/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 151 An open container of oil rests o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An open container of oil rests on the flatbed of a truck that is traveling along a horizontal road at \(55 \mathrm{mi} / \mathrm{hr}\). As the truck slows uniformly to a complete stop in \(5 \mathrm{s}\), what will be the slope of the oil surface during the period of constant deceleration?

Short Answer

Expert verified
The slope of the oil surface will be approximately 26.57 degrees.

Step by step solution

01

Identify Given Variables

The truck is initially moving at a speed of \(v_i = 55 \, \text{mi/hr}\), which we convert to meters per second (\(m/s\)) as follows: \(55 \, \text{mi/hr} \approx 24.5872 \, \text{m/s}\). The time \(t\) during which the truck slows down is \(5 \, \text{s}\). The final velocity \(v_f = 0 \, \text{m/s}\).
02

Calculate Deceleration

To find the slope of the oil surface, we need to calculate the deceleration. Use the equation \(a = \frac{v_f - v_i}{t} = \frac{0 - 24.5872}{5} = -4.91744 \, \text{m/s}^2\). The negative sign indicates deceleration.
03

Understand Water Surface Slope

The force causing the oil's surface to tilt is due to the lateral deceleration of the truck. The oil surface will have a slope, formed because the horizontal force (deceleration) and the gravitational force (vertical) work as vectors at right angles. The slope of the oil surface \(\theta\) is related to the tangent of this angle by \(\tan(\theta) = \frac{|a|}{g} \), where \(g = 9.81 \, \text{m/s}^2\).
04

Calculate the Slope of the Oil Surface

Substitute the values of \(|a|\) and \(g\) into the equation \(\tan(\theta) = \frac{4.91744}{9.81} = 0.50177\). The angle \(\theta\) can be found using \(\theta = \tan^{-1}(0.50177)\), which we calculate using a calculator: \(\theta \approx 26.57°\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Deceleration and Acceleration
In fluid mechanics, deceleration and acceleration refer to the change in the velocity of an object. Imagine a truck moving along a road. When it slows down, this is called deceleration. In our exercise, the truck was moving at speed and came to a halt in 5 seconds. Deceleration can be calculated using the formula \[ a = \frac{v_f - v_i}{t} \]where:
  • \(a\) is the acceleration (or deceleration if it's negative)
  • \(v_f\) is the final velocity
  • \(v_i\) is the initial velocity
  • \(t\) is the time over which the change occurs
In the truck example, it started at 55 mi/hr (about 24.58 m/s) and stopped in 5 seconds. By substituting these values, we found the deceleration to be approximately \(-4.92\, \text{m/s}^2\). The negative sign simply shows that the truck is slowing down.
Surface Slope
When a vehicle decelerates, the liquid inside it can behave in an interesting way. It can form a slope. The surface slope of a liquid, like oil in our example, will tilt because of the forces acting on it. This happens because deceleration creates a horizontal force that interacts with gravity, the vertical force.To understand the slope of the oil surface during the truck’s deceleration, we use the formula:\[ \tan(\theta) = \frac{|a|}{g} \]where:
  • \(\theta\) is the slope angle of the oil surface
  • \(|a|\) is the magnitude of the deceleration
  • \(g\) is the acceleration due to gravity (\(9.81 \text{m/s}^2\))
In this exercise, the calculated slope angle, \(\theta\), was approximately \(26.57^\circ\). This means the surface of the oil tilts at this angle relative to its original horizontal position during deceleration.
Tangential and Normal Forces
When considering objects in motion, especially in the context of fluid mechanics, it's crucial to understand how tangential and normal forces come into play. The tangential force acts along the path of motion, while the normal force acts perpendicular to it. In the example of the truck, because it is slowing down, the tangential force aligns with the direction of deceleration. This force arises due to the truck stopping, and it impacts how the liquid inside, like oil, behaves. On the other hand, the normal force is the gravitational pull acting perpendicular to the surface. It's what keeps the oil from floating away and maintains its weight. When these forces combine, they create the sloping effect of the oil surface observed in the truck. Understanding these forces is essential for comprehending how objects move and interact, especially in real-world environments like the decelerating truck scenario.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You partially fill a glass with water, place an index card on top of the glass, and then turn the glass upside down while holding the card in place. You can then remove your hand from the card and the card remains in place, holding the water in the glass. Explain how this works.

A 5 -gal, cylindrical open container with a bottom area of 120 in. \(^{2}\) is filled with glycerin and rests on the floor of an ele- vator. (a) Determine the fluid pressure at the bottom of the container when the elevator has an upward acceleration of \(3 \mathrm{ft} / \mathrm{s}^{2}\) (b) What resultant force does the container exert on the floor of the elevator during this acceleration? The weight of the container is negligible. (Note: 1 gal \(=231\) in. \(^{3}\) ).

An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown liquid is \(1.5 \mathrm{m}\) and the depth of the oil (specific weight \(=8.5 \mathrm{kN} / \mathrm{m}^{3}\) ) floating on top is \(5.0 \mathrm{m}\). A pressure gage connected to the bottom of the tank reads 65 kPa. What is the specific gravity of the unknown liquid?

On the suction side of a pump, a Bourdon pressure gage reads \(40 \mathrm{kPa}\) vacuum. What is the corresponding absolute pressure if the local atmospheric pressure is \(100 \mathrm{kPa}(\mathrm{abs}) ?\)

A 2 -ft-diameter hemispherical plexiglass "bubble" is to be used as a special window on the side of an above-ground swimming pool. The window is to be bolted onto the vertical wall of the pool and faces outward, covering a 2 -ft- diameter opening in the wall. The center of the opening is \(4 \mathrm{ft}\) below the surface. Determine the horizontal and vertical components of the force of the water on the hemisphere.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.