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Chapter 5: Problem 86

\(\bullet$$\bullet\) At night while it is dark, a driver inadvertently parks his car on a drawbridge. Some time later, the bridge must be raised to allow a boat to pass through. The coefficients of fric- tion between the bridge and the car's tires are \(\mu_{\mathrm{s}}=0.750\) and \(\mu_{\mathrm{k}}=0.550 .\) Start each part of your solution to this problem with a free-body diagram of the car. (a) At what angle will the car just start to slide? (b) If the bridge attendant sees the car suddenly start to slide and immediately turns off the bridge's motor, what will be the car's acceleration after it has begun to move?

Short Answer

Expert verified
(a) The car starts to slide at an angle \( \theta = \tan^{-1}(0.750) \). (b) After sliding begins, the car's acceleration is \( a = g(\sin \theta - 0.550 \cos \theta) \).

Step by step solution

01

Draw the Free-Body Diagram

Start by drawing the free-body diagram of the car on the inclined bridge. Include the gravitational force acting downward, the normal force perpendicular to the bridge's surface, and the static friction force preventing the car from sliding.
02

Analyze Forces When Car Just Starts to Slide

When the car starts to slide, the static friction force is at its maximum value. The forces along the plane include gravitational force component \( mg \sin(\theta) \) and static friction force \( f_s = \mu_s N \). Balance the forces to find \( \theta \).
03

Solve for the Critical Angle \( \theta \)

Using the equation \( f_s = mg \sin \theta = \mu_s N \) and \( N = mg \cos \theta \), equate \( mg \sin \theta = \mu_s mg \cos \theta \). Solve for \( \theta \) to get \( \tan(\theta) = \mu_s \). Therefore, \( \theta = \tan^{-1}(\mu_s) \).
04

Calculate the Angle

Substitute \( \mu_s = 0.750 \) into the equation \( \theta = \tan^{-1}(0.750) \) and calculate \( \theta \) using a calculator.
05

Analyze Forces as Car Validly Slides

After the car starts sliding, static friction converts to kinetic friction, \( f_k = \mu_k N \). The car's acceleration \( a \) down the incline can be found using \( f_k = ma = mg \sin \theta - \mu_k N \).
06

Solve for Acceleration \( a \)

Use the net force equation along the inclined plane: \( ma = mg \sin \theta - \mu_k mg \cos \theta \). Simplify to \( a = g(\sin \theta - \mu_k \cos \theta) \). Substitute known values to find \( a \).
07

Calculate Acceleration

Substitute \( \theta \) and \( \mu_k = 0.550 \) into the equation \( a = 9.81(\sin \theta - 0.550 \cos \theta) \) and solve for \( a \).

