A block is in limiting equilibrium on a rough horizontal surface. If the net contact force is 2 times the normal force, the coefficient of static friction is

Fnet =N2+f22N=N2+μ2N22N2=N21+μ2⇒2=μ2+1⇒μ2=1⇒μ=1

Fnet =N2+f22N=N2+μ2N22N2=N21+μ2⇒2=μ2+1⇒μ2=1⇒μ=1

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A block is in limiting equilibrium on a rough horizontal surface. If the net contact force is $\sqrt{2}$ times the normal force, the coefficient of static friction is

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$\frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{3}}$

0.5

answer is A.

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$F_{net} = \sqrt{N^{2} + f^{2}}$

$\sqrt{2} N = \sqrt{N^{2} + μ^{2} N^{2}}$

$2 N^{2} = N^{2} (1 + μ^{2})$

$\Rightarrow 2 = μ^{2} + 1 \Rightarrow μ^{2} = 1 \Rightarrow μ = 1$

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