3.10: Trends in Global Competition
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9. What are the trends in the global marketplace?
In this section, we will examine several underlying trends that will continue to propel the dramatic growth in world trade. These trends are market expansion, resource acquisition, and the emergence of China and India.
Market Expansion
The need for businesses to expand their markets is perhaps the most fundamental reason for the growth in world trade. The limited size of domestic markets often motivates managers to seek markets beyond their national frontiers. The economies of large-scale manufacturing demand big markets. Domestic markets, particularly in smaller countries like Denmark and the Netherlands, simply can’t generate enough demand. Nestlé was one of the first businesses to “go global” because its home country, Switzerland, is so small. Nestlé was shipping milk to 16 different countries as early as 1875. Today, hundreds of thousands of businesses are recognizing the potential rich rewards to be found in international markets.
Resource Acquisition
More and more companies are going to the global marketplace to acquire the resources they need to operate efficiently. These resources may be cheap or skilled labor, scarce raw materials, technology, or capital. Nike, for example, has manufacturing facilities in many Asian countries in order to use cheaper labor. Honda opened a design studio in southern California to put that “California flair” into the design of some of its vehicles. Large multinational banks such as Bank of New York and Citigroup have offices in Geneva, Switzerland. Geneva is the private banking center of Europe and attracts capital from around the globe.
The Emergence of China and India
China and India—two of the world’s economic powerhouses—are impacting businesses around the globe, in very different ways. The boom in China’s worldwide exports has left few sectors unscathed, be they garlic growers in California, jeans makers in Mexico, or plastic-mold manufacturers in South Korea. India’s impact has altered how hundreds of service companies from Texas to Ireland compete for billions of dollars in contracts.
The causes and consequences of each nation’s growth are somewhat different. China’s exports have boomed largely thanks to foreign investment: lured by low labor costs, big manufacturers have surged into China to expand their production base and push down prices globally. Now manufacturers of all sizes, making everything from windshield wipers to washing machines to clothing, are scrambling either to reduce costs at home or to outsource more of what they make in cheaper locales such as China and India. 39
Indians are playing invaluable roles in the global innovation chain. Hewlett-Packard, Cisco Systems, and other tech giants now rely on their Indian teams to devise software platforms and multimedia features for next-generation devices. Googleprincipal scientist Krishna Bharat set up the Google Bangalore lab complete with colorful furniture, exercise balls, and a Yamaha organ—like Google’s Mountain View, California, headquarters—to work on core search-engine technology. Indian engineering houses use 3-D computer simulations to tweak designs of everything from car engines and forklifts to aircraft wings for such clients as General MotorsCorp. and Boeing Co. Barring unforeseen circumstances, within five years India should vault over Germany as the world’s fourth-biggest economy. By mid-century, China should overtake the United States as number one. By then, China and India could account for half of global output. 40
ETHICS IN PRACTICE
The United Nations Sustainability Development Goals
Corporations like Albertson’s, Unilever, Kimberly Clark, and Siemens are starting to take action on the United Nations Sustainability Development Goals. For many years, through corporate social responsibility (CSR) programs, corporations have donated money and employee time to address various social and environmental problems, both globally and in their own backyards. The Carnegie Foundation and the Bill and Melinda Gates Foundation are examples of this commitment. While these efforts have achieved some progress in environmental protection, ethical business practices, building sustainable positive impacts, and economic development by organizations, they do require deeper and longer engagement. Because the benefits to corporations’ profitability are mostly peripheral, short-term impacts such as a drop in demand often mean that attention is drawn away from CSR programs to attending to immediate bottom-line issues.
In 2015, the United Nations member-nations adopted 17 resolutions aimed at ending poverty, ensuring sustainability, and ensuring prosperity for all. The aggressive goals were set to be met over the next 15 years.
- End poverty in all its forms everywhere.
- End hunger, achieve food security and improved nutrition, and promote sustainable agriculture.
- Ensure healthy lives and promote well-being for all at all ages.
- Ensure inclusive and equitable quality education and promote lifelong learning opportunities for all.
- Achieve gender equality and empower all women and girls.
- Ensure availability and sustainable management of water and sanitation for all.
- Ensure access to affordable, reliable, sustainable, modern energy for all.
- Promote sustained, inclusive, sustainable economic growth; full and productive employment; and decent work for all.
- Build resilient infrastructure, promote inclusive and sustainable industrialization, and foster innovation.
- Reduce inequality within and among countries.
- Make cities and human settlements inclusive, safe, resilient, and sustainable.
- Ensure sustainable consumption and production patterns.
- Take urgent action to combat climate change and its impacts.
- Conserve and sustainably use the oceans, seas, and marine resources for sustainable development.
- Protect, restore, and promote sustainable use of terrestrial ecosystems; sustainably manage forests; combat desertification and halt and reverse land degradation; and halt biodiversity loss.
- Promote peaceful and inclusive societies for sustainable development; provide access to justice for all; and build effective, accountable, inclusive institutions at all levels.
- Strengthen the means of implementation and revitalize the global partnership for sustainable development.
Companies like Albertson’s recognize that a robust CSR program can enhance a corporation’s reputation, which can indirectly boost the bottom line. They used number 14 on the United Nations Sustainability Development list in concert with World Oceans Day to announce that they as a company pledged to meet the U.N. goals. “We recognize that the wellbeing of people and the sustainability of our oceans are interdependent. As one of the largest U.S. retailers of seafood, we are committed to protecting the world’s oceans so they can remain a bountiful natural resource that contributes to global food security, the livelihoods of hard-working fishermen and the global economy,” said Buster Houston, Director of Seafood at Albertson’s Companies. The company is also committed to the concept of fair trade and was the first retailer to sell tuna with the fair trade seal.
Siemens, the German-based multinational, also supported the adoption of meeting the United Nations Sustainability Development goals, which they believe is based on their company values—responsible, excellent, innovative. They define sustainable development as the means to achieve profitable and long-term growth. In doing so, they align ourselves with the goals of the UN’s 2030 Agenda for Sustainable Development.
Critical Thinking Questions
- Why would companies pledge to meet the United Nations Sustainability Development goals when their competitors could ignore them in the name of greater, perhaps short-term, profits?
- Are you as a consumer more likely to purchase products from Albertson’s rather than another grocery chain that did not agree to the United Nations sustainability program? If you were working for a company deciding to purchase a large industrial component that was 10% more expensive than a competing product, would Siemens’s affirmation of meeting the United Nations Sustainability Development goals sway your decision? How would you explain the rationale for your decision?
Sources: Thane Kreiner, “Corporations and Social Entrepreneurship: A Shift?” https://www.scu.edu , accessed June 30, 2017; United Nations Sustainable Development website: http://www.un.org , accessed June 30, 2017; “Practicing Sustainability—in the Interest of Future Generations,” https://www.siemens.com , accessed June 30, 2017; “Albertsons Companies Commits to United Nations Sustainable Development Goals, Joins Influential Seafood Task Force,” Cision PR Newswire, http://www.prnewswire.com , June 6, 2017; Ingrid Embree, “How 17 Companies Are Tackling Sustainable Development Goals (and Your Company Can, Too),” Huffington Post, http://www.huffingtonpost.com , September 14, 2016.
An accelerating trend is that technical and managerial skills in both China and India are becoming more important than cheap assembly labor. China will stay dominant in mass manufacturing and is one of the few nations building multibillion-dollar electronics and heavy industrial plants. India is a rising power in software, design, services, and precision industry.
CONCEPT CHECK
- What trends will foster continued growth in world trade?
- Describe some of the ways businesses can take advantage of these trends to “go global.”


33 .
What is the objective and general form of a business card? Explain.
34 .
What are the similarities and differences between a brochure and a flyer?
35 .
What are some advantages of using mail merge?
36 .
Describe the process of inserting fields into the main document.
37 .
Describe the various content control fields used in creating fillable forms.
38 .
Explain the purpose of the combo box control.
39 .
Describe the process of submitting a template to Google. Why do you think Google allows only certain users to publish templates for others to use?
40 .
How do you use a Word template in Docs?
41 .
What are the advantages of Forms over the .dotx surveys you can produce in Word?
42 .
What is the objective of a public link?
43 .
What are the main differences between a Public link and an Anyone with the link shareable links?
44 .
How is Editing access different from Commenting access in a Doc?
45 .
Explain a potential concern you might have using a public link for collaboration on document.


4.1: Introduction to Activity-Based Costing
Companies must accurately determine the costs of their products and services to make sound management decisions, such as determining their selling prices to customers. A manufacturer’s product costs consist of direct materials, direct labor, and factory overhead. The materials and labor are direct costs that can be identified and traced to the product. Factory overhead, however, are indirect and must be allocated to the product cost on an estimated basis.
We have already looked at applying factory overhead to work in process using a predetermined factory overhead rate in the discussion of job order costing. We used a single factory overhead rate based on direct labor hours, total direct labor, or machine hours.
The following journal entry is an example of allocating $2,100 of factory overhead to work in process.
|
Account |
Debit |
Credit |
|
|
Work in Process |
2,100 |
▲ Work in Process is an asset (inventory) account that is increasing |
|
|
Factory Overhead |
2,100 |
▼ Factory Overhead is an expense account that is decreasing |
|
The issue we address now is how precise the estimate of $2,100 actually is. Three different ways of allocating factory overhead will be looked at, as follows:
- Single factory-wide rate – the same rate based on the same activity base for all departments
- Departmental rates – different rates for different departments, but a single rate within a department
- Activity-based costing – different rates for different processes and activities, regardless of department
An example of two different orders of custom wood furniture will be used to illustrate the three methods. Both the direct materials and direct labor costs are known for each job; only the factory overhead must be estimated. Assume jobs are manufactured in a factory with two departments, Cutting and Assembly.


1.5: Creating a Conflict
Conflict is the primary ingredient of drama and without it drama ceases to exist. Conflict is when two opposing forces collide. David and Goliath, Harry Potter and Voldemort, You and Your Future are all examples of opposing forces and these forces meet within the battlefield of story. Stories were invented to help us make sense of the world and find our place in it, and the world is full of conflict. Stories are how our ancestors taught the children in the tribe how to hunt, gather, be part of the community and find their way in life. The power of story still resonates with us today. We all have goals and needs and must navigate obstacles in order to achieve them. In fact you cannot exist without encountering conflict constantly throughout your day. In stories, the greater the conflict and the more impossible the obstacle, the more legendary the hero's journey is to its audience.
Conflict is the element that holds the audience’s attention throughout the play. The audience came to see someone deal with a problem. The conflict should build continuously throughout the play and create urgency with the characters involved. As an audience member you get to escape your reality and problems and witness another person's struggle in order to gain perspective on your current or future situations. It is for this reason that you must make sure that your conflict builds in intensity and reaches a satisfying conclusion.
