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A body of mass 2 kg is placed on a smooth horizontal surface. Two forces $F_{1} = 20 N a n d F_{2} = 10 \sqrt{3} N$ are acting on the body in directions making angle of $30^{0}$ and $60^{0}$ to the surface. The reaction of the surface on the body will be

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a

20N

b

25N

c

5N

d

zero

answer is D.

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$N + F_{1} Sin 30^{0} + F_{2} Sin 60^{0} = m g$
$\Rightarrow N + 20 \times \frac{1}{2} + 10 \sqrt{3} . \frac{\sqrt{3}}{2}$
$= 20$
$\Rightarrow N = 20 - 25 = - 5$
Normal reaction is coming negative means the body is lifted from the ground and hence N = 0

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