Continuing the crate scenario (weight 20 N, pulling force 4 N at 60° to the horizontal, normal reaction $R=20-2\sqrt3$ N, limiting equilibrium), find the exact coefficient of friction between the crate and the floor.

Resolving horizontally, $F=4\cos60°=2$ N. At limiting equilibrium, $F=\mu R$, so $\mu=\dfrac{2}{20-2\sqrt3}=\dfrac{1}{10-\sqrt3}$, and rationalising the denominator gives $\mu=\dfrac{10+\sqrt3}{97}$. Swapping sine and cosine, using $F=4\sin60°=2\sqrt3$ for the horizontal component instead of $4\cos60°$, gives $\dfrac{3+10\sqrt3}{97}$ after rationalising. Dividing by the crate's weight (20 N) instead of the actual normal reaction gives $\dfrac{2}{20}=\dfrac{1}{10}$. Multiplying $F$ and $R$ instead of dividing one by the other does not correspond to the friction law at all, giving $40-4\sqrt3$.