As soon as the conflict is resolved your audience will begin disengaging from your story and reentering their own lives. This is often a problem for new writers who create a conflict and then quickly resolve it thus terminating the audience's attention. It is possible to layer obstacles so that the audience stays engaged but you need to have a primary conflict for your story. Next you can have a series of smaller conflicts or obstacles that build tension as the story progresses, however you MUST introduce the next obstacle prior to the conclusion of the current obstacle. For example, if your main character needs to get to Las Vegas. Once they get to Las Vegas the audience feels that they have now completed the journey and will be furious if you introduce the new obstacle of getting to Denver once the characters have landed. Instead it is much more effective to have the characters discover while in flight that the venue they are traveling to or person they are pursuing has transferred to Denver, and they must adjust their plans. You have to stay ahead of your audience, if your audience knows what your characters are going to do, before they do it, your audience will be bored and you will lose them.
Conflict can take many forms, but is primarily broken into two main categories and each has several subcategories. The categories are as follows:
Internal
(Human vs. Self)
Mental
Emotional/Spiritual
Physical
External
(Human vs. Something Outside of Themselves)
Human Vs. Human
Human Vs. Nature
Human Vs. Supernatural
Human Vs. Society
Human Vs. Science
Process
List all of the types of conflict on the board. Vote as a group on the primary conflict of the story and then if it is external choose and internal conflict that the Hero must also overcome in order to gain the skill necessary to overcome the primary external conflict. Make sure discuss ideas as a group as to where each idea can go dramatically.


26.10: Moving Beyond the Nation- A Revolution
In our MANY complex discussions of what it means to be an American, as well as what it means for literature to be “nationalized”, a resounding theme has continuously taken dominance in these discussions. It seems as though the theme of freedom has rung true, as it continues to, throughout most of the literature we have read thus far. While most of our reading’s settings have taken place within our great nation, this specific narrative, while featuring an American character, does not take place within America’s borders, but within international waters. However, despite its setting, it seems as though Herman Melville’s “Benito Cereno” continues on with the theme of freedom.
While at first glance, this story seems to feature the meeting between an American captain and an ill-Spaniard captain, who has been given charge of 150 or so African slaves to be transported from one port to another, there is a much deeper message waiting to be picked up by the reader.
Don Benito, the Spaniard captain of the San Dominick, seems to have taken ill, left with very little of his own crew due to sickness sweeping the ship. This seems to leave him with the lone option of employing the help of the cargo that was meant to be slaves as his only crew. At least… this is the story he shares with Captain Amasa Delano, the American who offers his own men as well as supplies to alleviate some of his fellow captain’s suffering. However, something seems to be extremely off about Don Benito’s story, especially when Captain Delano notices that, “Suddenly, one of the black boys, enraged at a word dropped by one of his white companions, seized a knife, and though called to forbear by one of the oakum-pickers, struck the lad over the head, inflicting a gash from which blood flowed.” (15-16). At once, he explains to Don Benito that this would not fly had this occurred on his own ship- which is normal given the colonial time period in which this story takes place. However, Don Benito doesn’t seem to care- nor does he seem to care about much of anything else for that matter. This may probably be due to the fact that he is lying through his teeth, and has been taken hostage along with the rest of his Spanish crew by the African slaves in an act of defiance to insure their freedom.
Does this story sound familiar? Good… because it should, as we Americans were not always free citizens of our happy nation either. We earned our independence through the revolutionary war, in which we fought tooth in nail against the British, our captors, to insure our own freedom. And it seems as if our happy-ending of a narrative has spread from our own great nation to other nations as well- such as those being captured for enslavement in Africa.
So while these “captured African slaves” seem to be cast within this narrative as villainous, I am in strong support of the message they are sending in their theoretical takeover of both the San Dominick as well as of their own lives. They seem to be following America’s lead in a revolution of their very own. And thus, nationalizing literature has been taken to the next level- of literature that moves beyond the nation.


6.17: GIS and Location Decisions
- Describe the types of data a retailer can obtain from geographic information system software (GIS)
So you looked at your census data for your area, and you are still feeling pretty confident that you have chosen a good location for a retail business. It seems like the population will support your endeavor with a suitable income and education level, among other components. So what is next? Well, GIS might be your next step!
A geographic information system (GIS) is a system of hardware and software that can store, retrieve, map and analyze geographic data. This system can help an analyst visual information about customer buying behavior and demographics in a map format! So this process can help you to identify the boundaries of your trade area and isolate target customer groups.
How does Wendy’s restaurant use GIS?
You can view the transcript for “Wendy’s – John Crouse, Director of Real Estate Services” (opens in new window).
Typically, there is a user friendly interface so you can access and analyze data pretty easily. The output from these systems are maps, so you can visualize the implications of the date quickly and easily. Now, remember, the U.S. Census data is free to access, but it is typically outdated. GIS data on the other hand has a fee, but it is current and updated frequently!
Take a look at where GIS started.
You can view the transcript for “Intro to GIS Analysis” (opens in new window).
ESRI is a company that provides GIS data to businesses. Review ESRI’s website to see all the cool stuff you can do with data!
Nielsen is one company offering a wide range of tools that are useful to assess consumer demand in an area. Watch the short video to see what can be found through GIS analysis. GIS is an extremely useful tool to create effective and targeted marketing campaigns. GIS allows an overlay technique. You can purchase a bunch of data sets for an area, including population count, income, and competitor information. Then you can layer these bits of data on top of each other to see the data in a different way. It allows businesses to make decisions based on many factors rather than just one.
Imagine being able to pinpoint a demographic based on many different pieces of data allowing you to not only determine if a location is good to open, but then proceeding forward with data that allows you to directly market to your customer base! Now, you can see how paying for GIS data may be far more effective that just working with potentially outdated information from other sources!
Contributors and Attributions
- GIS and Location Decisions. Authored by : Freedom Learning Group. Provided by : Lumen Learning. License : CC BY: Attribution
- Wendy's - John Crouse, Director of Real Estate Services. Authored by : Esri Industries. Located at : https://youtu.be/j0iBhEEVPGc . License : All Rights Reserved . License Terms : Standard YouTube License
- Intro to GIS Analysis. Authored by : Marcos Luna. Located at : https://youtu.be/43pHfDv6wtQ . License : All Rights Reserved . License Terms : Standard YouTube License


2.4: Atomic Orbitals
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Skills to Develop
- Illustrate the general shape of atomic orbitals
- Identify the relationship between quantum numbers
The "old quantum mechanics" introduced the idea of quantization, and it was good for describing the position of spectroscopic lines of single electron atoms. But it couldn't predict how strong the lines were. The "new quantum mechanics" of Schrodinger and Heisenberg took the wave-particle duality to its logical conclusion. While Bohr's model was still "classical" in that there was a defined orbit or trajectory (a path that could be calculated) for the electron, Schrodinger and Heisenberg changed that. Basically, where is a wave? If a particle behaves as a wave, you can't point to the exact spot where it is. Also, if light energy is quantized, it turns out that you can't measure the path taken by an electron without changing the path. In microscopy, which is using microscopes to look at small things, you can't separate things that are closer together than the wavelength used. This is why we use X-ray diffraction and electron microscopes with very short wavelengths to look at atoms and molecules. So if you want to measure the electron's path with light, and measure it precisely on an atomic scale, you have to use a short wavelength of light. But if you use a short wavelength then it has a lot of energy (E = hν), enough to change the electron's direction. This is the basis of the Uncertainty Principle.
The result of this (and also of Schrodinger's 3-D, space-filling wavefunction) is that we no longer describe electrons using orbits, or defined paths. Instead we talk about orbitals , which are defined by wavefunctions Ψ(x,y,z) like the one calculated in the previous section. It turns out (although I think Schrodinger didn't exactly intend this at first) that Ψ(x,y,z) 2 is the probability of finding the "particle" at position (x,y,z).
Schrodinger found the standing waves or Ψ(x,y,z) for the electron in a hydrogen atom. These are much more complicated that the wavefunction we found because they are in 3-D and also have a potential energy to worry about. (But they aren't that complicated: you can look at them here .) There are an infinite number of functions that solve the equation, just like in the simple example. In the 1-D example, we used 1 quantum number n to determine the energies. As n increased, so did the energy, and so did the number of nodes (places where the amplitude is zero). In 3-D, the standing waves are specified by 3 quantum numbers. (Actually, Bohr and others were using 3 quantum numbers before Schrodinger published this, because it makes sense in 3-D.) The principal quantum number n corresponds roughly to the radius of the orbit in the Bohr model, or to the "most probable distance from the nucleus" in the orbital model. This also corresponds to the energy: the electron is lower energy when it is close to the nucleus, because of the Coulomb attraction. The second quantum number ℓ gives the shape. In the Bohr model this meant ellipse or circle. In Schrodinger's model the shapes are 3-D. You can look at them in detail on Wikipedia (scroll down to the table of images) or with an animated applet like this. This is also a good time to check out the "orbitals" tab on Ptable . Pick an element, then move over the arrows to see a picture of each orbital.
Finally, the third quantum number tells the orientation. In the Bohr model this meant what plane the orbit was in. In the figure, you can see that the last 3 orbitals have the same shape but point along different directions (x, y and z).
Notice that standard pictures of orbitals show most of them with 2 colors. This represents the phase of the orbital (whether Ψ is + or − at that point). This doesn't effect the energy or the probability of finding the particle there, but it is very important when atoms interact with each other, like forming bonds, which is what chemistry is all about! When waves interact with each other, it matters a lot whether the peak (high point) overlaps with the trough (low point) or not. This is true of water waves, sound waves, light waves and electron waves.
There are lots of ways of showing orbitals, but usually they just draw a surface so that you have 50% or 90% or whatever chance of finding the electron inside the surface. (Always, the probability of finding the electron becomes 0 as you get far enough from the nucleus.) The second orbital in the picture is shown as a slice, so you can see that the inside near the nucleus has the opposite phase as the outside. Wherever the color changes, there must be a node (Ψ passes through 0 as it goes from + or −). Thus the first orbital has no node, which makes sense because more nodes mean higher energy. The second has 1 spherical node. The next three have a planar (flat) node. If you look at more orbitals, you'll notice that the nodes keep increasing.
Here's the pattern of orbital shapes and quantum numbers.
- The principal quantum number n tells you how big the orbital is and what the energy is. It also tells you how many spherical nodes the orbital has. The values go from n = 1, 2, 3, 4, 5, 6, 7... they could keep going but we haven't found elements that need that many orbitals yet! The number of spherical nodes in the orbital is n − ℓ − 1.
- The angular momentum quantum number ℓ tells you the shape. If ℓ = 0, the orbital has spherical symmetry (it's round), and we say it is an "s orbital". If ℓ = 1, the orbital has one planar node, and it's called a "p orbital". See the figure below for d and f orbitals, with 2 and 3 planar nodes each. The possible values of ℓ are 0, 1, ... n − 1.
- The magnetic quantum number m ℓ tells the orientation of the orbital. The possible values are -ℓ, -ℓ + 1, ... 0, 1, ℓ − 1, ℓ. For instance, for the p orbitals, it can be -1, 0, 1. You can remember the number of orientations using the table below.
