150 \cos 45^{\circ} + 50 \sin 45^{\circ} \quad ...(ii)$Substituting $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$ in equation $(ii)$:$N = (150 + 50) 	imes \frac{1}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2} \, N$Substituting $N$ into equation $(i)$:$\mu (100\sqrt{2}) + 50 	imes \frac{1}{\sqrt{2}} = 150 	imes \frac{1}{\sqrt{2}}$$\mu (100\sqrt{2}) = \frac{100}{\sqrt{2}}$$\mu = \frac{100}{\sqrt{2} 	imes 100\sqrt{2}} = \frac{1}{2}$

$A$ block of mass $15 \, kg$ is resting on a rough inclined plane as shown in the figure. The block is tied by a horizontal string which has a tension of $50 \, N$. The minimum coefficient of friction between the surfaces of contact is $(g = 10 \, m/s^2)$.

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