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Chapter 9: Problem 30

When riding in a \(1300-\mathrm{kg}\) car, you go around a curve of radius \(59 \mathrm{~m}\) with a speed of \(16 \mathrm{~m} / \mathrm{s}\). The coefficient of static friction between the car and the road is \(0.88\). Assuming that the car doesn't skid, what is the force exerted on it by static friction?

Short Answer

Expert verified
The static friction force is approximately 5639.32 N.

Step by step solution

01

Identify the necessary formulas

To determine the force exerted by static friction, we need to identify the use of the centripetal force formula and the friction formula. The centripetal force needed to make a car go around a curve is given by: \[ F_c = \frac{mv^2}{r} \] where: \( m = 1300 \, \text{kg} \) (mass of the car), \( v = 16 \, \text{m/s} \) (velocity of the car), \( r = 59 \, \text{m} \) (radius of the curve) The maximum static friction force can be calculated using: \[ f_s = \mu_s \cdot N \] where: \( \mu_s = 0.88 \) (coefficient of static friction), \( N \) is the normal force (\( N = mg \), where \( g \) is the acceleration due to gravity, \( 9.81 \, \text{m/s}^2 \)).
02

Calculate the centripetal force

Substitute the given values into the centripetal force formula: \[ F_c = \frac{mv^2}{r} = \frac{1300 \times 16^2}{59} \] Do the calculations: \[ F_c = \frac{1300 \times 256}{59} = \frac{332800}{59} \approx 5639.32 \, \text{N} \] Thus, the required centripetal force is approximately \( 5639.32 \, \text{N} \).
03

Calculate the normal force

The normal force \( N \) is equal to the gravitational force, since the car is on a horizontal curve: \[ N = mg = 1300 \times 9.81 \] Calculate it: \[ N = 12753 \, \text{N} \].
04

Determine the maximum static friction force

Use the friction formula to find the maximum possible static friction force: \[ f_s = \mu_s \cdot N = 0.88 \times 12753 \] Calculate the force: \[ f_s = 11222.64 \, \text{N} \].
05

Compare forces and find static friction force

Since the car doesn't skid, the static friction provides exactly the centripetal force needed. Compare the calculated forces: - Maximum static friction force: \( 11222.64 \, \text{N} \) - Required centripetal force: \( 5639.32 \, \text{N} \)Since \( 5639.32 \, \text{N} \) is less than the maximum possible static friction force, the static friction force acting on the car is \( 5639.32 \, \text{N} \).

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Key Concepts

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

Centripetal Force
When you drive a car around a curve, the car needs a force directed towards the center of the curve to keep moving in a circle. This force is known as "centripetal force." It's crucial for making sure the car turns without skidding. The formula to calculate centripetal force is: \[ F_c = \frac{mv^2}{r} \] where:
  • \( m \) is the mass of the object, in this case, the car.
  • \( v \) is the velocity or speed at which the object is moving.
  • \( r \) is the radius of the curve.
Centripetal force is not a force on its own but results from other forces acting. For a car, this force is mainly provided by static friction between the car's tires and the road. When calculated, if this force is not enough or if the speed is too high, the car may skid off the curve.
Coefficient of Friction
Friction is the force that prevents surfaces from sliding over each other easily. The "coefficient of friction" is a number that represents this friction between two materials. For cars, the coefficient of static friction \( \mu_s \) is crucial because it relates to how much grip the tires have on the road.The formula used to calculate the maximum static friction force is:\[ f_s = \mu_s \cdot N \]where:
  • \( \mu_s \) is the coefficient of static friction. A higher value means more friction or grip.
  • \( N \) is the normal force, or the perpendicular contact force between the car and the road.
In this exercise, a coefficient of \(0.88\) indicates a strong frictional force, which helps keep the car on its path around the curve despite the high speed.
Normal Force
The "normal force" is a fundamental concept in physics. It is the force exerted by a surface to support the weight of an object resting on it. In simple terms, it's the force that keeps an object from "falling" through the surface it's resting on. For a car moving on a flat circular path, the normal force \( N \) is equal to the gravitational force acting on it:\[ N = mg \]where:
  • \( m \) is the mass of the object (the car in our example).
  • \( g \) is the acceleration due to gravity, approximately \(9.81 \, \text{m/s}^2\).
This force is crucial in calculating the maximum static friction, which, in turn, is essential for ensuring the car maintains a stable path around the curve without skidding. In this problem, the car weighs quite a bit, leading to a significant normal force, which supports a correspondingly high static friction force.

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