/*! 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 53 A \(1400 \mathrm{~kg}\) jet engi... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(1400 \mathrm{~kg}\) jet engine is fastened to the fuselage of a passenger jet by just three bolts (this is the usual practice). Assume that each bolt supports one-third of the load. (a) Calculate the force on each bolt as the plane waits in line for clearance to take off. (b) During flight, the plane encounters turbulence, which suddenly imparts an upward vertical acceleration of \(2.6 \mathrm{~m} / \mathrm{s}^{2}\) to the plane. Calculate the force on each bolt now.

Short Answer

Expert verified
The force on each bolt when the plane is stationary is 4573.33 N. The force on each bolt during turbulence is 5786.67 N.

Step by step solution

01

- Understand the problem

The jet engine is supported by three bolts, and each bolt equally shares the load of the engine. First, calculate the force on each bolt when the plane is stationary. Then, calculate the force when there is an additional upward acceleration due to turbulence.
02

- Calculate the weight of the jet engine

The weight of the jet engine can be found using the formula for gravitational force: $$F_{gravity} = m \times g$$ where: $$m = 1400 \text{ kg}$$ $$g = 9.8 \text{ m/s}^2$$. Substituting the values: $$F_{gravity} = 1400 \times 9.8 = 13720 \text{ N}$$.
03

- Find force on each bolt when the plane is stationary

Since the load is equally shared by 3 bolts, the force on each bolt is: $$F_{bolt} = \frac{F_{gravity}}{3}$$ Substituting the value for $$ F_{gravity} $$: $$F_{bolt} = \frac{13720}{3} = 4573.33 \text{ N}$$ Thus, the force on each bolt when the plane is stationary is 4573.33 N.
04

- Calculate the net force on each bolt due to additional acceleration

During turbulence, there is an additional upward vertical acceleration of 2.6 m/s^2. Calculate the additional force due to this acceleration using: $$F_{additional} = m \times a$$ where: $$a = 2.6 \text{ m/s}^2$$ Substituting the values: $$F_{additional} = 1400 \times 2.6 = 3640 \text{ N}$$
05

- Calculate total force on each bolt during turbulence

The total force on each bolt combines the gravitational force and the additional force due to turbulence: $$F_{total} = F_{gravity} + F_{additional}$$ Substituting the values found earlier: $$F_{total} = 13720 + 3640 = 17360 \text{ N}$$ Since this force is equally shared by 3 bolts: $$ F_{bolt\text{(turbulence)}} = \frac{F_{total}}{3}$$ Substituting the value for $$ F_{total} $$: $$F_{bolt\text{(turbulence)}} = \frac{17360}{3} = 5786.67 \text{ N}$$.Thus, the force on each bolt during turbulence is 5786.67 N.

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Key Concepts

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

Force Calculation
Understanding the forces that act on objects is an essential part of physics. The fundamental concept here involves calculating the force using Newton's second law, which states that force is the product of mass and acceleration.

Force is generally represented by the equation: \( F = m \times a \).

In our example, we first calculated the gravitational force on a 1400 kg jet engine using the formula: \( F_{\text{gravity}} = m \times g \), where:
  • \( m = 1400 \text{ kg} \)
  • \( g = 9.8 \text{ m/s}^2 \)
Substituting these values, we find \( F_{\text{gravity}} = 1400 \times 9.8 = 13720 \text{ N} \).

This calculated force is then distributed equally among three bolts holding the engine.
Gravitational Force
Gravitational force is a key concept in physics—it’s the force by which a planet or other body draws objects toward its center. The force is given by the equation: \( F_{\text{gravity}} = m \times g \).

In our situation, the mass (\( m \)) of the jet engine is 1400 kg, and the gravitational acceleration (\( g \)) is always 9.8 m/s² near Earth's surface.

So, the weight of the engine (gravitational force) is \(13720 \text{ N} \).

