The mass of the case is $m = 1200 mathrm{~kg}$. The radius of the corner is $r = 45 mathrm{~m}$. The car’s speed is $v= 12 mathrm{~m/s}$.

$textbf{Required: }$

Finding the minimum static friction coefficient.

Solve for the figure in example $9.8$:

According to Newton’s second law in the vertical direction:

$$
begin{align*}
sum F_{y} &= W – N \
&= 0 \
W &= N \
end{align*}
$$

According to Newton’s second law in the horizontal direction:

$$
begin{align*}
sum F_{x} &=m ~ a\
&= f_{s}
end{align*}
$$

As textbook mentions the centripetal acceleration is given by the following relation

$$
begin{align*}
a_{cp} &= dfrac{ v^{2} }{ r} \
&= dfrac{ left( 1 mathrm{~m/s} right)^{2} }{ 5 mathrm{~m} } \
&= 0.2 mathrm{~m/s^{2}}
end{align*}
$$

Substituting from the previous calculation, then we get

$$
begin{align*}
f_{s} &= m ~ a \
&= m ~ dfrac{ v^{2} }{ r} \
&= mu_{s} ~ N \
&= mu_{s} ~ W \
&= mu_{s} ~ m ~ g \
end{align*}
$$

Rearrange and solve for the minimum coefficient of static friction:

$$
begin{align*}
mu_{s} &= dfrac{ v^{2} }{ r ~ g } \
&= dfrac{ left( 12 mathrm{~m/s} right)^{2} }{ 45 mathrm{~m} times 9.8 mathrm{~m/s^{2}} } \
&= 0.327
end{align*}
$$

So, the minimum coefficient of static friction is $0.327$.

The mass of the car is $m = 1300 mathrm{~kg}$. The radius of the curved path is $r = 59 mathrm{~m}$. The car’s speed is $v = 16 mathrm{~m/s}$. The static friction coefficient is $mu_{s} = 0.88$.

$textbf{Required: }$

Finding the exerted force on the car by the static friction.

Solve for the figure in example $9.8$:

According to Newton’s second law in the vertical direction:

$$
begin{align*}
sum F_{y} &= W – N \
&= 0 \
W &= N \
end{align*}
$$

According to Newton’s second law in the horizontal direction:

$$
begin{align*}
f_{cp} &=m ~ a_{cp} \
&= f_{s}
end{align*}
$$

As Newton’s second law for circular motion, the centripetal force is given by the mass time the centripetal acceleration:

$$
begin{align*}
f_{cp} &= m ~ a_{cp} \
&= m ~ dfrac{ v^{2} }{ r} \
&= 1.3 times 10^{3} mathrm{~kg} times dfrac{ left( 16 mathrm{~m/s} right)^{2} }{ 59 mathrm{~m} } \
&= 5.64 times 10^{3} mathrm{~N}
end{align*}
$$

So, the exerted force on the car by the static friction. is $5.64 times 10^{3} mathrm{~N}$.