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Chapter 3: Problem 9

The inexperienced driver of an all-wheel-drive car applies too much throttle as he attempts to accelerate from rest up the slippery 10 -percent incline. The result is wheel spin at all four tires, each of which has the same gripping ability. Determine the vehicle acceleration for the conditions of \((a)\) light snow, \(\mu_{k}=\) 0.12 and \((b)\) ice \(, \mu_{k}=0.05\).

Short Answer

Expert verified
Acceleration: (a) Light snow: 0.075 m/s虏, (b) Ice: -0.561 m/s虏.

Step by step solution

01

Understand Given Information

We have a car on a 10% incline, which means the incline angle \( \theta \) such that \( \tan(\theta) = 0.1 \). The coefficient of kinetic friction \( \mu_k \) is given for two scenarios: light snow \( \mu_k = 0.12 \) and ice \( \mu_k = 0.05 \).
02

Free Body Diagram and Forces Analysis

The forces acting on the car are its weight \( mg \), the normal force \( N \), and the frictional force. The frictional force provides the forward acceleration, calculated from \( f_{friction} = \mu_k N \). The weight component along the incline is \( mg \sin(\theta) \), and perpendicular to the incline is \( mg \cos(\theta) \).
03

Calculate Normal Force

The normal force \( N \) can be calculated from the balance of forces perpendicular to the incline. Thus, \( N = mg \cos(\theta) \).
04

Calculate Frictional Force

The frictional force is given by \( f_{friction} = \mu_k N = \mu_k mg \cos(\theta) \).
05

Set up the Equation of Motion

The net force along the incline which provides the acceleration is the difference between the frictional force and the downhill component of the weight: \( f_{net} = \mu_k mg \cos(\theta) - mg \sin(\theta) \). This net force equals \( ma \) by Newton's second law.
06

Solve for Acceleration

The acceleration \( a \) is given by the equation \( a = \mu_k g \cos(\theta) - g \sin(\theta) \). Substitute \( g = 9.81 \text{ m/s}^2 \) and calculate for each \( \mu_k \).
07

Calculate for Light Snow Condition

For light snow, \( \mu_k = 0.12 \). Therefore, \( a = 0.12 \times 9.81 \times \cos(\theta) - 9.81 \times \sin(\theta) \). Using \( \tan(\theta) = 0.1 \), find \( \cos(\theta) = \frac{1}{\sqrt{1+0.1^2}} \approx 0.995 \) and \( \sin(\theta) = \frac{0.1}{\sqrt{1+0.1^2}} \approx 0.1 \). Thus, \( a \approx (0.12 \times 9.81 \times 0.995) - (9.81 \times 0.1) \approx 0.075 \text{ m/s}^2 \).
08

Calculate for Ice Condition

For ice, \( \mu_k = 0.05 \). Therefore, \( a = 0.05 \times 9.81 \times \cos(\theta) - 9.81 \times \sin(\theta) \). Using the same \( \cos(\theta) \) and \( \sin(\theta) \) values, \( a \approx (0.05 \times 9.81 \times 0.995) - (9.81 \times 0.1) \approx -0.561 \text{ m/s}^2 \).

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Key Concepts

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

Kinetic Friction
Kinetic friction is a force that acts between moving surfaces. When a vehicle attempts to move up an inclined plane, like a hill, the interaction between the tires and the surface plays a significant role. This interaction can be quantified by the coefficient of kinetic friction, denoted as \( \mu_k \). It represents how much frictional force opposes the motion between the surfaces. In our scenario, two coefficients are given: for light snow, \( \mu_k = 0.12 \), and for ice, \( \mu_k = 0.05 \). The lower the coefficient, the slipperier the surface.

Kinetic friction can be harnessed to help move the vehicle forward by providing the necessary grip for acceleration. However, insufficient friction, like on ice, can lead to wheel spin due to inadequate traction, affecting the vehicle's performance.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, other than horizontal. Here, it's a 10% incline, meaning it rises 1 unit for every 10 horizontal units, defined by \( \tan(\theta) = 0.1 \). This incline causes gravitational force to split into two components: one parallel to the incline and one perpendicular. Knowing these components is crucial as they impact the vehicle's motion.

The parallel component (\( mg \sin(\theta) \)) propels the vehicle downward, resisting upward motion, while the perpendicular component (\( mg \cos(\theta) \)), contributes to the normal force pressing the vehicle against the incline.
  • Incline Angle: Determines the difficulty of climbing the plane.
  • Components of Weight: Calculated using sine and cosine of the angle \( \theta \).
Newton's Second Law
Isaac Newton's second law of motion is a fundamental concept in dynamics. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The equation is given by \( F_{net} = ma \), where \( F_{net} \) is the net force, \( m \) is the mass, and \( a \) is the acceleration.