Understanding gravitational force is essential for calculating the load distributions and the effects of additional forces like turbulence on the bolts supporting the engine.
Newton's Second Law
Newton's second law explains how the velocity of an object changes when it is subjected to an external force. Mathematically, it is represented as: \( F = m \times a \), where:
  • \( F \) is the force applied
  • \( m \) is the mass of the object
  • \( a \) is the acceleration
In our example, we need to find the additional force introduced by turbulence.

With an additional acceleration (\( a \)) of 2.6 m/s² and mass (\( m \)) of 1400 kg, the force calculation is: \( F_{\text{additional}} = 1400 \times 2.6 = 3640 \text{ N} \).

This additional force, when combined with the gravitational force, helps us to determine the new force each bolt must support.
Turbulence Effects
Turbulence can introduce sudden and unpredictable forces to an aircraft. When the jet encounters turbulence, it experiences additional vertical acceleration.

In our exercise, this acceleration (\( a \)) is 2.6 m/s². To understand the impact of turbulence, we calculate the additional force using Newton’s second law: \( F_{\text{additional}} = m \times a \).

Here, \( m = 1400 \text{ kg} \) and \( a = 2.6 \text{ m/s}^2 \), so \( F_{\text{additional}} = 3640 \text{ N} \).

This additional force adds to the gravitational force, increasing the total force experienced by each bolt.
Load Distribution
Load distribution is the process by which the total applied load is shared or divided among the supporting elements—in this case, the bolts.

Initially, when the plane is stationary, the total gravitational force (\( F_{\text{gravity}} \)) of 13720 N is divided equally among three bolts, which means each bolt supports \( \frac{13720}{3} = 4573.33 \text{ N} \).

During turbulence, the force changes due to additional acceleration. We calculated the new total force as \( 17360 \text{ N} \), and each bolt then supports \( \frac{17360}{3} = 5786.67 \text{ N} \).

Understanding load distribution allows for safer designs and better predictions on material stress and potential failures.

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Most popular questions from this chapter

An initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed \(1100 \mathrm{~N}\). The coefficient of static friction between the box and the floor is \(0.35 .\) (a) What should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand, and (b) what is the weight of the sand and box in that situation?

calculate the ratio of the drag force on a passenger jet flying with a speed of \(1000 \mathrm{~km} / \mathrm{h}\) at an altitude of \(10 \mathrm{~km}\) to the drag force on a prop-driven transport flying at half the speed and half the altitude of the jet. At \(10 \mathrm{~km}\) the density of air is \(0.38 \mathrm{~kg} / \mathrm{m}^{3}\), and at \(5.0 \mathrm{~km}\) it is \(0.67 \mathrm{~kg} / \mathrm{m}^{3}\). Assume that the airplanes have the same effective cross-sectional area and the same drag coefficient \(C\).

A \(40 \mathrm{~kg}\) skier comes directly down a frictionless ski slope that is inclined at an angle of \(10^{\circ}\) with the horizontal while a strong wind blows parallel to the slope. Determine the magnitude and direction of the force of the wind on the skier if (a) the magnitude of the skier's velocity is constant, (b) the magnitude of the skier's velocity is increasing at a rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2} .\) and \((\mathrm{c})\) the magnitude of the skier's velocity is increasing at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\).

A bicyclist travels in a circle of radius \(25.0 \mathrm{~m}\) at a constant speed of \(9.00 \mathrm{~m} / \mathrm{s}\). The bicycle-rider mass is \(85.0 \mathrm{~kg} .\) Calculate the magnitudes of (a) the force of friction on the bicycle from the road and (b) the total force on the bicycle from the road.

A worker pushes horizontally on a \(35 \mathrm{~kg}\) crate with a force of magnitude \(110 \mathrm{~N}\). The coefficient of static friction between the crate and the floor is \(0.37\). (a) What is the frictional force on the crate from the floor? (b) What is the maximum magnitude \(f_{\max }^{\text {stat }}\) of the static frictional force under the circumstances? (c) Does the crate move? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull that will allow the first worker's \(110 \mathrm{~N}\) push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?

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