A geometric pattern for remembering the number of each type of orbitals and the possible values of the quantum numbers.


 Please note that this reading deals with all seven movements of the piece. On the listening exam, you will only be responsible for movements I, IV, and VII.
Introduction
Wachet auf, ruft uns die Stimme (Awake, calls the voice to us), BWV 140 , also known as Sleepers Wake , is a church cantata by Johann Sebastian Bach. He composed the chorale cantatain Leipzig for the 27th Sunday after Trinity and first performed it on 25 November 1731. It is based on the hymn “ Wachet auf, ruft uns die Stimme ” (1599) by Philipp Nicolai. Movement 4 of the cantata is the base for the first of Bach’s Schübler Chorales, BWV 645. The cantata is a late addition to Bach’s cycle of chorale cantatas, featuring additional poetry for two duets of Jesus and the Soul which expand the theme of the hymn.
Scoring and Structure
The cantata in seven movements is scored for three soloists (soprano, tenor, and bass), a four-part choir, horn, two oboes, taille, violino piccolo, two violins, viola, and basso continuo.
 (When will you come, my salvation?)
- Chorale (tenor): Zion hört die Wächter singen (Zion hears the watchmen singing)
- Recitative: So geh herein zu mir (So come in with me)
- Aria (soprano, bass): Mein Freund ist mein! (My friend is mine!)
- Chorale: Gloria sei dir gesungen (May Gloria be sung to you)
Music
The first movement is a chorale fantasia based on the first verse of the chorale, a common feature of Bach’s earlier chorale cantatas. It is in E-flat major. The cantus firmus is sung by the soprano. The orchestra plays independent material mainly based on two motifs: a dotted rhythm and an ascending scale “with syncopated accent shifts.” The lower voices add in unusually free polyphonic music images such as the frequent calls “wach auf!” (wake up!) and “wo, wo?” (where, where?), and long melismas in a fugato on “Halleluja.”
The second movement is a recitative for tenor as a narrator who calls the “Töchter Zions” (daughters of Zion). In the following duet with obbligato violino piccolo, the soprano represents the Soul and the bass is the vox Christi (voice of Jesus).
The fourth movement, based on the second verse of the chorale, is written in the style of a chorale prelude, with the phrases of the chorale, sung as a cantus firmus by the tenors (or by the tenor soloist), entering intermittently against a famously lyrical melody played in unison by the violins (without the violino piccolo) and the viola, accompanied by the basso continuo. Bach later transcribed this movement for organ (BWV 645), and it was subsequently published along with five other transcriptions Bach made of his cantata movements as the Schübler Chorales.
The fifth movement is a recitative for bass, accompanied by the strings. It pictures the unity of the bridegroom and the “chosen bride.” The sixth movement is another duet for soprano and bass with obbligato oboe. This duet, like the third movement, is a love duet between the soprano Soul and the bass Jesus. Alfred Dürr describes it as giving “expression to the joy of the united pair,” showing a “relaxed mood” in “artistic intensity.”
The closing chorale is a four-part setting of the third verse of the hymn. The high pitch of the melody is doubled by a violino piccolo an octave higher, representing the bliss of the “heavenly Jerusalem.”
Contributors and Attributions
- Authored by : Elliott Jones. Provided by : Santa Ana College. Located at : http://www.sac.edu .  Provided by : Wikipedia. Located at : https://en.Wikipedia.org/wiki/Wachet_auf,_ruft_uns_die_Stimme,_BWV_140 . License : CC BY-SA: Attribution-ShareAlike


26.4: Newborn Loss
By the end of this section, you will be able to:
- Describe the causes of neonatal deaths
- Identify ways to debrief with fellow coworkers and providers after a neonatal death
- Identify the nurse’s role in supporting a family grieving after a neonatal death
A neonatal death is when a baby dies within the first 28 days of life. The occurrence in the United States is about 4 in 1,000 babies (less than 1 percent) (National Center for Health Statistics, 2023). Death of a newborn at any time is a traumatic event and requires specialized interdisciplinary care for families. In this section, the causes of neonatal deaths and the nurse’s role in supporting the grieving family will be reviewed.
Causes of Neonatal Mortality
According to the World Health Organization ([WHO], 2022), in 2020, rates of neonatal death varied widely depending on where the infant was born; for example, the area of sub-Saharan Africa represented 43 percent of all neonatal deaths, and Central and Southern Asia represented 36 percent (WHO, 2022). In the United States, neonatal mortality rates were higher in Black (non-Hispanic), American Indian, Alaskan Native, and Pacific Islander persons, and in the states of Mississippi, Louisiana, North Carolina, South Carolina, and Arkansas (United Health Foundation, n.d.). Most neonatal deaths occur within the first 24 hours of birth or the first week of life.
The most common causes of neonatal loss are prematurity, low birth weight, and congenital anomalies (CDC, 2023b). A pregnancy, labor, or birth with complications also increases the risk for neonatal loss, especially that due to asphyxia and infection . Refer to Chapter 25 Care of the Newborn at Risk for more information on causes of neonatal loss.
Preterm Birth
Preterm birth, birth occurring prior to 37 weeks’ gestation, is a significant cause of neonatal mortality. Preterm newborns are usually of low birth weight and have more health problems than term babies. In the United States in 2022, 1 in 10 infants was born preterm (Martin et al., 2022). These infants’ brain, liver, and lungs lack the intrauterine development that occurs in the last few weeks and months of pregnancy and, therefore, have developmental delays, respiratory problems, feeding difficulties, and higher mortality (CDC, 2023c).
Newborns born prior to 34 weeks (Figure 26.2) will most likely suffer from respiratory distress syndrome due to lack of surfactant , which allows the lungs to inflate and deflate easily (March of Dimes, 2017). Intraventricular hemorrhage, or bleeding in the brain, is seen in preterm babies and is a common cause of mortality (March of Dimes, 2017). Preterm babies can also suffer from necrotizing enterocolitis (tissue death in the intestines); this causes diarrhea, feeding problems, and an edematous abdomen. This is a serious complication that can lead to death (March of Dimes, 2017). Some maternal complications can cause preterm birth, such as preeclampsia, placental abruption, and premature rupture of membranes.
Childbirth Complications
Research suggests that timely treatment of preterm birth, intrapartum complications, and infection could result in fewer neonatal deaths (Lawn et al., 2023). Delay in decision making during labor, delay in presenting to the hospital, and delay in adequate prenatal care all contribute to neonatal mortality (Lawn et al., 2023). Common intrapartum complications resulting in neonatal mortality include birth trauma, such as intracranial hemorrhage and skull fracture ; placental problems, such as placental abruption, cord avulsion, vasa previa, and cord accidents; and maternal complications, such as uterine rupture, preeclampsia/eclampsia, injury, and substance use (Tesfay et al., 2022).
Infections
Chorioamnionitis is the infection of the bag of water affecting the uterus and the fetus. Group B streptococcus (GBS) is the cause of most neonatal invasive infections and is associated with a high morbidity and mortality rate. Most GBS infections occur within the first week of life and are acquired during birth but can occur weeks later. Between 20 and 30 percent of pregnant persons are colonized with GBS, but only 1 to 2 percent of neonates will develop the infection (Mynarek et al., 2021). GBS can be a cause of preterm birth, leading to neonatal mortality (Mynarek et al., 2021).
Neonatal sepsis can be caused by a maternal infection. During pregnancy, an infection can cross the placenta; during labor, it can ascend the vaginal canal to the uterus and fetus (Figure 26.3), and after birth, an infection can be acquired from breast-feeding (Pace & Yanowitz, 2022). Neonatal sepsis can be caused by a bacterium, virus, or fungus with a presentation of fever, tachypnea, lethargy, hypothermia, and poor feeding (Pace & Yanowitz, 2022). The age of the neonate can drastically change the chance of survival, with a fatality rate of 20 percent in preterm neonates and 2 percent in term neonates (Pace & Yanowitz, 2022).
Herpes simplex virus (HSV) infection occurs in 10 per 100,000 live births (Mahant et al., 2019). HSV can infect the neonate during pregnancy or delivery and has a mortality rate of 4.1 to 7.3 percent (Mahant et al., 2019). Symptoms of neurologic infection, usually occurring after 2 weeks of life, include seizures, poor feeding, lethargy, bulging fontanelles, temperature irregularities, and skin lesions; disseminated disease occurs after 10 days of life, with symptoms of respiratory failure, encephalitis, multiorgan failure, and death (Mahant et al., 2019). Another organism causing neonatal mortality is E. coli. E. coli can cause pneumonia, sepsis, and meningitis and is transmitted mostly as a nosocomial infection (Lai et al., 2021). Neonates with fever or temperature instability can be readmitted for possible sepsis.
Birth Defects
Birth defects can be responsible for neonatal deaths. Cardiac defects can be treated with surgery; however, more complex defects may not be treatable and can cause neonatal death (March of Dimes, 2017). Lung defects can occur because of underdevelopment, which can be caused by prematurity or oligohydramnios , as lungs cannot expand and develop with low amniotic fluid. Neurologic defects such as anencephaly, that is, lack of development of the brain and skull, are many times the cause of stillbirth; however, some anencephalic infants are born but die shortly after birth (March of Dimes, 2017).
Caring for the Caregiver
Many parents or caregivers blame themselves and experience extreme guilt in cases of newborn death, especially in the case of inheritable genetic diseases. Grief has been described as a loss of an attachment, and the attachment between a parent and child is one of the strongest of all human attachments (Duncan, 2018). Coping after the loss of an infant can be influenced by culture, religion, family dynamics, and social support. The nurse should assess these influences in order to provide referrals and arrange for additional support.
Nurses who care for these families need support themselves. Debriefing with other nurses, managers, and social workers is important for self-care. Nurses can suffer from grief and depression. When they recognize these symptoms, they must reach out for help in order to care for themselves. Being supported by peers allows nurses to work through their feelings and ideally prevent grief from turning into depression.
Cultural Considerations
The process of grieving can be influenced by a family’s culture, such as their values, beliefs, and religion (Arslan & Buldukoğlu, 2019). Grief rituals can range from wearing black clothing for a certain period to avoiding weddings or other celebrations while grieving to isolating from others (Arslan & Buldukoğlu, 2019). Culture can influence the desire for a burial or cremation. Beliefs can also determine if the body will be viewed in an open casket. The nurse can assess for cultural beliefs to provide culturally competent care. The nurse can contact the patient’s church or religious group to help plan for burial or cremation.
Death Rituals Practiced in Certain Religions
In Judaism, the body is not embalmed, nor will the service include an open casket (Lowey, 2015). In Catholicism, a priest performs a funeral mass at the church followed by internment (Lowey, 2015). In Protestant denominations, cremation or burial is acceptable, and funerals may be held at a funeral home or church. For followers of Islam, burial occurs as soon as possible, and the body of the deceased is buried facing Mecca (Lowey, 2015). The nurse can assess for cultural beliefs and provide culturally competent care.
