Question

A concrete highway curve of radius $70m$is banked at a $15°$angle. What is the maximum speed with which a $1500kg$ rubber tired car can take this curve without sliding?

Short answer

The maximum sped the car can move without sliding is $34.47m/s$.

Step-by-step solution

Given Information

A concrete highway curve of radius $70m$ is banked at a $15°$ angle.

Explanation

We need to draw a graph for analyze the situation.

From the graph we can obtain the expression for the horizontal direction;

$\sum F_x:\frac{m{v}_m^2}{r}=N\sin \theta +F\cos \theta$

Where,

$v_m$: The maximum force required

$N$: Normal force

$r$ : Radius of the curve

$m$: mass of the car

role="math" localid="1648181750837" $\theta$: Banking angle

Frictional force role="math" localid="1648181740479" $F={\mu }_s\mathrm{N}$

Therefore,

$N\left(\sin \theta +{\mu }_s\cos \theta \right)=\frac{m{v}_m^2}{r}$................(a)

Expression for vertical force

Expression for vertical force:

$\sum F_y:N\cos \theta =mg+F\sin \theta$

$\Rightarrow N\left(\cos \theta -{\mu }_s\sin \theta \right)=mg$.................(b)

Dividing expression (a) by expression (b), we get

$\frac{v_m^2}{rg}=\frac{\sin \theta +{\mu }_s\cos \theta }{\cos \theta -{\mu }_s\sin \theta }$

$\Rightarrow v_m^2=\frac{rg\left(\sin \theta +{\mu }_s\cos \theta \right)}{\cos \theta -{\mu }_s\sin \theta }$

$\Rightarrow v_m^2=\frac{70\times 9.8\left(\sin {15}^{\circ }+1\times \cos {15}^{\circ }\right)}{\cos {15}^{\circ }-1\times \sin {15}^{\circ }}$

$\Rightarrow v_m^2=1188.1868$

Final Answer

Therefore, the maximum speed the care can move without sliding is $34.47m/s$.