A block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient of friction is $\mu$. The force required to just push it up the inclined plane is $3$ times the force required to just prevent it from sliding down. If we define $N=10 \ \mu$, then $N$ is
IIT 2011, Advanced
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$F_1=3 F_2$
$mg \left(\sin 45^{\circ}+\mu \cos 45^{\circ}\right)=3 mg \left(\sin 45^{\circ}-\mu \cos 45^{\circ}\right)$
$\frac{1}{\sqrt{2}}(1+\mu)=\frac{3}{\sqrt{2}}(1-\mu)$
$\mu=0.5$
$\therefore \quad N=10 \mu$
$=10 \times 0.5$
$N=5$
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