Referrals for Help
Validated depression screening should be done. Depending on the health-care facility, nurses can refer families to bereavement resources or consult the hospital’s social worker or bereavement committee to obtain referrals for families. Cognitive behavior therapy (CBT) and counseling can be helpful to those families experiencing a newborn death. Grief counseling is another therapy that can be helpful. Support groups are available to provide professional and social support to families experiencing similar grief. Medications such as antidepressants and anxiolytics can be used to assist with coping.
Mommies Enduring Neonatal Death (MEND) is a support group providing resources and support to families who have lost a child to miscarriage, stillbirth, and infant loss.
Helping Families Understand the Grief Process
Parents dealing with the loss of a newborn can experience a lack of desire for life and daily activities. They may feel anger, sorrow, and shock . Physical symptoms, such as chest pain, fatigue, stomachache, and digestive issues, can result from grief (Arslan & Buldukoğlu, 2019). Table 26.4 lists more symptoms of grief. Families should be provided reassurance that these feelings and symptoms are part of the normal grief process. Nurses should explain that there is no timeline on grieving, and everyone grieves in their own way. Suggestions for dealing with the grief process include the following:
- Acknowledge the loss and pain.
- Understand that grief can trigger many unexpected emotions.
- Know that grief is unique to the person.
- Find people who care and support you.

- Recognize signs of depression.
(Smith et al., 2023)
| Type | Symptoms |
|---|---|
| Emotional |
Shock
Disbelief Sadness Guilt Fear Anger |
| Physical |
Fatigue
Nausea Compromised immunity Weight changes Aches/pains Insomnia |
The nurse will explain that grief is different from depression. Depression is despair and extreme sadness that interferes with daily activities, with frequent thoughts of death, worthlessness, or suicide, and does not lessen with time (Schimelpfening & Gans, 2023). Grief can decrease with support and time but can also recur during special situations such as the birthday or anniversary of a loved one; depression does not change depending on the circumstances (Schimelpfening & Gans, 2023).
Helping Siblings
Helping a sibling understand the death of a newborn is difficult. Children’s understanding of death depends on their age, development, their ability to think abstractly, their other experiences with loss, and their cognitive development (Arslan & Buldukoğlu, 2019). Young siblings who experience loss can exhibit anxiety and have behavior issues; older children can show signs of depression (Arslan & Buldukoğlu, 2019).
Supporting a sibling through a newborn death requires honesty and openness. Parents should acknowledge the sibling’s grief, be honest about their own grief, and help the sibling keep the memory of the newborn alive (Support for Siblings after a Neonatal Death, n.d.). Give the sibling permission to cry and grieve, explain that the death was not anyone’s fault, and reassure them their parents are okay, as children many times fear other loved ones dying (Support for Siblings after a Neonatal Death, n.d.). Counseling specifically for siblings and children is available and referrals can be provided by the nurse.


45.1: Population Demography
Skills to Develop
- Describe how ecologists measure population size and density
- Describe three different patterns of population distribution
- Use life tables to calculate mortality rates
- Describe the three types of survivorship curves and relate them to specific populations
Populations are dynamic entities. Populations consist all of the species living within a specific area, and populations fluctuate based on a number of factors: seasonal and yearly changes in the environment, natural disasters such as forest fires and volcanic eruptions, and competition for resources between and within species. The statistical study of population dynamics, demography , uses a series of mathematical tools to investigate how populations respond to changes in their biotic and abiotic environments. Many of these tools were originally designed to study human populations. For example, life tables , which detail the life expectancy of individuals within a population, were initially developed by life insurance companies to set insurance rates. In fact, while the term “demographics” is commonly used when discussing humans, all living populations can be studied using this approach.
Population Size and Density
The study of any population usually begins by determining how many individuals of a particular species exist, and how closely associated they are with each other. Within a particular habitat, a population can be characterized by its population size ( N ) , the total number of individuals, and its population density , the number of individuals within a specific area or volume. Population size and density are the two main characteristics used to describe and understand populations. For example, populations with more individuals may be more stable than smaller populations based on their genetic variability, and thus their potential to adapt to the environment. Alternatively, a member of a population with low population density (more spread out in the habitat), might have more difficulty finding a mate to reproduce compared to a population of higher density. As is shown in Figure \(\PageIndex{1}\), smaller organisms tend to be more densely distributed than larger organisms.
Art Connection
As this graph shows, population density typically decreases with increasing body size. Why do you think this is the case?
Population Research Methods
The most accurate way to determine population size is to simply count all of the individuals within the habitat. However, this method is often not logistically or economically feasible, especially when studying large habitats. Thus, scientists usually study populations by sampling a representative portion of each habitat and using this data to make inferences about the habitat as a whole. A variety of methods can be used to sample populations to determine their size and density. For immobile organisms such as plants, or for very small and slow-moving organisms, a quadrat may be used (Figure \(\PageIndex{2}\)). A quadrat is a way of marking off square areas within a habitat, either by staking out an area with sticks and string, or by the use of a wood, plastic, or metal square placed on the ground. After setting the quadrats, researchers then count the number of individuals that lie within their boundaries. Multiple quadrat samples are performed throughout the habitat at several random locations. All of this data can then be used to estimate the population size and population density within the entire habitat. The number and size of quadrat samples depends on the type of organisms under study and other factors, including the density of the organism. For example, if sampling daffodils, a 1 m 2 quadrat might be used whereas with giant redwoods, which are larger and live much further apart from each other, a larger quadrat of 100 m 2 might be employed. This ensures that enough individuals of the species are counted to get an accurate sample that correlates with the habitat, including areas not sampled.
For mobile organisms, such as mammals, birds, or fish, a technique called mark and recapture is often used. This method involves marking a sample of captured animals in some way (such as tags, bands, paint, or other body markings), and then releasing them back into the environment to allow them to mix with the rest of the population; later, a new sample is collected, including some individuals that are marked (recaptures) and some individuals that are unmarked (Figure \(\PageIndex{3}\)).
Using the ratio of marked and unmarked individuals, scientists determine how many individuals are in the sample. From this, calculations are used to estimate the total population size. This method assumes that the larger the population, the lower the percentage of tagged organisms that will be recaptured since they will have mixed with more untagged individuals. For example, if 80 deer are captured, tagged, and released into the forest, and later 100 deer are captured and 20 of them are already marked, we can determine the population size ( N ) using the following equation:
Using our example, the population size would be estimated at 400.
Therefore, there are an estimated 400 total individuals in the original population.
There are some limitations to the mark and recapture method. Some animals from the first catch may learn to avoid capture in the second round, thus inflating population estimates. Alternatively, animals may preferentially be retrapped (especially if a food reward is offered), resulting in an underestimate of population size. Also, some species may be harmed by the marking technique, reducing their survival. A variety of other techniques have been developed, including the electronic tracking of animals tagged with radio transmitters and the use of data from commercial fishing and trapping operations to estimate the size and health of populations and communities.
Species Distribution
In addition to measuring simple density, further information about a population can be obtained by looking at the distribution of the individuals. Species dispersion patterns (or distribution patterns) show the spatial relationship between members of a population within a habitat at a particular point in time. In other words, they show whether members of the species live close together or far apart, and what patterns are evident when they are spaced apart.
Individuals in a population can be more or less equally spaced apart, dispersed randomly with no predictable pattern, or clustered in groups. These are known as uniform, random, and clumped dispersion patterns, respectively (Figure \(\PageIndex{4}\)). Uniform dispersion is observed in plants that secrete substances inhibiting the growth of nearby individuals (such as the release of toxic chemicals by the sage plant Salvia leucophylla, a phenomenon called allelopathy) and in animals like the penguin that maintain a defined territory. An example of random dispersion occurs with dandelion and other plants that have wind-dispersed seeds that germinate wherever they happen to fall in a favorable environment. A clumped dispersion may be seen in plants that drop their seeds straight to the ground, such as oak trees, or animals that live in groups (schools of fish or herds of elephants). Clumped dispersions may also be a function of habitat heterogeneity. Thus, the dispersion of the individuals within a population provides more information about how they interact with each other than does a simple density measurement. Just as lower density species might have more difficulty finding a mate, solitary species with a random distribution might have a similar difficulty when compared to social species clumped together in groups.
Demography
While population size and density describe a population at one particular point in time, scientists must use demography to study the dynamics of a population. Demography is the statistical study of population changes over time: birth rates, death rates, and life expectancies. Each of these measures, especially birth rates, may be affected by the population characteristics described above. For example, a large population size results in a higher birth rate because more potentially reproductive individuals are present. In contrast, a large population size can also result in a higher death rate because of competition, disease, and the accumulation of waste. Similarly, a higher population density or a clumped dispersion pattern results in more potential reproductive encounters between individuals, which can increase birth rate. Lastly, a female-biased sex ratio (the ratio of males to females) or age structure (the proportion of population members at specific age ranges) composed of many individuals of reproductive age can increase birth rates.
In addition, the demographic characteristics of a population can influence how the population grows or declines over time. If birth and death rates are equal, the population remains stable. However, the population size will increase if birth rates exceed death rates; the population will decrease if birth rates are less than death rates. Life expectancy is another important factor; the length of time individuals remain in the population impacts local resources, reproduction, and the overall health of the population. These demographic characteristics are often displayed in the form of a life table.
Life Tables
Life tables provide important information about the life history of an organism. Life tables divide the population into age groups and often sexes, and show how long a member of that group is likely to live. They are modeled after actuarial tables used by the insurance industry for estimating human life expectancy. Life tables may include the probability of individuals dying before their next birthday (i.e., their mortality rate ), the percentage of surviving individuals dying at a particular age interval, and their life expectancy at each interval. An example of a life table is shown in Table \(\PageIndex{1}\) from a study of Dall mountain sheep, a species native to northwestern North America. Notice that the population is divided into age intervals (column A). The mortality rate (per 1000), shown in column D, is based on the number of individuals dying during the age interval (column B) divided by the number of individuals surviving at the beginning of the interval (Column C), multiplied by 1000.
For example, between ages three and four, 12 individuals die out of the 776 that were remaining from the original 1000 sheep. This number is then multiplied by 1000 to get the mortality rate per thousand.
