Problem 42
Question
(II) How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius \(85 \mathrm{~m}\) at a speed of \(95 \mathrm{~km} / \mathrm{h} ?\)
Step-by-Step Solution
Verified Answer
The coefficient of static friction must be at least 0.833.
1Step 1: Convert the speed to meters per second
The given speed is \(95 \text{ km/h}\). We need to convert this to meters per second (m/s) to use it in our calculations.\[\text{Speed in m/s} = \frac{95 \times 1000}{60 \times 60} = 26.39 \text{ m/s}\]
2Step 2: Calculate the centripetal force required
The centripetal force \(F_c\) needed for a car to move in a circle is given by the formula \( F_c = \frac{mv^2}{r} \), where \(m\) is the mass, \(v\) is the speed, and \(r\) is the radius of the curve. We will introduce the mass variable \(m\) for the car, but it will cancel out later, so we don't need to know its exact value yet.
3Step 3: Relate centripetal force to static friction
The frictional force \(f_s\) provides the necessary centripetal force. Hence, \(f_s = F_c\). Also, \(f_s = \mu_s N \), where \(\mu_s\) is the coefficient of static friction and \(N = mg\) is the normal force. Thus, \( \mu_s mg = \frac{mv^2}{r} \).
4Step 4: Solve for the coefficient of static friction \( \mu_s \)
We equate the static frictional force equation: \( \mu_s mg = \frac{mv^2}{r} \) to find \( \mu_s \). Cancel out \(m\) from both sides:\[\mu_s = \frac{v^2}{rg}\].
5Step 5: Substitute known values
Now, substitute the known values: \( v = 26.39 \text{ m/s} \), \(r = 85 \text{ m}\), and gravitational acceleration \( g = 9.81 \text{ m/s}^2 \) into the equation: \[\mu_s = \frac{(26.39)^2}{85 \times 9.81}\]
6Step 6: Calculate \( \mu_s \)
Substitute the values into the formula to find the coefficient of static friction \( \mu_s \):\[\mu_s = \frac{695.712}{834.85} = 0.8334\]
Key Concepts
Centripetal ForceNormal ForceStatic Frictional Force
Centripetal Force
When a car moves around a curve, it needs to stay on its path without sliding off. This is where centripetal force comes in, acting as the "inward" force required to keep the car moving in its circular path. The formula for centripetal force is:
- \(F_c = \frac{mv^2}{r}\)
- \(F_c\) is the centripetal force
- \(m\) is the mass of the car
- \(v\) is the velocity of the car
- \(r\) is the radius of the curve
Normal Force
Normal force is the force exerted by a surface perpendicular to the object resting upon it. For a car traveling on a level road, this force is directed upward, counterbalancing the gravitational pull downward. For a car of mass \(m\) on such a horizontal surface, the normal force \(N\) can be represented as:
- \(N = mg\)
Static Frictional Force
Static friction is the frictional force that prevents two surfaces from sliding past each other. In the context of a car on a curve, it's this friction that provides the necessary grip for the tires to not slip on the road.The formula relating static frictional force \(f_s\) to normal force \(N\) includes the coefficient of static friction \(\mu_s\):
- \(f_s = \mu_s N\)
- \(f_s = F_c\)
- \(\mu_s mg = \frac{mv^2}{r}\)
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