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

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(A) The block is in equilibrium. Let the angle of inclination be $	heta = 45^{\circ}$. The mass of the block is $m = 15\, kg$,so the weight is $W = m

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) The block is in equilibrium. Let the angle of inclination be $	heta = 45^{\circ}$. The mass of the block is $m = 15\, kg$,so the weight is $W = mg = 15 	imes 10 = 150\, N$.We resolve the forces along the inclined plane ($x$-axis) and perpendicular to it ($y$-axis).The forces acting on the block are:$1$. Weight $(mg = 150\, N)$ acting vertically downwards.$2$. Normal reaction $(N)$ perpendicular to the plane.$3$. Tension $(T = 50\, N)$ acting horizontally.$4$. Frictional force $(f = \mu N)$ acting down the plane to oppose the tendency of motion.Resolving forces along the $x$-axis (parallel to the plane):$\Sigma F_x = 0 \implies T \cos 45^{\circ} = mg \sin 45^{\circ} + f$$50 \cos 45^{\circ} = 150 \sin 45^{\circ} - f$$f = 150 \sin 45^{\circ} - 50 \cos 45^{\circ} = (150 - 50) 	imes \frac{1}{\sqrt{2}} = \frac{100}{\sqrt{2}} = 50\sqrt{2}\, N$.Resolving forces along the $y$-axis (perpendicular to the plane):$\Sigma F_y = 0 \implies N = mg \cos 45^{\circ} + T \sin 45^{\circ}$$N = 150 \cos 45^{\circ} + 50 \sin 45^{\circ} = (150 + 50) 	imes \frac{1}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2}\, N$.Using $f = \mu N$:$\mu = \frac{f}{N} = \frac{50\sqrt{2}}{100\sqrt{2}} = \frac{50}{100} = \frac{1}{2}$.

$A$ 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 coefficient of friction between the surfaces of contact is $(g = 10\, m/s^2)$.

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