In the following arrangement, an object of mass 1.2 kg is at rest at point p. If N and the F are the reaction and the force of friction, then $(g = 10 m s^{- 1})$ .

N = 6 N, F = 6 \sqrt{3} N

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N = 3 N, F = 3 \sqrt{3} N

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N = 6 N, F = 3 N

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N = 6 \sqrt{3} N, F = 6 N

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Solution

The correct option is D $N = 6 \sqrt{3} N, F = 6 N$
$N = m g c o s 30 = 1.2 \times 10 \times \frac{\sqrt{3}}{2} = 6 \sqrt{3} N$
$F = m g s i n 30 = 1.2 \times 10 \times \frac{1}{2} = 6 N$

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