Answer
The minimum coefficient of static friction is $~~0.50$
Work Step by Step
The box will roll about a rotation axis at one of the bottom edges of the box.
Let $L$ be the length of each side of the box.
To find the minimum required force, we can assume that the torque due to the applied force is equal and opposite to the torque due to the box's weight:
$L~F = (\frac{L}{2})(mg)$
$F = \frac{mg}{2}$
$F = \frac{890~N}{2}$
$F = 445~N$
So that the box does not slide, the force of static friction must be equal and opposite to the applied force. We can find the minimum coefficient of static friction:
$mg~\mu_s = F$
$\mu_s = \frac{F}{mg}$
$\mu_s = \frac{445~N}{890~N}$
$\mu_s = 0.50$
The minimum coefficient of static friction is $~~0.50$
