We don't want the crate to slip, we are essential to see when the force of deceleration is equal to the force of friction.
Let:
m = mass of the crate
a = acceleration (in this case deceleration of the truck)
g = acceleration due to gravity, 9.8 m/s^2
μ = coefficient of friction, 0.73
Then
ma = μmg
a = μg
a = 0.75 * 9.8 m/s^2
a = 7.35 m/s^2
Therefore, the driver can securely decelerate at about 7.2 m/s^2, any tougher
and the crate will probably slide in contradiction of the cab.