A body of mass 1, kg lies on a smooth incline | vedclass

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(A) To find the normal reaction $N$,we resolve the forces perpendicular to the inclined plane.The forces acting on the body are:$1$. The weight of the

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$10 \sqrt{2}\, N$

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(A) To find the normal reaction $N$,we resolve the forces perpendicular to the inclined plane.The forces acting on the body are:$1$. The weight of the body $mg$ acting vertically downwards. Its component perpendicular to the inclined plane is $mg \cos 45^{\circ}$.$2$. The horizontal force $F = 10\, N$. Its component perpendicular to the inclined plane is $F \sin 45^{\circ}$.Since the body is in equilibrium in the direction perpendicular to the plane,the normal reaction $N$ must balance these components:$N = mg \cos 45^{\circ} + F \sin 45^{\circ}$Given $m = 1\, kg$,$g = 10\, m/s^2$,$F = 10\, N$,and $	heta = 45^{\circ}$:$N = (1)(10) \cos 45^{\circ} + 10 \sin 45^{\circ}$$N = 10 \left( \frac{1}{\sqrt{2}} ight) + 10 \left( \frac{1}{\sqrt{2}} ight)$$N = \frac{10}{\sqrt{2}} + \frac{10}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10 \sqrt{2}\, N$.

$A$ body of mass $1\, kg$ lies on a smooth inclined plane. The body is subjected to a horizontal force $F = 10\, N$ as shown in the figure. The magnitude of the net normal reaction on the body is:

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