A 13552.2 N car traveling at 52.2 km/h rounds a curve of radius 1.68 
102 m..
What is the minimum coefficient of static friction (μs) between the tires and the road that will allow the car to round the curve safely?
+33614 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.
+33614 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.
