5. When a road is dry, a particular car can safely navigate a turn with a 155 m radius of curvature at a maximum speed of 26.0 m/s without slipping. What is the coefficient of static friction between the tires and the road?
5. When a road is dry, a particular car can safely navigate a turn with a 155 m radius of curvature at a maximum speed of 26.0 m/s without slipping. What is the coefficient of static friction between the tires and the road?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Physics Problem: Coefficient of Static Friction**
**Problem Statement:**
When a road is dry, a particular car can safely navigate a turn with a 155 m radius of curvature at a maximum speed of 26.0 m/s without slipping. What is the coefficient of static friction between the tires and the road?
**Solution:**
To find the coefficient of static friction (\(\mu_s\)), we can use the formula for centripetal force, which must be provided by the frictional force for the car to navigate the turn safely:
\[ f_{\text{friction}} = \mu_s \cdot F_{\text{normal}} = \mu_s \cdot (m \cdot g) \]
The centripetal force (\(F_{\text{centripetal}}\)) necessary for the car to navigate the turn is given by:
\[ F_{\text{centripetal}} = \frac{m \cdot v^2}{r} \]
Where:
- \(m\) is the mass of the car
- \(v\) is the speed of the car (26.0 m/s)
- \(r\) is the radius of curvature (155 m)
- \(g\) is the acceleration due to gravity (approximately 9.8 m/s²)
Since \(f_{\text{friction}} = F_{\text{centripetal}}\), we have:
\[ \mu_s \cdot (m \cdot g) = \frac{m \cdot v^2}{r} \]
The masses (\(m\)) cancel out, leaving:
\[ \mu_s \cdot g = \frac{v^2}{r} \]
Solving for \(\mu_s\):
\[ \mu_s = \frac{v^2}{r \cdot g} \]
Plug in the values:
\[ \mu_s = \frac{(26.0 \, \text{m/s})^2}{155 \, \text{m} \cdot 9.8 \, \text{m/s}^2} \]
\[ \mu_s = \frac{676 \, \text{m}^2/\text{s}^2}{1519 \, \text{m}^2/\text{s}^2} \]
\[ \mu_s = 0.445 \]
Thus, the coefficient of static](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcd0a9c9c-b0c8-47bb-88dc-3ac08f167320%2Ff6c1b331-a06e-4e7f-ac62-fbeae9113c13%2F2f8j8w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Physics Problem: Coefficient of Static Friction**
**Problem Statement:**
When a road is dry, a particular car can safely navigate a turn with a 155 m radius of curvature at a maximum speed of 26.0 m/s without slipping. What is the coefficient of static friction between the tires and the road?
**Solution:**
To find the coefficient of static friction (\(\mu_s\)), we can use the formula for centripetal force, which must be provided by the frictional force for the car to navigate the turn safely:
\[ f_{\text{friction}} = \mu_s \cdot F_{\text{normal}} = \mu_s \cdot (m \cdot g) \]
The centripetal force (\(F_{\text{centripetal}}\)) necessary for the car to navigate the turn is given by:
\[ F_{\text{centripetal}} = \frac{m \cdot v^2}{r} \]
Where:
- \(m\) is the mass of the car
- \(v\) is the speed of the car (26.0 m/s)
- \(r\) is the radius of curvature (155 m)
- \(g\) is the acceleration due to gravity (approximately 9.8 m/s²)
Since \(f_{\text{friction}} = F_{\text{centripetal}}\), we have:
\[ \mu_s \cdot (m \cdot g) = \frac{m \cdot v^2}{r} \]
The masses (\(m\)) cancel out, leaving:
\[ \mu_s \cdot g = \frac{v^2}{r} \]
Solving for \(\mu_s\):
\[ \mu_s = \frac{v^2}{r \cdot g} \]
Plug in the values:
\[ \mu_s = \frac{(26.0 \, \text{m/s})^2}{155 \, \text{m} \cdot 9.8 \, \text{m/s}^2} \]
\[ \mu_s = \frac{676 \, \text{m}^2/\text{s}^2}{1519 \, \text{m}^2/\text{s}^2} \]
\[ \mu_s = 0.445 \]
Thus, the coefficient of static
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