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

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(A) The forces acting on the block of mass $m = 15\,kg$ are:$1$. Weight $mg = 15 	imes 10 = 150\,N$ acting vertically downwards.$2$. Normal force $N_

Vedclass · Accepted Answer

$1/2$

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(A) The forces acting on the block of mass $m = 15\,kg$ are:$1$. Weight $mg = 15 	imes 10 = 150\,N$ acting vertically downwards.$2$. Normal force $N_1$ perpendicular to the inclined plane.$3$. Tension $T = 50\,N$ acting horizontally.$4$. Friction force $f$ acting up the incline.Resolving forces perpendicular to the inclined plane:$N_1 = mg \cos(45^{\circ}) + T \sin(45^{\circ})$$N_1 = 150 	imes \frac{1}{\sqrt{2}} + 50 	imes \frac{1}{\sqrt{2}} = \frac{200}{\sqrt{2}}\,N$Resolving forces parallel to the inclined plane:$f + T \cos(45^{\circ}) = mg \sin(45^{\circ})$$f = 150 	imes \frac{1}{\sqrt{2}} - 50 	imes \frac{1}{\sqrt{2}} = \frac{100}{\sqrt{2}}\,N$Using the relation $f = \mu N_1$:$\frac{100}{\sqrt{2}} = \mu 	imes \frac{200}{\sqrt{2}}$$\mu = \frac{100}{200} = \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 up by a horizontal string which has a tension of $50\,N$. The coefficient of friction between the surface of contact is: $(g = 10\,m/s^2)$

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