As can be seen from the mortality rate data (column D), a high death rate occurred when the sheep were between 6 and 12 months old, and then increased even more from 8 to 12 years old, after which there were few survivors. The data indicate that if a sheep in this population were to survive to age one, it could be expected to live another 7.7 years on average, as shown by the life expectancy numbers in column E.
| Age interval (years) | Number dying in age interval out of 1000 born | Number surviving at beginning of age interval out of 1000 born | Mortality rate per 1000 alive at beginning of age interval | Life expectancy or mean lifetime remaining to those attaining age interval |
|---|---|---|---|---|
| 0-0.5 | 54 | 1000 | 54.0 | 7.06 |
| 0.5-1 | 145 | 946 | 153.3 | -- |
| 1-2 | 12 | 801 | 15.0 | 7.7 |
| 2-3 | 13 | 789 | 16.5 | 6.8 |
| 3-4 | 12 | 776 | 15.5 | 5.9 |
| 4-5 | 30 | 764 | 39.3 | 5.0 |
| 5-6 | 46 | 734 | 62.7 | 4.2 |
| 6-7 | 48 | 688 | 69.8 | 3.4 |
| 7-8 | 69 | 640 | 107.8 | 2.6 |
| 8-9 | 132 | 571 | 231.2 | 1.9 |
| 9-10 | 187 | 439 | 426.0 | 1.3 |
| 10-11 | 156 | 252 | 619.0 | 0.9 |
| 11-12 | 90 | 96 | 937.5 | 0.6 |
| 12-13 | 3 | 6 | 500.0 | 1.2 |
| 13-14 | 3 | 3 | 1000 | 0.7 |
Survivorship Curves
Another tool used by population ecologists is a survivorship curve , which is a graph of the number of individuals surviving at each age interval plotted versus time (usually with data compiled from a life table). These curves allow us to compare the life histories of different populations (Figure \(\PageIndex{5}\)). Humans and most primates exhibit a Type I survivorship curve because a high percentage of offspring survive their early and middle years—death occurs predominantly in older individuals. These types of species usually have small numbers of offspring at one time, and they give a high amount of parental care to them to ensure their survival. Birds are an example of an intermediate or Type II survivorship curve because birds die more or less equally at each age interval. These organisms also may have relatively few offspring and provide significant parental care. Trees, marine invertebrates, and most fishes exhibit a Type III survivorship curve because very few of these organisms survive their younger years; however, those that make it to an old age are more likely to survive for a relatively long period of time. Organisms in this category usually have a very large number of offspring, but once they are born, little parental care is provided. Thus these offspring are “on their own” and vulnerable to predation, but their sheer numbers assure the survival of enough individuals to perpetuate the species.
Summary
Populations are individuals of a species that live in a particular habitat. Ecologists measure characteristics of populations: size, density, dispersion pattern, age structure, and sex ratio. Life tables are useful to calculate life expectancies of individual population members. Survivorship curves show the number of individuals surviving at each age interval plotted versus time.
Art Connections
Figure \(\PageIndex{1}\): As this graph shows, population density typically decreases with increasing body size. Why do you think this is the case?
- Answer
-
Smaller animals require less food and other resources, so the environment can support more of them.
Footnotes
- 1 Data Adapted from Edward S. Deevey, Jr., “Life Tables for Natural Populations of Animals,” The Quarterly Review of Biology 22, no. 4 (December 1947): 283-314.
Glossary
- demography
- statistical study of changes in populations over time
- life table
- table showing the life expectancy of a population member based on its age
- mark and recapture
- technique used to determine population size in mobile organisms
- mortality rate
- proportion of population surviving to the beginning of an age interval that die during the age interval
- population density
- number of population members divided by the area or volume being measured
- population size ( N )
- number of population members in a habitat at the same time
- quadrat
- square made of various materials used to determine population size and density in slow moving or stationary organisms
- species dispersion pattern
- (also, species distribution pattern) spatial location of individuals of a given species within a habitat at a particular point in time
- survivorship curve
- graph of the number of surviving population members versus the relative age of the member


1.17: Strategic Planning in Retail
- Classify the general steps of strategic planning in retail
- Company strengths and weaknesses
- Personal values of the key implementers (i.e., management and the board)
- Industry opportunities and threats
- Broader societal expectations
The first two elements relate to factors internal to the company (i.e., the internal environment), while the latter two relate to factors external to the company (i.e., the external environment). These elements are considered throughout the strategic planning process.
For example, many retailers are trying to respond to customer migration from brick-and-mortar stores to online retailers. What situational, monetary, or product mix is drawing customers away from traditional stores? Imagine a local grocery store chain is trying to compete with Amazon pantry. The local store decides their goal is venture into the world of e-commerce by integrating an online website for customers to order products online, but can pick up their orders at the store. How did this local chain come to this solution and how would they go about it?
Here is a quick look at the steps in the strategic retail planning process.
- Define the business mission
- Conduct a situation audit. This can include a market attractiveness analysis, competitor analysis, self-analysis
- Identify strategic opportunities
- Evaluate strategic alternatives
- Establish specific objectives and allocate resources
- Develop a retail mix to implement strategy
- Evaluate performance and make adjustments
Here is a video that walks you through the first five steps of the strategic retail planning process. Afterwards we will discuss how to develop the retail mix to implement the retail strategy as well as evaluating performance.
You can view the transcript for “What is Strategic Planning? Process, Model, and Steps” (opens in new window).
Contributors and Attributions
- Strategic planning. Provided by : Wikipedia. Located at : https://en.wikipedia.org/wiki/Strategic_planning#Strategic_planning_vs._financial_planning . License : CC BY-SA: Attribution-ShareAlike
- Business Strategy/Overview of Strategic Planning. Provided by : Wikibooks. Located at : https://en.wikibooks.org/wiki/Business_Strategy/Marketing_Plans_and_Strategies . License : CC BY-SA: Attribution-ShareAlike


1.7: Weekend assignment 1
Watch this song by Lebanese Nancy Agram . She is singing to Egypt.
A repeated phrase in the song :
أنا مصري وأبويا مصري I am Egyptian and my father is Egyptian.
https://www.youtube.com/watch?v=mjBmjTJ7a3s
Speaking : Tell me about yourself
Using old and new vocabulary , present information about yourself extending to family members and friends. You can use the following guide.
|
My name __________. I am___________(nationality). I am from___________(city) in______________(state) and I live in _________________(city) . I am a student at Michigan State University. I study_______. I work in _______/ do not work.
My father, his name is_________. He is ……(nationality) He works in _________ My mother, her name is _______She is ……(nationality) She works in ________. I have_____________(brother/sister). My brother/sister lives in ———– and works ————– My friend, his/her name . He/she studies ——————- and lives ——–, and works/does not work in ————– |
Your task
1-Collect all the vocabulary you will need to complete this narrative.
2-Draft the sentences in your notebook taking care of spelling and structure.
3-Rehearse several times until you can almost say them from memory.
4-Record your narrative in audio and submit to the Dropbox
5-Keep what have written handy so we can recall them easily in class when needed.


Chapter 9 Review Exercises
Chapter Review Exercises
Solve Quadratic Equations Using the Square Root Property
In the following exercises, solve using the Square Root Property.
- \(y^{2}=144\)
- \(n^{2}-80=0\)
- \(4 a^{2}=100\)
- \(2 b^{2}=72\)
- \(r^{2}+32=0\)
- \(t^{2}+18=0\)
- \(\frac{2}{3} w^{2}-20=30\)
- \(5 c^{2}+3=19\)
- Answer
-
1. \(y=\pm 12\)
3. \(a=\pm 5\)
5. \(r=\pm 4 \sqrt{2} i\)
7. \(w=\pm 5 \sqrt{3}\)
In the following exercises, solve using the Square Root Property.
- \((p-5)^{2}+3=19\)
- \((u+1)^{2}=45\)
- \(\left(x-\frac{1}{4}\right)^{2}=\frac{3}{16}\)
- \(\left(y-\frac{2}{3}\right)^{2}=\frac{2}{9}\)
- \((n-4)^{2}-50=150\)
- \((4 c-1)^{2}=-18\)
- \(n^{2}+10 n+25=12\)
- \(64 a^{2}+48 a+9=81\)
- Answer
-
1. \(p=-1,9\)
3. \(x=\frac{1}{4} \pm \frac{\sqrt{3}}{4}\)
5. \(n=4 \pm 10 \sqrt{2}\)
7. \(n=-5 \pm 2 \sqrt{3}\)
Solve Quadratic Equations by Completing the Square
In the following exercises, complete the square to make a perfect square trinomial. Then write the result as a binomial squared.
- \(x^{2}+22 x\)
- \(m^{2}-8 m\)
- \(a^{2}-3 a\)
- \(b^{2}+13 b\)
- Answer
-
1. \((x+11)^{2}\)
3. \(\left(a-\frac{3}{2}\right)^{2}\)
In the following exercises, solve by completing the square.
- \(d^{2}+14 d=-13\)
- \(y^{2}-6 y=36\)
- \(m^{2}+6 m=-109\)
- \(t^{2}-12 t=-40\)
- \(v^{2}-14 v=-31\)
- \(w^{2}-20 w=100\)
- \(m^{2}+10 m-4=-13\)
- \(n^{2}-6 n+11=34\)
- \(a^{2}=3 a+8\)
- \(b^{2}=11 b-5\)
- \((u+8)(u+4)=14\)
- \((z-10)(z+2)=28\)
- Answer
-
1. \(d=-13,-1\)
3. \(m=-3 \pm 10 i\)
5. \(v=7 \pm 3 \sqrt{2}\)
7. \(m=-9,-1\)
9. \(a=\frac{3}{2} \pm \frac{\sqrt{41}}{2}\)
11. \(u=-6 \pm 2 \sqrt{2}\)
Solve Quadratic Equations of the Form \(ax^{2}+bx+c=0\) by Completing the Square
In the following exercises, solve by completing the square.
- \(3 p^{2}-18 p+15=15\)
- \(5 q^{2}+70 q+20=0\)
- \(4 y^{2}-6 y=4\)
- \(2 x^{2}+2 x=4\)
- \(3 c^{2}+2 c=9\)
- \(4 d^{2}-2 d=8\)
- \(2 x^{2}+6 x=-5\)
- \(2 x^{2}+4 x=-5\)
- Answer
-
1. \(p=0,6\)
3. \(y=-\frac{1}{2}, 2\)
5. \(c=-\frac{1}{3} \pm \frac{2 \sqrt{7}}{3}\)
7. \(x=\frac{3}{2} \pm \frac{1}{2} i\)
In the following exercises, solve by using the Quadratic Formula.
- \(4 x^{2}-5 x+1=0\)
- \(7 y^{2}+4 y-3=0\)
- \(r^{2}-r-42=0\)
- \(t^{2}+13 t+22=0\)
- \(4 v^{2}+v-5=0\)
- \(2 w^{2}+9 w+2=0\)
- \(3 m^{2}+8 m+2=0\)
- \(5 n^{2}+2 n-1=0\)
- \(6 a^{2}-5 a+2=0\)
- \(4 b^{2}-b+8=0\)
- \(u(u-10)+3=0\)
- \(5 z(z-2)=3\)
- \(\frac{1}{8} p^{2}-\frac{1}{5} p=-\frac{1}{20}\)
- \(\frac{2}{5} q^{2}+\frac{3}{10} q=\frac{1}{10}\)
- \(4 c^{2}+4 c+1=0\)
- \(9 d^{2}-12 d=-4\)
- Answer
-
1. \(x=\frac{1}{4}, 1\)
3. \(r=-6,7\)
5. \(v=\frac{-1 \pm \sqrt{21}}{8}\)
7. \(m=\frac{-4 \pm \sqrt{10}}{3}\)
9. \(a=\frac{5}{12} \pm \frac{\sqrt{23}}{12} i\)
11. \(u=5 \pm \sqrt{21}\)
13. \(p=\frac{4 \pm \sqrt{5}}{5}\)
15. \(c=-\frac{1}{2}\)
In the following exercises, determine the number of solutions for each quadratic equation.
