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Chapter 5: Problem 42

(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 \(95 \mathrm{~km} / \mathrm{h} ?\)

Short Answer

Expert verified
The coefficient of static friction must be at least 0.833.

Step by step solution

01

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.\[\text{Speed in m/s} = \frac{95 \times 1000}{60 \times 60} = 26.39 \text{ m/s}\]
02

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.
03

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, \( \mu_s mg = \frac{mv^2}{r} \).
04

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:\[\mu_s = \frac{v^2}{rg}\].
05

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: \[\mu_s = \frac{(26.39)^2}{85 \times 9.81}\]
06

Calculate \( \mu_s \)

Substitute the values into the formula to find the coefficient of static friction \( \mu_s \):\[\mu_s = \frac{695.712}{834.85} = 0.8334\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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:
  • \(F_c = \frac{mv^2}{r}\)
Here:
  • \(F_c\) 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
In many cases, the mass \(m\) appears on both sides of the equation and can be canceled out, making calculations simpler when we just need to solve for other variables like speed or friction. This centripetal force is vital in ensuring the vehicle stays on its circular track.
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:
  • \(N = mg\)
Here, \(g\) is the acceleration due to gravity, which is approximately \(9.81\, \text{m/s}^2\) on Earth. This normal force plays a crucial role because it impacts how much friction the tires can exert on the road鈥攅ssential for preventing the car from skidding. Without the normal force, static friction wouldn't be present to provide the centripetal force needed to safely maneuver around curves.
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 \(\mu_s\):
  • \(f_s = \mu_s N\)
In our exercise, static friction delivers the centripetal force, so we have:
  • \(f_s = F_c\)
  • \(\mu_s mg = \frac{mv^2}{r}\)
After canceling the mass \(m\) (as it appears on both sides), you can solve for \(\mu_s\) to find out just how "sticky" or "grippy" the road needs to be. The coefficient of static friction \(\mu_s\) informs us about the minimum friction necessary to keep the car from sliding off the curve while maintaining its speed.

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Most popular questions from this chapter

(II) The design of a new road includes a straight stretch that is horizontal and flat but that suddenly dips down a steep hill at \(22^{\circ} .\) The transition should be rounded with what minimum radius so that cars traveling \(95 \mathrm{~km} / \mathrm{h}\) will not leave the road (Fig. \(5-45) ?\)

The sides of a cone make an angle \(\phi\) with the vertical. A small mass \(m\) is placed on the inside of the cone and the cone, with its point down, is revolved at a frequency \(f\) (revolutions per second) about its symmetry axis. If the coefficient of static friction is \(\mu_{\mathrm{s}}\), at what positions on the cone can the mass be placed without sliding on the cone? (Give the maximum and minimum distances, \(r\), from the axis).

(II) A sports car crosses the bottom of a valley with a radius of curvature equal to \(95 \mathrm{~m}\). At the very bottom, the normal force on the driver is twice his weight. At what speed was the car traveling?

A car drives at a constant speed around a banked circular track with a diameter of 127 \(\mathrm{m}\) . The motion of the car can be described in a coordinate system with its origin at the center of the circle. At a particular instant the car's accel- eration in the horizontal plane is given by $$\vec{\mathbf{a}}=(-15.7 \hat{\mathbf{i}}-23.2 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2}$$ (a) What is the car's speed? (b) Where \((x\) and \(y)\) is the car at this instant?

A 72 -kg water skier is being accelerated by a ski boat on a flat \((\) "glassy") lake. The coefficient of kinetic friction between the skier's skis and the water surface is \(\mu_{k}=0.25\) (Fig. 55 ). (a) What is the skier's acceleration if the rope pulling the skier behind the boat applies a horizontal tension force of magnitude \(F_{T}=240 \mathrm{N}\) to the skier \(\left(\theta=0^{\circ}\right) ?\) (b) What is the skier's horizontal acceleration if the rope pulling the skier exerts a force of \(F_{T}=240 \mathrm{N}\) on the skier at an upward angle \(\theta=12^{\circ} ?\) (c) Explain why the skier's acceleration in part \((b)\) is greater than that in part \((a)\) .
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