**– Without causing the car to skid off, calculate the friction Force action on the car while taking the turn.**

This question aims to find the **friction force** acting on the car while it is taking a **turn on an unbanked curve**.

The basic concept behind **friction force** is the **centrifugal force** that is acting on the car away from the center of the curve while taking a turn. When a car takes a turn with a certain velocity, it experiences a **centripetal acceleration** $a_c$.

To keep the car moving without skidding off, a **static frictional force** $F_f$ must act towards the center of the curve, which is always equal and opposite to the **centrifugal force.**

We know that **Centripetal Acceleration** is $a_c$.

\[a_c= \frac{v^2}{r}\]

As per **Newton’s Second Law of Motion:**

\[F_f=ma_c\]

By multiplying both side with mass $m$, we get:

\[F_f=ma_c= \frac{mv^2}{r}\]

Where:

**$F_f=$ Friction force**

**$m=$ Mass of Object**

**$v=$Velocity of Object**

**$r=$ Radius of Curve or Circular path**

## Expert Answer

Given As:

**Mass of Car $m=1500kg$**

**Velocity of Car $v=15\dfrac{m}{s}$**

**Radius of Curve $r=50m$**

**Friction Force $F_f=?$**

As we know that when the car is taking a turn, a **static frictional force** $F-f$ is required to act towards the center of the curve in order to oppose the** centrifugal force** and prevent the car from skidding off.

We know that **Friction Force** $F_f$ is calculated as follows:

\[F_f= \frac{mv^2}{r} \]

Substituting the values from the given data:

\[F_f= \frac{1500kg\times{(15\dfrac{m}{s})}^2}{50m} \]

\[F_f= 6750\frac{kgm}{s^2}\]

As we know that **SI Unit** of **Force** is **Newton** $N$:

\[1N=1 \frac{kgm}{s^2}\]

Hence:

\[F_f=6750N\]

## Numerical Result

The **Friction Force** $F_f$ acting on the car while taking a turn and preventing it from skidding off is $6750N$.

## Example

A **car weighing** $2000kg$, moving at $96.8 \dfrac{km}{h}$, travels around a circular curve of **radius** $182.9m$ on a flat country road. Calculate the **Friction Force** action on the car while taking the turn without slipping.

Given As:

**Mass of Car $m=2000kg$**

**Velocity of Car $v=96.8\dfrac{km}{h}$**

**Radius of Curve $r=182.9m$**

**Friction Force $F_f=?$**

Converting the **velocity** into $\dfrac{m}{s}$

\[v=96.8\frac{km}{h}=\dfrac{96.8\times1000}{60 \times60}\dfrac{m}{s} \]

\[v=26.89\dfrac{m}{s} \]

Now by using the concept of **Frictional Force** acting on bodies which are moving in a curved path, we know that **Friction Force** $F_f$ is calculated as follows:

\[F_f= \frac{mv^2}{r}\]

Substituting the values from the given data:

\[F_f= \frac{2000kg\times{(26.89\dfrac{m}{s})}^2}{182.9m}\]

\[F_f=7906.75\dfrac{kgm}{s^2} \]

As we know that **SI Unit** of **Force** is **Newton** $N$:

\[1N=1 \frac{kgm}{s^2}\]

Hence:

\[F_f=7906.75N\]

Hence, the **Friction Force** $F_f$ acting on the car while taking a turn and preventing it from slipping is $7906.75N$.