-
- \(9 x^{2}-6 x+1=0\)
- \(3 y^{2}-8 y+1=0\)
- \(7 m^{2}+12 m+4=0\)
- \(5 n^{2}-n+1=0\)
-
- \(5 x^{2}-7 x-8=0\)
- \(7 x^{2}-10 x+5=0\)
- \(25 x^{2}-90 x+81=0\)
- \(15 x^{2}-8 x+4=0\)
- Answer
-
1.
- \(1\)
- \(2\)
- \(2\)
- \(2\)
In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve.
-
- \(16 r^{2}-8 r+1=0\)
- \(5 t^{2}-8 t+3=9\)
- \(3(c+2)^{2}=15\)
-
- \(4 d^{2}+10 d-5=21\)
- \(25 x^{2}-60 x+36=0\)
- \(6(5 v-7)^{2}=150\)
- Answer
-
1.
- Factor
- Quadratic Formula
- Square Root
Solve Equations in Quadratic Form
In the following exercises, solve.
- \(x^{4}-14 x^{2}+24=0\)
- \(x^{4}+4 x^{2}-32=0\)
- \(4 x^{4}-5 x^{2}+1=0\)
- \((2 y+3)^{2}+3(2 y+3)-28=0\)
- \(x+3 \sqrt{x}-28=0\)
- \(6 x+5 \sqrt{x}-6=0\)
- \(x^{\frac{2}{3}}-10 x^{\frac{1}{3}}+24=0\)
- \(x+7 x^{\frac{1}{2}}+6=0\)
- \(8 x^{-2}-2 x^{-1}-3=0\)
- Answer
-
1. \(x=\pm \sqrt{2}, x=\pm 2 \sqrt{3}\)
3. \(x=\pm 1, x=\pm \frac{1}{2}\)
5. \(x=16\)
7. \(x=64, x=216\)
9. \(x=-2, x=\frac{4}{3}\)
Solve Applications of Quadratic Equations
In the following exercises, solve by using the method of factoring, the square root principle, or the Quadratic Formula. Round your answers to the nearest tenth, if needed.
- Find two consecutive odd numbers whose product is \(323\).
- Find two consecutive even numbers whose product is \(624\).
- A triangular banner has an area of \(351\) square centimeters. The length of the base is two centimeters longer than four times the height. Find the height and length of the base.
- Julius built a triangular display case for his coin collection. The height of the display case is six inches less than twice the width of the base. The area of the of the back of the case is \(70\) square inches. Find the height and width of the case.
- A tile mosaic in the shape of a right triangle is used as the corner of a rectangular pathway. The hypotenuse of the mosaic is \(5\) feet. One side of the mosaic is twice as long as the other side. What are the lengths of the sides? Round to the nearest tenth.
Figure 9.E.1
6. A rectangular piece of plywood has a diagonal which measures two feet more than the width. The length of the plywood is twice the width. What is the length of the plywood’s diagonal? Round to the nearest tenth.
7. The front walk from the street to Pam’s house has an area of \(250\) square feet. Its length is two less than four times its width. Find the length and width of the sidewalk. Round to the nearest tenth.
8. For Sophia’s graduation party, several tables of the same width will be arranged end to end to give serving table with a total area of \(75\) square feet. The total length of the tables will be two more than three times the width. Find the length and width of the serving table so Sophia can purchase the correct size tablecloth . Round answer to the nearest tenth.
9. A ball is thrown vertically in the air with a velocity of \(160\) ft/sec. Use the formula \(h=-16 t^{2}+v_{0} t\) to determine when the ball will be \(384\) feet from the ground. Round to the nearest tenth.
10. The couple took a small airplane for a quick flight up to the wine country for a romantic dinner and then returned home. The plane flew a total of \(5\) hours and each way the trip was \(360\) miles. If the plane was flying at \(150\) mph, what was the speed of the wind that affected the plane?
11.  The trip was \(4\) miles each way and the current was difficult. If Roy kayaked at a speed of \(5\) mph, what was the speed of the current?
12. Two handymen can do a home repair in \(2\) hours if they work together. One of the men takes \(3\) hours more than the other man to finish the job by himself. How long does it take for each handyman to do the home repair individually?
- Answer
-
2. Two consecutive even numbers whose product is \(624\) are \(24\) and \(26\), and \(−24\) and \(−26\).
4. The height is \(14\) inches and the width is \(10\) inches.
6. The length of the diagonal is \(3.6\) feet.
8. The width of the serving table is \(4.7\) feet and the length is \(16.1\) feet.
10. The speed of the wind was \(30\) mph.
12. One man takes \(3\) hours and the other man \(6\) hours to finish the repair alone.
Graph Quadratic Functions Using Properties
In the following exercises, graph by plotting point.
- Graph \(y=x^{2}-2\)
- Graph \(y=-x^{2}+3\)
- Answer
-
2.
In the following exercises, determine if the following parabolas open up or down.
-
- \(y=-3 x^{2}+3 x-1\)
- \(y=5 x^{2}+6 x+3\)
-
- \(y=x^{2}+8 x-1\)
- \(y=-4 x^{2}-7 x+1\)
- Answer
-
2.
- Up
- Down
In the following exercises, find
- The equation of the axis of symmetry
-
The vertex
- \(y=-x^{2}+6 x+8\)
- \(y=2 x^{2}-8 x+1\)
- Answer
-
2. \(x=2\) ; \((2,-7)\)
In the following exercises, find the \(x\)- and \(y\)-intercepts.
- \(y=x^{2}-4x+5\)
- \(y=x^{2}-8x+15\)
- \(y=x^{2}-4x+10\)
- \(y=-5x^{2}-30x-46\)
- \(y=16x^{2}-8x+1\)
- \(y=x^{2}+16x+64\)
- Answer
-
2. \(\begin{array}{l}{y :(0,15)} \\ {x :(3,0),(5,0)}\end{array}\)
4. \(\begin{array}{l}{y :(0,-46)} \\ {x : \text { none }}\end{array}\)
6. \(\begin{array}{l}{y :(0,-64)} \\ {x :(-8,0)}\end{array}\)
Graph Quadratic Functions Using Properties
In the following exercises, graph by using its properties.
- \(y=x^{2}+8 x+15\)
- \(y=x^{2}-2 x-3\)
- \(y=-x^{2}+8 x-16\)
- \(y=4 x^{2}-4 x+1\)
- \(y=x^{2}+6 x+13\)
- \(y=-2 x^{2}-8 x-12\)
- Answer
-
2.
4.
6.
In the following exercises, find the minimum or maximum value.
- \(y=7 x^{2}+14 x+6\)
- \(y=-3 x^{2}+12 x-10\)
- Answer
-
2. The maximum value is \(2\) when \(x=2\).
In the following exercises, solve. Rounding answers to the nearest tenth.
- A ball is thrown upward from the ground with an initial velocity of \(112\) ft/sec. Use the quadratic equation \(h=-16 t^{2}+112 t\) to find how long it will take the ball to reach maximum height, and then find the maximum height.
- A daycare facility is enclosing a rectangular area along the side of their building for the children to play outdoors. They need to maximize the area using \(180\) feet of fencing on three sides of the yard. The quadratic equation \(A=-2 x^{2}+180 x\) gives the area, \(A\), of the yard for the length, \(x\), of the building that will border the yard. Find the length of the building that should border the yard to maximize the area, and then find the maximum area.
- Answer
-
2. The length adjacent to the building is \(90\) feet giving a maximum area of \(4,050\) square feet.
Graph Quadratic Functions Using Transformations
In the following exercises, graph each function using a vertical shift.
- \(g(x)=x^{2}+4\)
- \(h(x)=x^{2}-3\)
- Answer
-
2.
In the following exercises, graph each function using a horizontal shift.
- \(f(x)=(x+1)^{2}\)
- \(g(x)=(x-3)^{2}\)
- Answer
-
2.
In the following exercises, graph each function using transformations.
- \(f(x)=(x+2)^{2}+3\)
- \(f(x)=(x+3)^{2}-2\)
- \(f(x)=(x-1)^{2}+4\)
- \(f(x)=(x-4)^{2}-3\)
- Answer
-
2.
4.
In the following exercises, graph each function.
- \(f(x)=2x^{2}\)
- \(f(x)=-x^{2}\)
- \(f(x)=\frac{1}{2} x^{2}\)
- Answer
-
2.
In the following exercises, rewrite each function in the \(f(x)=a(x-h)^{2}+k\) form by completing the square.
- \(f(x)=2 x^{2}-4 x-4\)
- \(f(x)=3 x^{2}+12 x+8\)
- Answer
-
1. \(f(x)=2(x-1)^{2}-6\)
In the following exercises,
- Rewrite each function in \(f(x)=a(x−h)^{2}+k\) form
-
Graph it by using transformations
- \(f(x)=3 x^{2}-6 x-1\)
- \(f(x)=-2 x^{2}-12 x-5\)
- \(f(x)=2 x^{2}+4 x+6\)
- \(f(x)=3 x^{2}-12 x+7\)
- Answer
-
1.
- \(f(x)=3(x-1)^{2}-4\)
-
Figure 9.E.13
3.
- \(f(x)=2(x+1)^{2}+4\)
-
Figure 9.E.14
In the following exercises,
- Rewrite each function in \(f(x)=a(x−h)^{2}+k\) form
-
Graph it using properties
- \(f(x)=-3 x^{2}-12 x-5\)
- \(f(x)=2 x^{2}-12 x+7\)
- Answer
-
1.
- \(f(x)=-3(x+2)^{2}+7\)
-
Figure 9.E.15
In the following exercises, write the quadratic function in \(f(x)=a(x−h)^{2}+k\) form.
-
Figure 9.E.16 -
Figure 9.E.17
- Answer
-
1. \(f(x)=(x+1)^{2}-5\)
Solve Quadratic Inequalities
In the following exercises, solve graphically and write the solution in interval notation.
- \(x^{2}-x-6>0\)
- \(x^{2}+4 x+3 \leq 0\)
- \(-x^{2}-x+2 \geq 0\)
- \(-x^{2}+2 x+3<0\)
- Answer
-
1.
-
Figure 9.E.18 - \((-\infty,-2) \cup(3, \infty)\)
3.
-
Figure 9.E.19 - \([-2,1]\)
-
In the following exercises, solve each inequality algebraically and write any solution in interval notation.
