A maximum of 3N ~force can be applied on& a& block, down the...

Block does not move upto a maximum applied force of $3N$down the inclined plane.

Equation of motion is $3 + mgsin30 ° = F$

$3 + mgsin30 ° = \mu mgcos30 °$ & & & & &....(i)

Similarly, block also does not move upto a maximum applied force of $12N$ up the plane.

Now, equation of motion is

$mgsin30 ° + F = 12$

or, $mgsin30 ° + \mu mgcos30 ° = 12$ & & & & &...(ii)

Solving Eqs. (i) and (ii), we have

$mgsin30 ° = \dfrac{9}{2}$ and $\mu mgcos30 ° = \dfrac{15}{2}$

Hence, dividing

$∴ \dfrac{1}{μ}\tan 30 ° = \dfrac{9 / 2}{15 / 2} = \dfrac{3}{5}$

$⇒ \mu = \dfrac{5}{3\sqrt{3}}$