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Chapter 13: Problem 7

Determine the maximum theoretical speed that may be achieved over a distance of 110 m by a car starting from rest assuming there is no slipping. The coefficient of static friction between the tires and pavement is 0.75, and 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume (a) front-wheel drive, (b) rear-wheel drive.

Short Answer

Expert verified
Maximum speed is approximately 28.0 m/s for front-wheel drive and 23.3 m/s for rear-wheel drive.

Step by step solution

01

Understanding Forces

First, we need to calculate the maximum friction force since it gives the maximum possible accelerating force. The force of friction is found using the formula: \[ F_{f} = \mu_s \times N \]where \(\mu_s = 0.75\) is the coefficient of static friction, and \(N\) is the normal force which is equal to the weight distributed over the driving wheels.
02

Calculate Normal Forces

The total weight \( W \) of the car is distributed as 60% over the front tires and 40% over the rear tires. So for a front-wheel-drive:\[ N_{front} = 0.6 \times W \]And for a rear-wheel-drive:\[ N_{rear} = 0.4 \times W \]
03

Determine Maximum Frictional Force

Using the expression from Step 1, calculate the maximum frictional force for both scenarios:- For front-wheel drive: \[ F_{f,max,front} = \mu_s \times N_{front} = 0.75 \times 0.6W = 0.45W \]- For rear-wheel drive: \[ F_{f,max,rear} = \mu_s \times N_{rear} = 0.75 \times 0.4W = 0.3W \]
04

Apply Newton's Second Law

Determine the acceleration using Newton's second law, \( F = ma \). So:- For front-wheel drive:\[ 0.45W = ma \] leading to \( a_{front} = \frac{0.45W}{m} = 0.45g \)- For rear-wheel drive:\[ 0.3W = ma \] leading to \( a_{rear} = \frac{0.3W}{m} = 0.3g \) where \( g \approx 9.81 \, \text{m/s}^2 \).
05

Use Kinematic Equation for Maximum Speed

Using the kinematic equation, \( v^2 = u^2 + 2as \), where \(u = 0\), \(s = 110\, \text{m}\):- For front-wheel drive:\[ v_{max,front} = \sqrt{2 \times 0.45g \times 110} \]- For rear-wheel drive:\[ v_{max,rear} = \sqrt{2 \times 0.3g \times 110} \]
06

Calculate Values

Now calculate using \( g = 9.81 \, \text{m/s}^2 \):- For front-wheel drive:\[ v_{max,front} = \sqrt{2 \times 0.45 \times 9.81 \times 110} \approx 28.0\, \text{m/s} \]- For rear-wheel drive:\[ v_{max,rear} = \sqrt{2 \times 0.3 \times 9.81 \times 110} \approx 23.3\, \text{m/s} \]

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Key Concepts

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

Coefficient of Static Friction
The coefficient of static friction is a crucial value when it comes to understanding how vehicles grip the road. It represents the friction between the car's tires and the road without sliding or slipping happening. This coefficient is denoted as \[ \mu_s \]and varies based on road conditions and tire material. In the exercise, the coefficient of static friction is 0.75, which means that the frictional force is 75% of the normal force exerted by the tires on the road.
  • This value indicates the maximum grip before sliding.
  • It helps to understand how much speed and acceleration a vehicle can handle before losing traction.
Having a good knowledge of this coefficient in vehicle dynamics helps engineers design safer and more efficient automobiles.
Newton's Second Law
One of the fundamental laws of physics, Newton's second law, is expressed in the formula:\[ F = ma \]where \(F\) is the force applied to an object, \(m\) is the mass of the object, and \(a\) is the acceleration.In this context, the force in question is the frictional force preventing the tires from slipping. By rearranging the equation to solve for acceleration \(a\), we get:\[ a = \frac{F}{m} \]For our exercise, this law is applied to determine how much acceleration the car can have.
  • Higher frictional force leads to better acceleration.
  • The distribution of weight affects which wheels can provide more force to accelerate.
Understanding and applying Newton's second law allows us to predict how a car will behave under different driving conditions.
Kinematic Equations
Kinematic equations describe motion without considering the forces that cause it. In our problem, we use one such equation to calculate the maximum speed of the car:\[ v^2 = u^2 + 2as \]Where:
  • \(v\) = final velocity
  • \(u\) = initial velocity (0 since the car starts from rest)
  • \(a\) = acceleration
  • \(s\) = distance (110 m in our problem)
The equation helps us to understand the relationship between acceleration and the distance traveled when calculating final velocity.
  • This method calculates speed without immediately involving force or friction.
  • It's a tool to predict how fast a car can theoretically go over a given distance.
By using these equations, we can assess the performance of different wheel drive configurations.
Wheel Drive Systems
Wheel drive systems determine how a vehicle's engine power is distributed to its wheels. In the exercise, we compare a front-wheel drive against a rear-wheel drive setup.
  • In front-wheel drive (FWD), power is sent to the front wheels. It typically offers better traction under light loads because the weight of the engine is over the driving wheels.
  • Rear-wheel drive (RWD) sends power to the back wheels, often providing better handling and acceleration performance once the car is in motion, especially in performance vehicles.
Understanding the impact of these systems helps when considering vehicle dynamics and how they relate to forces such as friction and acceleration. Each setup has its advantages and ideal conditions for use; for instance, FWD offers better traction in slippery conditions, while RWD can be superior in dry conditions with higher speeds. In our specific problem:
  • The FWD vehicle can reach a higher speed (28 m/s) because more weight, and hence more friction, is applied to the driving front wheels.
  • The RWD vehicle achieves a slightly lower speed (23.3 m/s) due to less weight on the driving rear wheels.
With these insights, we can see how engineering choices influence a car's performance according to the wheels tasked with driving it.

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Most popular questions from this chapter

A satellite is projected into space with a velocity \(\mathbf{v}_{0}\) at a distance \(r_{0}\) from the center of the earth by the last stage of its launching rocket. The velocity \(\mathbf{v}_{0}\) was designed to send the satellite into a circular orbit of radius \(r_{0}\). However, owing to a malfunction of control, the satellite is not projected horizontally but at an angle \(\alpha\) with the horizontal and, as a result, is propelled into an elliptic orbit. Determine the maximum and minimum values of the distance from the center of the earth to the satellite.

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A missile is fired from the ground with an initial velocity \(v_{0}\) forming an angle \(\phi_{0}\) with the vertical. If the missile is to reach a maximum altitude equal to \(\alpha R,\) where \(R\) is the radius of the earth, \((a)\) show that the required angle \(\phi_{0}\) is defined by the relation $$\sin \phi_{0}=(1+\alpha) \sqrt{1-\frac{\alpha}{1+\alpha}\left(\frac{v_{\mathrm{esc}}}{v_{0}}\right)^{2}}$$ where \(v_{\mathrm{esc}}\) is the escape velocity, ( \(b\) ) determine the range of allowable values of \(v_{0}\).
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