A body of mass 8kg is in limiting equilibrium over an inclined plane of inclination 30º. If the inclination is made 60º, the minimum force required to prevent the body from sliding down is (g = 10ms^–2)

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$\frac{80}{\sqrt{3}} N$

$40 \sqrt{3} N$

$\frac{40}{\sqrt{3}} N$

80 N

answer is B.

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Detailed Solution

When the inclination is 30

f = $μ$ N = $μ$ mgcos $θ$
$μ$ mg cos(30) = mgsin(30)
$μ$ $\sqrt{3}$ /2 = 1/2
$μ$ = 1/ $\sqrt{3}$

When the inclination is 60
f = $μ$ N
= N/ $\sqrt{3}$
= mgcos(60)/ $\sqrt{3}$

= $\frac{1}{2 \sqrt{3}} m g$

$\frac{1}{2 \sqrt{3}} m g$ + f = mg $\frac{\sqrt{3}}{2}$

f = $\frac{m g}{\sqrt{3}}$

= 80/ $\sqrt{3}$

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