A 13552.2 N car traveling at 52.2 km/h rounds a curve of radius 1.68 multiplied by 102 m.. What is the minimum coefficient of static frictio

Frictional force is given by  $$F=\mu W$$  where μ is the coefficient of friction and W is the weight (13552.2N here).

The centripetal force required to prevent slipping is  $$F=m\frac{v^2}{r}$$  where m is the mass (=W/g), v is speed and r is radius.

Equating these two forces we have:  $$\mu W = \frac{W}{g}\frac{v^2}{r}$$   so   $$\mu = \frac{v^2}{gr}$$   

v = 52.2*10³/3600 m/s, r = 1.68*102m and g = 9.81m/s². 

$${\frac{{\left({\frac{{\mathtt{52\,200}}}{{\mathtt{3\,600}}}}\right)}^{{\mathtt{2}}}}{\left({\mathtt{9.81}}{\mathtt{\,\times\,}}{\mathtt{1.68}}{\mathtt{\,\times\,}}{\mathtt{102}}\right)}} = {\mathtt{0.125\: \!071\: \!265\: \!339\: \!299\: \!2}}$$

So   $$\mu = 0.125$$   approx.

Thanks Alan.