A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 15 m/s Will the car follow the curve, or will it skid? Assume: (a) the pavement is dry and the coefficient of static friction is \mu _s=0.60; (b) the pavement is icy and \mu _s=0.25.

Solution:

The forces on the car are gravity mg downward, the normal force F_N exerted upward by the road, and a horizontal friction force due to the road. They are shown in the free-body diagram of the car below. The car will follow the curve if the maximum static friction force is greater than the mass times the centripetal acceleration.

Part A

In the vertical direction (y) there is no acceleration. Newton’s second law tells us that the normal force on the car is equal to the weight mg since the road is flat:

\begin{align*}
\sum_{}^{}F_y & =m\ a_c\\
\\
F_N\ -\ mg & =0\\
\\
F_N&=mg\\
\\
F_N &=\left( 1000\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)\\
\\
F_N &=9810\  \text{N}
\end{align*}

In the horizontal direction the only force is friction, and we must compare it to the force needed to produce the centripetal acceleration to see if it is sufficient. The net horizontal force required to keep the car moving in a circle around the curve is

\begin{align*}
\sum_{}^{}F_c & =m\ a_c\\
\\
 & =m\cdot \frac{v^2}{r}\\
\\
& =\left( 1000\ \text{kg} \right)\cdot \frac{\left( 15\ \text{m/s} \right)^2}{50\ \text{m}}\\
\\
&=4500\ \text{N}
\end{align*}

Now we compute the maximum total static friction force (the sum of the friction forces acting on each of the four tires) to see if it can be large enough to provide a safe centripetal acceleration. For (a), \mu _s=0.60, and the maximum friction force attainable is

\begin{align*}
\sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\
\\
&=\left( 0.60 \right)\left( 9810\ \text{N} \right)\\
\\
&=5886\ \text{N}
\end{align*}

Since a force of only 4500 N is needed, and that is, in fact, how much will be exerted by the road as a static friction force, the car can follow the curve.

Part B

The maximum static friction force possible is

\begin{align*}
\sum_{}^{}F_{fr_{max}}& =\mu _s \ F_N\\
\\
&=\left( 0.25 \right)\left( 9810\ \text{N} \right)\\
\\
&=2452.5\ \text{N}
\end{align*}

The car will skid because the ground cannot exert sufficient force (4500 N is needed) to keep it moving in a curve of radius 50 m at a speed of 54 km/h.