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Chapter 3: Problem 8

In a mall, a shopper rides up an escalator between floors. At the top of the escalator, the shopper turns right and walks \(9.00 \mathrm{m}\) to a store. The magnitude of the shopper's displacement from the bottom of the escalator to the store is \(16.0 \mathrm{m} .\) The vertical distance between the floors is \(6.00 \mathrm{m} .\) At what angle is the escalator inclined above the horizontal?

Short Answer

Expert verified
The escalator is inclined at approximately 22.11掳 above the horizontal.

Step by step solution

01

Understand the Problem

We are given the vertical distance ( 6.00 m), the horizontal distance from the top of the escalator to the store (9.00 m), and the total displacement from the bottom of the escalator to the store (16.0 m). We need to find the angle of inclination of the escalator above the horizontal.
02

Visualize the Scenario

Visualize a right triangle where the vertical leg is the vertical distance (6.00 m), the horizontal leg combines the horizontal distance to the store (9.00 m) and the distance covered horizontally by the escalator, forming the base, and the hypotenuse is the material displacement (16.0 m).
03

Use the Pythagorean Theorem

Let the length of the escalator be represented by the hypotenuse of the triangle: \( \sqrt{x^2 + 6^2} = 16.0 \) where \( x \) is the horizontal distance covered by the escalator. Solve this equation to find \( x \).
04

Solve for Horizontal Distance of the Escalator

Rearrange the Pythagorean equation: \( x^2 + 6^2 = 16.0^2 \) Simplify it: \( x^2 + 36 = 256 \) \( x^2 = 220 \) \( x = \sqrt{220} \approx 14.83 \text{ m} \).
05

Calculate the Angle of Inclination

Use the tangent function: \( \tan(\theta) = \frac{\text{vertical distance}}{\text{horizontal distance of escalator}} = \frac{6.00}{14.83} \) \( \theta = \tan^{-1}\left(\frac{6.00}{14.83}\right) \).
06

Determine the Angle

Calculate the angle using a calculator: \( \theta \approx \tan^{-1}(0.405) \approx 22.11^\circ \).

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Key Concepts

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

Angle of Inclination
The angle of inclination, often denoted by \( \theta \), is an important concept when analyzing the motion on an inclined surface, like an escalator. In this problem, the angle of inclination refers to how steep the escalator is relative to the horizontal floor. To find this angle, we use trigonometry, particularly the tangent function.
  • Trigonometric Definitions: The tangent of an angle in a right triangle is the ratio of the opposite side (vertical rise) to the adjacent side (horizontal run).
  • Practical Application: Knowing the angle can help in calculating forces acting on an object on the incline, such as gravity and friction.
By rearranging the trigonometric equation \( \tan(\theta) = \frac{\text{vertical distance}}{\text{horizontal distance}} \), we can solve for \( \theta \). This provides a practical understanding of how steep the escalator is in real-world terms.
Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle in geometry. It is used to relate the sides of a right triangle. The theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.In this exercise, we use the Pythagorean Theorem to find one of the sides of the triangle formed by the escalator's path.
  • Equation Form: \( c^2 = a^2 + b^2 \), where \( c \) is the hypotenuse, and \( a \) and \( b \) are the triangle's other sides.
  • Solution Steps: The given displacement \(16.0 \text{ m}\) represents the hypotenuse, and you need to find the horizontal distance covered by the escalator using \( x^2 + 6^2 = 16.0^2 \).
The Pythagorean Theorem allows us to solve for unknown distances, an essential step when dealing with inclined planes or other common physics problems.
Escalator Physics
Escalator physics involves studying the mechanics of moving at a constant speed on an inclined plane, as the everyday action of using an escalator can be analyzed using basic physics principles. In escalator-related problems:
  • Inclined Plane Concept: The movement along an escalator can be broken down into horizontal and vertical components. This helps analyze the motion using vectors.
  • Displacement Consideration: Displacement combines direction and distance moved. The escalator gives a seamless transition from one floor to another, showcasing effective displacement utilization.
Understanding the components of escalator motion helps students comprehend broader concepts like balancing forces and efficient travel paths. This knowledge can then be applied to similar scenarios, illustrating the relevance of physics in everyday scenarios.

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Most popular questions from this chapter

In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.050 s, during which time it experiences an acceleration of \(340 \mathrm{m} / \mathrm{s}^{2} .\) The ball is launched at an angle of \(51^{\circ}\) above the ground. Determine the horizontal and vertical components of the launch velocity.

You are in a hot-air balloon that, relative to the ground, has a velocity of \(6.0 \mathrm{m} / \mathrm{s}\) in a direction due east. You see a hawk moving directly away from the balloon in a direction due north. The speed of the hawk relative to you is \(2.0 \mathrm{m} / \mathrm{s} .\) What are the magnitude and direction of the hawk's velocity relative to the ground? Express the directional angle relative to due east.

Relative to the ground, a car has a velocity of \(16.0 \mathrm{m} / \mathrm{s}\), directed due north. Relative to this car, a truck has a velocity of \(24.0 \mathrm{m} / \mathrm{s}\), directed \(52.0^{\circ}\) north of east. What is the magnitude of the truck's velocity relative to the ground?

On a spacecraft, two engines are turned on for 684 s at a moment when the velocity of the craft has \(x\) and \(y\) components of \(v_{0 x}=4370 \mathrm{m} / \mathrm{s}\) and \(v_{0 y}=6280 \mathrm{m} / \mathrm{s} .\) While the engines are firing, the craft undergoes a displacement that has components of \(x=4.11 \times 10^{6} \mathrm{m}\) and \(y=6.07 \times 10^{6} \mathrm{m} .\) Find the \(x\) and \(y\) components of the craft's acceleration.

The earth moves around the sun in a nearly circular orbit of radius \(1.50 \times 10^{11} \mathrm{m} .\) During the three summer months (an elapsed time of \(\left.7.89 \times 10^{6} \mathrm{s}\right),\) the earth moves one-fourth of the distance around the sun. (a) What is the average speed of the earth? (b) What is the magnitude of the average velocity of the earth during this period?
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