- \(x^{2}-6 x+8<0\)
- \(x^{2}+x>12\)
- \(x^{2}-6 x+4 \leq 0\)
- \(2 x^{2}+7 x-4>0\)
- \(-x^{2}+x-6>0\)
- \(x^{2}-2 x+4 \geq 0\)
- Answer
-
1. \((2,4)\)
3. \([3-\sqrt{5}, 3+\sqrt{5}]\)
5. no solution
Practice Test
- Use the Square Root Property to solve the quadratic equation \(3(w+5)^{2}=27\).
- Use Completing the Square to solve the quadratic equation \(a^{2}-8 a+7=23\).
- Use the Quadratic Formula to solve the quadratic equation \(2 m^{2}-5 m+3=0\).
- Answer
-
1. \(w=-2, w=-8\)
3. \(m=1, m=\frac{3}{2}\)
Solve the following quadratic equations. Use any method.
- \(2 x(3 x-2)-1=0\)
- \(\frac{9}{4} y^{2}-3 y+1=0\)
- Answer
-
2. \(y=\frac{2}{3}\)
Use the discriminant to determine the number and type of solutions of each quadratic equation.
- \(6 p^{2}-13 p+7=0\)
- \(3 q^{2}-10 q+12=0\)
- Answer
-
2. \(2\) complex
Solve each equation.
- \(4 x^{4}-17 x^{2}+4=0\)
- \(y^{\frac{2}{3}}+2 y^{\frac{1}{3}}-3=0\)
- Answer
-
2. \(y=1, y=-27\)
For each parabola, find
- Which direction it opens
- The equation of the axis of symmetry
- The vertex
- The \(x\) - and \(y\)-intercepts
-
The maximum or minimum value
- \(y=3 x^{2}+6 x+8\)
- \(y=-x^{2}-8 x+16\)
- Answer
-
2.
- down
- \(x=-4\)
- \((-4,0)\)
- \(y: (0,16); x: (-4,0)\)
- minimum value of \(-4\) when \(x=0\)
Graph each quadratic function using intercepts, the vertex, and the equation of the axis of symmetry.
- \(f(x)=x^{2}+6 x+9\)
- \(f(x)=-2 x^{2}+8 x+4\)
- Answer
-
2.
In the following exercises, graph each function using transformations.
- \(f(x)=(x+3)^{2}+2\)
- \(f(x)=x^{2}-4 x-1\)
- Answer
-
2.
Figure 9.E.21
In the following exercises, solve each inequality algebraically and write any solution in interval notation.
- \(x^{2}-6 x-8 \leq 0\)
- \(2 x^{2}+x-10>0\)
- Answer
-
2. \(\left(-\infty,-\frac{5}{2}\right) \cup(2, \infty)\)
Model the situation with a quadratic equation and solve by any method.
- Find two consecutive even numbers whose product is \(360\).
- The length of a diagonal of a rectangle is three more than the width. The length of the rectangle is three times the width. Find the length of the diagonal. (Round to the nearest tenth.)
- Answer
-
2. A water balloon is launched upward at the rate of \(86\) ft/sec. Using the formula \(h=-16 t^{2}+86 t\) find how long it will take the balloon to reach the maximum height, and then find the maximum height. Round to the nearest tenth.


19.1: Conditions for Static Equilibrium
- Identify the physical conditions of static equilibrium.
- Draw a free-body diagram for a rigid body acted on by forces.
- Explain how the conditions for equilibrium allow us to solve statics problems.
We say that a rigid body is in equilibrium when both its linear and angular acceleration are zero relative to an inertial frame of reference. This means that a body in equilibrium can be moving, but if so, its linear and angular velocities must be constant. We say that a rigid body is in static equilibrium when it is at rest in our selected frame of reference . Notice that the distinction between the state of rest and a state of uniform motion is artificial—that is, an object may be at rest in our selected frame of reference, yet to an observer moving at constant velocity relative to our frame, the same object appears to be in uniform motion with constant velocity. Because the motion is relative , what is in static equilibrium to us is in dynamic equilibrium to the moving observer, and vice versa. Since the laws of physics are identical for all inertial reference frames, in an inertial frame of reference, there is no distinction between static equilibrium and equilibrium.
According to Newton’s second law of motion, the linear acceleration of a rigid body is caused by a net force acting on it, or
\[\sum_{k} \vec{F}_{k} = m \vec{a}_{CM} \ldotp \label{12.1}\]
Here, the sum is of all external forces acting on the body, where m is its mass and \(\vec{a}_{CM}\) is the linear acceleration of its center of mass (a concept we discussed in Linear Momentum and Collisions on linear momentum and collisions). In equilibrium, the linear acceleration is zero. If we set the acceleration to zero in Equation \ref{12.1}, we obtain the following equation:
The first equilibrium condition for the static equilibrium of a rigid body expresses translational equilibrium:
\[\sum_{k} \vec{F}_{k} = \vec{0} \ldotp \label{12.2}\]
The first equilibrium condition, Equation \ref{12.2}, is the equilibrium condition for forces, which we encountered when studying applications of Newton’s laws.
This vector equation is equivalent to the following three scalar equations for the components of the net force:
\[\sum_{k} F_{kx} = 0,\; \sum_{k} F_{ky} = 0,\; \sum_{k} F_{kz} = 0 \ldotp \label{12.3}\]
Although we have not talked about this yet, there is an analogous formula to Equation \ref{12.1} for rotational motion, which looks like:
\[\sum_{k} \vec{\tau}_{k} = I \vec{\alpha} \ldotp \label{12.4}\]
That quantity \(\vec{\alpha}\) is the angular acceleration, so this equation says that the sum of all the torques on a body is equal to the moment of inertia times the angular acceleration. We will pick up on that in the next chapter, right now we just need to know it's like the acceleration, but for circular motion. In equilibrium, this rotational acceleration is zero. By setting to zero the right-hand side of Equation \ref{12.4}, we obtain the second equilibrium condition:
The second equilibrium condition for the static equilibrium of a rigid body expresses rotational equilibrium:
\[\sum_{k} \vec{\tau}_{k} = \vec{0} \ldotp \label{12.5}\]
The second equilibrium condition, Equation \ref{12.5}, is the equilibrium condition for torques that we encountered when we studied rotational dynamics. It is worth noting that this equation for equilibrium is generally valid for rotational equilibrium about any axis of rotation (fixed or otherwise). Again, this vector equation is equivalent to three scalar equations for the vector components of the net torque:
\[\sum_{k} \tau_{kx} = 0,\; \sum_{k} \tau_{ky} = 0,\; \sum_{k} \tau_{kz} = 0 \ldotp \label{12.6}\]
The second equilibrium condition means that in equilibrium, there is no net external torque to cause rotation about any axis. The first and second equilibrium conditions are stated in a particular reference frame. The first condition involves only forces and is therefore independent of the origin of the reference frame. However, the second condition involves torque, which is defined as a cross product, \(\vec{\tau}_{k} = \vec{r}_{k} \times \vec{F}_{k}\), where the position vector \(\vec{r}_{k}\) with respect to the axis of rotation of the point where the force is applied enters the equation. Therefore, torque depends on the location of the axis in the reference frame. However, when rotational and translational equilibrium conditions hold simultaneously in one frame of reference, then they also hold in any other inertial frame of reference, so that the net torque about any axis of rotation is still zero. The explanation for this is fairly straightforward.
Suppose vector \(\vec{R}\) is the position of the origin of a new inertial frame of reference S′ in the old inertial frame of reference S. From our study of relative motion, we know that in the new frame of reference S′, the position vector \(\vec{r}′_{k}\) of the point where the force \(\vec{F}_{k}\) is applied is related to \(\vec{r}_{k}\) via the equation
\[\vec{r}'_{k} = \vec{r}_{k} - \vec{R} \ldotp\]
Now, we can sum all torques \(\vec{\tau}'_{k} = \vec{r}'_{k} \times \vec{F}_{k}\) of all external forces in a new reference frame, S':
\[\sum_{k} \vec{\tau}'_{k} = \sum_{k} \vec{r}'_{k} \times \vec{F}_{k} = \sum_{k} (\vec{r}_{k} - \vec{R}) \times \vec{F}_{k} = \sum_{k} \vec{r}_{k} \times \vec{F}_{k} - \sum_{k} \vec{R} \times \vec{F}_{k} = \sum \vec{\tau}_{k} - \vec{R} \times \sum_{k} \vec{F}_{k} = \vec{0} \ldotp\]
In the final step in this chain of reasoning, we used the fact that in equilibrium in the old frame of reference, S, the first term vanishes because of Equation \ref{12.5} and the second term vanishes because of Equation \ref{12.2}. Hence, we see that the net torque in any inertial frame of reference S′ is zero, provided that both conditions for equilibrium hold in an inertial frame of reference S.
The practical implication of this is that when applying equilibrium conditions for a rigid body, we are free to choose any point as the origin of the reference frame. Our choice of reference frame is dictated by the physical specifics of the problem we are solving. In one frame of reference, the mathematical form of the equilibrium conditions may be quite complicated, whereas in another frame, the same conditions may have a simpler mathematical form that is easy to solve. The origin of a selected frame of reference is called the pivot point.
In the most general case, equilibrium conditions are expressed by the six scalar equations (Equations \ref{12.3} and \ref{12.6}). For planar equilibrium problems with rotation about a fixed axis, which we consider in this chapter, we can reduce the number of equations to three. The standard procedure is to adopt a frame of reference where the z-axis is the axis of rotation. With this choice of axis, the net torque has only a z-component, all forces that have non-zero torques lie in the xy-plane, and therefore contributions to the net torque come from only the x- and y-components of external forces. Thus, for planar problems with the axis of rotation perpendicular to the xy-plane, we have the following three equilibrium conditions for forces and torques:
\[F_{1x} + F_{2x} + \cdots + F_{Nx} = 0 \label{12.7}\]
\[F_{1y} + F_{2y} + \cdots + F_{Ny} = 0 \label{12.8}\]
\[\tau_{1} + \tau_{2} + \cdots + \tau_{N} = 0 \label{12.9}\]
where the summation is over all N external forces acting on the body and over their torques. In Equation \ref{12.9}, we simplified the notation by dropping the subscript z, but we understand here that the summation is over all contributions along the z-axis, which is the axis of rotation. In Equation \ref{12.9}, the z-component of torque \(\vec{\tau}_{k}\) from the force \(\vec{F}_{k}\) is
\[\tau_{k} = r_{k} F_{k} \sin \theta \label{12.10}\]
where r k is the length of the lever arm of the force and F k is the magnitude of the force (as you saw when we studied Angular Momentum ). The angle \(\theta\) is the angle between vectors \(\vec{r}_{k}\) and \(\vec{F}_{k}\), measuring from vector \(\vec{r}_{k}\) to vector \(\vec{F}_{k}\) in the counterclockwise direction (Figure \(\PageIndex{1}\)). When using Equation \ref{12.10}, we often compute the magnitude of torque and assign its sense as either positive (+) or negative (−), depending on the direction of rotation caused by this torque alone. In Equation \ref{12.9}, net torque is the sum of terms, with each term computed from Equation \ref{12.10}, and each term must have the correct sense . Similarly, in Equation \ref{12.7}, we assign the + sign to force components in the + x-direction and the − sign to components in the − x-direction. The same rule must be consistently followed in Equation \ref{12.8}, when computing force components along the y-axis.
