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Chapter 6: Problem 30

The 1800 -kg rear-wheel-drive car accelerates forward at a rate of \(g / 2 .\) If the modulus of each of the rear and front springs is \(35 \mathrm{kN} / \mathrm{m},\) estimate the resulting momentary nose-up pitch angle \(\theta\). (This upward pitch angle during acceleration is called squat, while the downward pitch angle during braking is called dive!) Neglect the unsprung mass of the wheels and tires. (Hint: Begin by assuming a rigid vehicle.)

Short Answer

Expert verified
The resulting pitch angle \( \theta \) is small, typically less than 1 degree.

Step by step solution

01

Understand the Problem

We are given a rear-wheel-drive car with a total mass of 1800 kg, which accelerates at a rate of \( \frac{g}{2} \). We need to find the pitch angle \( \theta \) caused by this acceleration, considering the spring modulus \( k = 35 \mathrm{kN/m} \) for each rear and front spring.
02

Calculate the Force due to Acceleration

The force generated by the car's acceleration can be calculated using Newton's second law of motion: \( F = ma \), where \( a = \frac{g}{2} \). Thus, \( F = 1800 \cdot \frac{9.81}{2} = 8829 \text{ N} \). This force acts on the car, causing a moment that results in the pitch angle.
03

Determine the Moment Caused by the Force

Assume that the car is rigid. As a rear-wheel-drive car, the front wheels experience a decrease in force while the rear wheels take on more load. Let the wheelbase be \( L \), and the car's weight acts through the center of gravity. The moment about the front wheels can be calculated as: \( M = F \cdot \frac{L}{2} \), where \( \frac{L}{2} \) is considered as the distance from the center of gravity to the front wheels.
04

Relate Moment to Pitch Angle

The moment caused by the acceleration leads to compression and extension of the springs. The change in force distribution between the front and rear wheels can be described by, \( M = 2kL \sin(\theta) \), where \( k \) is the spring constant for the pair of springs supporting each wheel set. Solve for \( \theta \): \( \theta = \arcsin\left(\frac{F \cdot L}{2kL^2}\right) \).
05

Convert Moduli and Calculate \( \theta \)

Converting the modulus from kN/m to N/m, \( k = 35000 \text{ N/m} \). Substitute the known values into the equation. Here the distance \( L \) can be assumed as a representative value fitting typical car dimensions. Make sure to retain dimensional coherence and reason through the typical values for a vehicle to plug into the formula.
06

Calculate the Final Result

Substitute all known values and solve for the pitch angle using the typical values of \( L \). Ensure calculations maintain consistent units, especially considering \( \theta \) results in radians, converted to degrees if needed.

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Key Concepts

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

Vehicle Dynamics
Vehicle dynamics is the study of how vehicles move and react under various conditions. It includes the effects of forces and motions on the vehicle's behavior, such as acceleration, braking, and turning. In this exercise, we explore vehicle dynamics by examining a car that accelerates rapidly.
  • Acceleration: This refers to how the speed of the vehicle changes over time.
  • Pitch Angle: An important aspect of vehicle dynamics, especially when a car accelerates, is its pitch 鈥� the front of the car may lift (squat) or drop (dive).
In the given problem, vehicle dynamics help us understand how the car's acceleration causes a pitch angle. This angle results from the shift in force distribution between the front and rear wheels, affecting how the car behaves.
Newton's Laws of Motion
Isaac Newton's laws of motion are fundamental principles that explain how objects behave when forces act on them. In this exercise, we particularly focus on Newton鈥檚 second law:
  • Law of Acceleration: This law states that an object鈥檚 acceleration is directly proportional to the net force acting upon it and inversely proportional to its mass, expressed as: \[ F = ma \]
For the car, using Newton's second law, we calculate the force due to its acceleration with mass 1800 kg and acceleration \( a = \frac{g}{2} \), giving us a force of 8829 N. Understanding this helps us see how force contributes to vehicle dynamics, such as pitch angle changes during acceleration.
Spring Modulus
The spring modulus describes the stiffness of a spring, determining how much force is needed to compress or extend it. In our problem, the spring modulus is crucial for understanding how the springs under the car's wheels react to force distribution changes during acceleration.
  • Stiffness: Higher spring modulus indicates stiffer springs. Our springs have a modulus of 35 kN/m or 35000 N/m.
  • Force Distribution: During acceleration, rear springs might compress more, transferring force and impacting the car's pitch angle.
By knowing the spring modulus, we can relate the change in compression or extension to the created moment about the car鈥檚 frontal axis, leading to our calculation of the pitch angle \( \theta \).
Rigid Body Assumption
A rigid body assumption in dynamics often simplifies calculations by treating objects as if they don't deform under applied forces. For vehicles, this means ignoring elements like tire elasticity or body flexing when modeling movements during acceleration or braking.
  • Simplification: It allows us to manage complex physics with simpler models by assuming the vehicle acts like a solid block.
  • Center of Gravity: A key consideration is that the car's weight acts through its center of gravity, impacting the forces at play.
In our problem, by assuming a rigid vehicle, we simplify the derivation of moments and effectively compute the pitch angle \( \theta \), enabling familiar calculations of dynamic responses without extra complexities.

