Suggested languages for you:

Americas

Europe

Q6.3-30 PE

Expert-verifiedFound in: Page 191

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a \(100{\rm{ m}}\) radius curve banked at \(15.0^\circ \). (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at \(20.0{\rm{ km}}/{\rm{h}}\)?**

- The ideal speed to take the turn is \(16.2{\rm{ m}}/{\rm{s}}\).
- Minimum coefficient of friction is \(0.27\).

**The coefficient of friction is defined as the ratio of frictional force to the normal reaction force. It is a scalar and dimensionless quantity with no units.**

The ideal speed to take a safe turn is,

\(v = \sqrt {rg\tan \theta } \)

Here, \(r\) is the radius of the curved road, \(g\) is the acceleration due to gravity, and \(\theta \) is the ideal banking angle.

Substitute \(100{\rm{ m}}\) for \(r\), \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\), and \(15.0^\circ \) for \(\theta \),

\(\begin{aligned}{}v = \sqrt {\left( {100{\rm{ m}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \tan \left( {15.0^\circ } \right)} \\ = 16.2{\rm{ m}}/{\rm{s}}\end{aligned}\)

Hence, the ideal speed to take the turn is \(16.2{\rm{ m}}/{\rm{s}}\).

The coefficient of friction is,

\(\mu = \frac{{{v^2}}}{{rg}}\)

Substitute \(16.2{\rm{ m}}/{\rm{s}}\) for \(v\), \(100{\rm{ m}}\) for \(r\), \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\),

\(\begin{aligned}{}\mu = \frac{{{{\left( {16.2{\rm{ m}}/{\rm{s}}} \right)}^2}}}{{\left( {100{\rm{ m}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = 0.27\end{aligned}\)

Hence, the minimum coefficient of friction is \(0.27\).

94% of StudySmarter users get better grades.

Sign up for free