– 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 $7906.75N$.