To find the smallest radius of curvature the truck can take while going 2.9 m/s, we'll use the formula that relates the centripetal force to the force of static friction:

$$ F_c = F_s $$

$$ \frac{mv^2}{r}= \mu_sn$$

>>Solving for r we have:

$$r= \frac{mv^2}{\mu_sn}=\frac{(96\text{ kg})(2.9\text{ m/s})^2}{(0.14)(96\text{ kg})(9.81\text{ m/s}^2)}$$

$$r=\frac{793.72 \text{ m}^2/\text{s}^2}{941.76 \text{ m/s}^2}$$

$$ \boxed{r= \bf{0.842 \text{ m} } }$$