In our inclined plane problem, Newton's second law helps us understand how the forces of friction and gravity interact. The friction force seeks to move the vehicle uphill, opposing the gravitational pull. By setting the net force expression to equal \( ma \), we can solve for acceleration \( a \). This approach shows us how different surfaces (snow and ice) affect vehicle dynamics.
Acceleration Calculation
Acceleration is the rate of change of velocity of an object. In our case, it's crucial to determine how quickly the vehicle can accelerate up the incline under different surface conditions.

We compute acceleration using the relationship: \( a = \mu_k g \cos(\theta) - g \sin(\theta) \). Here, \( \mu_k \) varies based on surface type, \( g \) is the acceleration due to gravity (\( 9.81 \text{ m/s}^2 \)), \( \cos(\theta) \) and \( \sin(\theta) \) are the trigonometric functions derived from the incline angle.
  • Light Snow: Substituting \( \mu_k = 0.12 \) gives us a slower acceleration of \( 0.075 \text{ m/s}^2 \).
  • Ice: With \( \mu_k = 0.05 \), the acceleration is actually negative, \( -0.561 \text{ m/s}^2 \), indicating a deceleration because the friction can't overcome gravity.

These calculations underscore the variance in performance due to changing friction levels on different surfaces.

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Most popular questions from this chapter

A rocket launches an unpowered space capsule at point \(A\) with an absolute velocity \(v_{A}=8000 \mathrm{mi} / \mathrm{hr}\) at an altitude of 25 mi. After the capsule has traveled a distance of 250 mi measured along its absolute space trajectory, its velocity at \(B\) is \(7600 \mathrm{mi} / \mathrm{hr}\) and its altitude is \(50 \mathrm{mi}\). Determine the average resistance \(P\) to motion in the rarified atmosphere. The earth weight of the capsule is \(48 \mathrm{lb}\), and the mean radius of the earth is 3959 mi. Consider the center of the earth fixed in space.

A spacecraft \(m\) is heading toward the center of the moon with a velocity of \(2000 \mathrm{mi} / \mathrm{hr}\) at a distance from the moon's surface equal to the radius \(R\) of the moon. Compute the impact velocity \(v\) with the surface of the moon if the spacecraft is unable to fire its retro-rockets. Consider the moon fixed in space. The radius \(R\) of the moon is \(1080 \mathrm{mi}\), and the acceleration due to gravity at its surface is \(5.32 \mathrm{ft} / \mathrm{sec}^{2}\).

The velocity of a 1.2 -kg particle is given by \(\mathbf{v}=\) \(1.5 t^{3} \mathbf{i}+\left(2.4-3 t^{2}\right) \mathbf{j}+5 \mathbf{k},\) where \(\mathbf{v}\) is in meters per second and the time \(t\) is in seconds. Determine the linear momentum G of the particle, its magnitude \(G,\) and the net force \(\mathbf{R}\) which acts on the particle when \(t=2\) s.

A jet-propelled airplane with a mass of \(10 \mathrm{Mg}\) is flying horizontally at a constant speed of \(1000 \mathrm{km} / \mathrm{h}\) under the action of the engine thrust \(T\) and the equal and opposite air resistance \(R\). The pilot ignites two rocket-assist units, each of which develops a forward thrust \(T_{0}\) of \(8 \mathrm{kN}\) for \(9 \mathrm{s}\). If the velocity of the airplane in its horizontal flight is \(1050 \mathrm{km} / \mathrm{h}\) at the end of the \(9 \mathrm{s}\), calculate the timeaverage increase \(\Delta R\) in air resistance. The mass of the rocket fuel used is negligible compared with that of the airplane.

A spacecraft moving in a west-to-east equatorial orbit is observed by a tracking station located on the equator. If the spacecraft has a perigee altitude \(H=150 \mathrm{km}\) and velocity \(v\) when directly over the station and an apogee altitude of \(1500 \mathrm{km},\) determine an expression for the angular rate \(p\) (relative to the earth) at which the antenna dish must be rotated when the spacecraft is directly overhead. Compute \(p .\) The angular velocity of the earth is \(\omega=0.7292\left(10^{-4}\right) \mathrm{rad} / \mathrm{s}\).
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