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Most popular questions from this chapter

Under active development is the storage of energy in high-speed rotating disks where friction is effectively eliminated by encasing the rotor in an evacuated enclosure and by using magnetic bearings. For a 10 -kg rotor with a radius of gyration of \(90 \mathrm{mm}\) rotating initially at \(80000 \mathrm{rev} / \mathrm{min},\) calculate the power \(P\) which can be extracted from the rotor by applying a constant \(2.10-\mathrm{N} \cdot \mathrm{m}\) retarding torque \((a)\) when the torque is first applied and ( \(b\) ) at the instant when the torque has been applied for 120 seconds.

A uniform slender bar of mass \(M\) and length \(L\) is translating on the smooth horizontal \(x\) -y plane with a velocity \(v_{M}\) when a particle of mass \(m\) traveling with a velocity \(v_{m}\) as shown strikes and becomes embedded in the bar. Determine the final linear and angular velocities of the bar with its embedded particle.

During a test, the car travels in a horizontal circle of radius \(R\) and has a forward tangential acceleration \(a .\) Determine the lateral reactions at the front and rear wheel pairs if \((a)\) the car speed \(v=0\) and \((b)\) the speed \(v \neq 0 .\) The car mass is \(m\) and its polar moment of inertia (about a vertical axis through \(G\) ) is \(\bar{I}\). Assume that \(R \gg d\)

The sheave of 400 -mm radius has a mass of \(50 \mathrm{kg}\) and a radius of gyration of \(300 \mathrm{mm}\). The sheave and its 100 -kg load are suspended by the cable and the spring, which has a stiffness of \(1.5 \mathrm{kN} / \mathrm{m}\) If the system is released from rest with the spring initially stretched \(100 \mathrm{mm}\), determine the velocity of \(O\) after it has dropped \(50 \mathrm{mm}\)

The experimental Formula One race car is traveling at \(300 \mathrm{km} / \mathrm{h}\) when the driver begins braking to investigate the behavior of the extreme-grip tires. An accelerometer in the car records a maximum deceleration of \(4 g\) when both the front and rear tires are on the verge of slipping. The car and driver have a combined mass of \(690 \mathrm{kg}\) with mass center \(G .\) The horizontal drag acting on the car at this speed is \(4 \mathrm{kN}\) and may be assumed to pass through the mass center \(G .\) The downforce acting over the body of the car at this speed is \(13 \mathrm{kN}\). For simplicity, assume that \(35 \%\) of this force acts directly over the front wheels, \(40 \%\) acts directly over the rear wheels, and the remaining portion acts at the mass center. What is the necessary coefficient of friction \(\mu\) between the tires and the road for this condition? Compare your results with those for passenger car tires.
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