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Key Concepts

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

Free-Body Diagram
When solving physics problems, especially ones involving forces, a Free-Body Diagram (FBD) is incredibly useful. It provides a way to visually represent all the forces acting on an object, helping you see the problem more clearly.
In our exercise, imagine a car parked on an inclined drawbridge. The FBD would include:
  • The gravitational force acting downward, often represented as an arrow pointing straight down labeled \( mg \).
  • The normal force perpendicular to the surface of the bridge, shown as an arrow pointing out from the surface.
  • The static friction force preventing the car from sliding down, pointing up the inclined plane.
Each force must be accounted for to solve the problem effectively. FBDs help organize information and are the basis for setting up equations to solve for variables like angles or acceleration.
Static Friction
Static friction is the force that keeps an object at rest when a force is applied. It acts in the opposite direction to the applied force, preventing movement.
In our example, static friction holds the car in place on the inclined bridge. It can be calculated with: \[ f_s = \mu_s N \]where \( \mu_s \) is the coefficient of static friction (0.750 in this case), and \( N \) is the normal force.
Static friction will resist motion until its maximum value is exceeded, at which point the car will start to slide. Understanding static friction is essential as it determines the critical angle at which sliding begins.
Kinetic Friction
Once the car starts moving, static friction transitions to kinetic friction. This type of friction acts between surfaces in relative motion, usually less than static friction.
The force of kinetic friction can be calculated using:\[ f_k = \mu_k N \]Here, \( \mu_k \) is the coefficient of kinetic friction, which is 0.550 for this exercise. Kinetic friction is less effective at preventing motion than static friction, resulting in the car accelerating down the incline.
It's crucial to grasp the concept of kinetic friction, as it plays a significant role in determining the car's acceleration after sliding begins.
Critical Angle
The critical angle is the point at which an object starts to move from its static position. It鈥檚 vital for understanding when an object transitions from static to kinetic friction.
To find this angle, use the equilibrium equation for forces parallel to the incline:\[ \tan(\theta) = \mu_s \]Substituting the static friction coefficient, you solve for \( \theta \):\[ \theta = \tan^{-1}(0.750) \]Calculating this gives you the angle at which the car starts to slide. Knowing the critical angle helps predict at what point motion begins.
Acceleration Calculation
After the car begins to slide, it accelerates due to gravity and kinetic friction. Acceleration can be calculated by analyzing the net forces acting on the car.
The formula to find acceleration down the incline is:\[ a = g(\sin \theta - \mu_k \cos \theta) \]where \( g \) is the acceleration due to gravity (9.81 m/s虏). Substitute the values for \( \theta \) and \( \mu_k \) into the formula to compute \( a \).
This calculation is crucial for understanding the dynamics once the car starts moving, offering insights into the behavior of forces in motion.

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Most popular questions from this chapter

\(\bullet$$\bullet\) A large fish hangs from a spring balance supported from the roof of an elevator. (a) If the elevator has an upward accel- eration of 2.45 \(\mathrm{m} / \mathrm{s}^{2}\) and the balance reads \(60.0 \mathrm{N},\) what is the true weight of the fish? (b) Under what circumstances will the balance read 35.0 \(\mathrm{N} ?\) (c) What will the balance read if the ele- vator cable breaks?

\(\bullet$$\bullet\) Force during a jump. An average person can reach a maximum height of about 60 \(\mathrm{cm}\) when jumping straight up from a crouched position. During the jump itself, the person's body from the knees up typically rises a distance of around 50 \(\mathrm{cm} .\) To keep the calculations simple and yet get a reason- able result, assume that the entire body rises this much during the jump. (a) With what initial speed does the person leave the ground to reach a height of 60 \(\mathrm{cm} ?\) (b) Make a free-body diagram of the person during the jump. (c) In terms of this jumper's weight \(W,\) what force does the ground exert on him or her during the jump?

\(\bullet\) A box of bananas weighing 40.0 \(\mathrm{N}\) rests on a horizontal sur- face. The coefficient of static friction between the box and the surface is \(0.40,\) and the coefficient of kinetic friction is \(0.20 .\) (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box? (b) What is the magnitude of the friction force if a monkey applies a hor- izontal force of 6.0 \(\mathrm{N}\) to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started?

\(\bullet$$\bullet \mathrm{A} 75,600 \mathrm{N}\) spaceship comes in for a vertical landing. From an initial speed of \(1.00 \mathrm{km} / \mathrm{s},\) it comes to rest in 2.00 \(\mathrm{min}\) . with uniform acceleration. (a) Make a free-body diagram of this ship as it is coming in. (b) What braking force must its rockets provide? Ignore air resistance.

\(\bullet\) An adventurous archaeologist crosses between two rock cliffs by slowly going hand over hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope (Figure 5.38 ). The rope will break if the tension in it exceeds \(2.50 \times 10^{4} \mathrm{N} .\) Our hero's mass is 90.0 \(\mathrm{kg}\) . (a) If the angle \(\theta\) is \(10.0^{\circ}\) , find the tension in the rope. Start with a free-body diagram of the archaeologist. (b) What is the small- est value the angle \(\theta\) can have if the rope is not to break?
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