View this demonstration to see two forces act on a rigid square in two dimensions. At all times, the static equilibrium conditions given by Equation \ref{12.7} through Equation \ref{12.9} are satisfied. You can vary magnitudes of the forces and their lever arms and observe the effect these changes have on the square.
In many equilibrium situations, one of the forces acting on the body is its weight. In free-body diagrams, the weight vector is attached to the center of gravity of the body. For all practical purposes, the center of gravity is identical to the Center of Mass . Recall that the CM has a special physical meaning: When an external force is applied to a body at exactly its CM, the body as a whole undergoes translational motion and such a force does not cause rotation.
When the CM is located off the axis of rotation, a net gravitational torque occurs on an object. Gravitational torque is the torque caused by weight. This gravitational torque may rotate the object if there is no support present to balance it. The magnitude of the gravitational torque depends on how far away from the pivot the CM is located. For example, in the case of a tipping truck (Figure \(\PageIndex{2}\)), the pivot is located on the line where the tires make contact with the road’s surface. If the CM is located high above the road’s surface, the gravitational torque may be large enough to turn the truck over. Passenger cars with a low-lying CM, close to the pavement, are more resistant to tipping over than are trucks.
If you tilt a box so that one edge remains in contact with the table beneath it, then one edge of the base of support becomes a pivot. As long as the center of gravity of the box remains over the base of support, gravitational torque rotates the box back toward its original position of stable equilibrium. When the center of gravity moves outside of the base of support, gravitational torque rotates the box in the opposite direction, and the box rolls over. View this demonstration to experiment with stable and unstable positions of a box.
A passenger car with a 2.5-m wheelbase has 52% of its weight on the front wheels on level ground, as illustrated in Figure 12.4. Where is the CM of this car located with respect to the rear axle?
Strategy
We do not know the weight w of the car. All we know is that when the car rests on a level surface, 0.52w pushes down on the surface at contact points of the front wheels and 0.48w pushes down on the surface at contact points of the rear wheels. Also, the contact points are separated from each other by the distance d = 2.5 m. At these contact points, the car experiences normal reaction forces with magnitudes F F = 0.52w and F R = 0.48w on the front and rear axles, respectively. We also know that the car is an example of a rigid body in equilibrium whose entire weight w acts at its CM. The CM is located somewhere between the points where the normal reaction forces act, somewhere at a distance x from the point where F R acts. Our task is to find x. Thus, we identify three forces acting on the body (the car), and we can draw a free-body diagram for the extended rigid body, as shown in Figure \(\PageIndex{4}\).
We are almost ready to write down equilibrium conditions Equation \ref{12.7} through Equation \ref{12.9} for the car, but first we must decide on the reference frame. Suppose we choose the x-axis along the length of the car, the yaxis vertical, and the z-axis perpendicular to this xy-plane. With this choice we only need to write Equation \ref{12.7} and Equation \ref{12.9} because all the y-components are identically zero. Now we need to decide on the location of the pivot point. We can choose any point as the location of the axis of rotation (z-axis). Suppose we place the axis of rotation at CM, as indicated in the free-body diagram for the car. At this point, we are ready to write the equilibrium conditions for the car.
Solution
Each equilibrium condition contains only three terms because there are N = 3 forces acting on the car. The first equilibrium condition, Equation \ref{12.7}, reads
\[+F_{F} - w + F_{R} = 0 \ldotp \label{12.11}\]
This condition is trivially satisfied because when we substitute the data, Equation \ref{12.11} becomes +0.52w − w + 0.48w = 0. The second equilibrium condition, Equation \ref{12.9}, reads
\[\tau_{F} + \tau_{w} + \tau_{R} = 0 \label{12.12}\]
where \(\tau_{F}\) is the torque of force F F , \(\tau_{w}\) is the gravitational torque of force w, and \(\tau_{R}\) is the torque of force F R . When the pivot is located at CM, the gravitational torque is identically zero because the lever arm of the weight with respect to an axis that passes through CM is zero. The lines of action of both normal reaction forces are perpendicular to their lever arms, so in Equation \ref{12.10}, we have |sin \(\theta\)| = 1 for both forces. From the free-body diagram, we read that torque \(\tau_{F}\) causes clockwise rotation about the pivot at CM, so its sense is negative; and torque \(\tau_{R}\) causes counterclockwise rotation about the pivot at CM, so its sense is positive. With this information, we write the second equilibrium condition as
\[-r_{F} F_{F} + r_{R} F_{R} = 0 \ldotp \label{12.13}\]
With the help of the free-body diagram, we identify the force magnitudes F R = 0.48w and F F = 0.52w, and their corresponding lever arms r R = x and r F = d − x. We can now write the second equilibrium condition, Equation \ref{12.13}, explicitly in terms of the unknown distance x:
\[-0.52(d - x)w + 0.48xw = 0 \ldotp \label{12.14}\]
Here the weight w cancels and we can solve the equation for the unknown position x of the CM. The answer is x = 0.52d = 0.52(2.5 m) = 1.3 m. Solution Choosing the pivot at the position of the front axle does not change the result. The free-body diagram for this pivot location is presented in Figure 12.6. For this choice of pivot point, the second equilibrium condition is
\[-r_{w} w + r_{R} F_{R} = 0 \ldotp \label{12.15}\]
When we substitute the quantities indicated in the diagram, we obtain
\[-(d - x)w + 0.48dw = 0 \ldotp \label{12.16}\]
The answer obtained by solving Equation \ref{12.13} is, again, x = 0.52d = 1.3 m.
Significance
This example shows that when solving static equilibrium problems, we are free to choose the pivot location. For different choices of the pivot point we have different sets of equilibrium conditions to solve. However, all choices lead to the same solution to the problem.
Solve Example 12.1 by choosing the pivot at the location of the rear axle.
Explain which one of the following situations satisfies both equilibrium conditions: (a) a tennis ball that does not spin as it travels in the air; (b) a pelican that is gliding in the air at a constant velocity at one altitude; or (c) a crankshaft in the engine of a parked car.
A special case of static equilibrium occurs when all external forces on an object act at or along the axis of rotation or when the spatial extension of the object can be disregarded. In such a case, the object can be effectively treated like a point mass. In this special case, we need not worry about the second equilibrium condition, Equation \ref{12.9}, because all torques are identically zero and the first equilibrium condition (for forces) is the only condition to be satisfied. The free-body diagram and problem-solving strategy for this special case were outlined in Newton’s Laws of Motion . You will see a typical equilibrium situation involving only the first equilibrium condition in the next example.
View this demonstration to see three weights that are connected by strings over pulleys and tied together in a knot. You can experiment with the weights to see how they affect the equilibrium position of the knot and, at the same time, see the vector-diagram representation of the first equilibrium condition at work.
A small pan of mass 42.0 g is supported by two strings, as shown in Figure 12.7. The maximum tension that the string can support is 2.80 N. Mass is added gradually to the pan until one of the strings snaps. Which string is it? How much mass must be added for this to occur?
Strategy
This mechanical system consisting of strings, masses, and the pan is in static equilibrium. Specifically, the knot that ties the strings to the pan is in static equilibrium. The knot can be treated as a point; therefore, we need only the first equilibrium condition. The three forces pulling at the knot are the tension \(\vec{T}_{1}\) in the 5.0-cm string, the tension \(\vec{T}_{2}\) in the 10.0-cm string, and the weight \(\vec{w}\) of the pan holding the masses. We adopt a rectangular coordinate system with the y-axis pointing opposite to the direction of gravity and draw the free-body diagram for the knot (see Figure 12.8). To find the tension components, we must identify the direction angles \(\alpha_{1}\) and \(\alpha_{2}\) that the strings make with the horizontal direction that is the x-axis. As you can see in Figure 12.7, the strings make two sides of a right triangle. We can use the Pythagorean theorem to solve this triangle, shown in Figure 12.8, and find the sine and cosine of the angles \(\alpha_{1}\) and \(\alpha_{2}\). Then we can resolve the tensions into their rectangular components, substitute in the first condition for equilibrium (Equation \ref{12.7} and Equation \ref{12.8}), and solve for the tensions in the strings. The string with a greater tension will break first.
Solution
The weight w pulling on the knot is due to the mass M of the pan and mass m added to the pan, or w = (M + m)g. With the help of the free-body diagram in Figure 12.8, we can set up the equilibrium conditions for the knot:
in the x-direction, \[-T_{1x} + T_{2x} = 0\] n the y-direction, \[+T_{1y} + T_{2y} - w = 0 \ldotp\]
From the free-body diagram, the magnitudes of components in these equations are
\[\begin{split} T_{1x} & = T_{1} \cos \alpha_{1} = \frac{T_{1}}{\sqrt{5}},\quad T_{1y} = T_{1} \sin \alpha_{1} = \frac{2T_{1}}{\sqrt{5}} \\ T_{2x} & = T_{2} \cos \alpha_{2} = \frac{2T_{2}}{\sqrt{5}},\quad T_{2y} = T_{2} \sin \alpha_{2} = \frac{T_{2}}{\sqrt{5}} \ldotp \end{split}\]
We substitute these components into the equilibrium conditions and simplify. We then obtain two equilibrium equations for the tensions:
in x-direction, \[T_{1} = 2T_{2}\]
in y-direction, \[\frac{2T_{1}}{\sqrt{5}} + \frac{T_{2}}{\sqrt{5}} = (M + m)g \ldotp\]
The equilibrium equation for the x-direction tells us that the tension T 1 in the 5.0-cm string is twice the tension T 2 in the 10.0-cm string. Therefore, the shorter string will snap. When we use the first equation to eliminate T 2 from the second equation, we obtain the relation between the mass m on the pan and the tension T 1 in the shorter string:
\[\frac{2.5T_{1}}{\sqrt{5}} = (M + m)g \ldotp\]
The string breaks when the tension reaches the critical value of T 1 = 2.80 N. The preceding equation can be solved for the critical mass m that breaks the string:
\[m = \frac{2.5}{\sqrt{5}} \frac{T_{1}}{g} - M = \frac{2.5}{\sqrt{5}} \frac{2.80\; N}{9.8\; m/s^{2}} - 0.042\; kg = 0.277\; kg = 277.0\; g \ldotp\]
Significance
Suppose that the mechanical system considered in this example is attached to a ceiling inside an elevator going up. As long as the elevator moves up at a constant speed, the result stays the same because the weight w does not change. If the elevator moves up with acceleration, the critical mass is smaller because the weight of M + m becomes larger by an apparent weight due to the acceleration of the elevator. Still, in all cases the shorter string breaks first.