Question

A $1500kg$car takes a $50-m$-radius unbanked curve at localid="1649866624550" $15m/s$.What is the size of the friction force on the car?

Short answer

The friction force will be$f=6.8\times {10}^3\mathrm{N}$.

Step-by-step solution

Given Information

mass of the car is$1500kg$

car takes a $50-m$-radius unbanked curve

speed of the car during the unbanked curve islocalid="1647793818214" $15m/s$.

Explanation

The car's velocity is tangent to the circle of motion, and its acceleration is called the centripetal acceleration and it points toward the center of the circle. This acceleration is given by equation in the form

$\overrightarrow{a}=\frac{v^2}{r}\left(I\right)$

Where $r$is the radius of the circle and $v$is the velocity of the car. At a uniform circular motion, the velocity vector has a tangential component and the acceleration vector has a radial component. From Newton 's first law, the car has a net force exerted on it as it doesn't move with a constant velocity. So, using the expression of the acceleration from equation $\left(1\right)$, we get the net force by

localid="1649866873086" ${\overrightarrow{F}}_{\text{net}}=m\overrightarrow{a}=\frac{m{v}^2}{r}\left(2\right)$

Explanation

This force has a direction toward the center of the circular motion. The car moves in a circle with a radius $r=50\mathrm{m}$. Now, we plug the values for $m,v$and $r$into equation $\left(2\right)$to get $F_{\text{net}}$

$F_{\text{net}}=\frac{m{v}^2}{r}$

$=\frac{(1500\mathrm{kg})(15\mathrm{m}/\mathrm{s}{)}^2}{50\mathrm{m}}$

$\approx 6.8\times {10}^3\mathrm{N}$.

This force is an identifiable agent and it is one of our familiar forces such as friction, normal or tension forces/ The car moves on a road in a circle motion, there is no tension force between the car and the center of the circle, also there is no normal force, so the static friction causes this force. Therefore, the friction force will be

$f=6.8\times {10}^3\mathrm{N}$.

Final Answer

The friction force will be$f=6.8\times {10}^3\mathrm{N}$.